On Completions, neat atom structures, and omitting types
Tarek Sayed Ahmed
Department of Mathematics
Faculty of Science
Giza
Egypt.
1304.1149.tex

\documentclass[12pt]{article}
\usepackage{amsmath,amssymb,amsfonts,amsthm}
%\thispagestyle{empty}
%\pagestyle{empty}
%\pagenumbering{}
\setlength{\parindent}{18pt}
\textwidth14cm
\textheight21.6cm
%
\newenvironment{myabstract}{\par\noindent
{\bf Abstract . } \small }
{\par\vskip8pt minus3pt\rm}
%
\newcounter{item}[section]
\newcounter{kirshr}
\newcounter{kirsha}
\newcounter{kirshb}
%
\newenvironment{enumroman}{\setcounter{kirshr}{1}
\begin{list}{(\roman{kirshr})}{\usecounter{kirshr}} }{\end{list}}
%
\newenvironment{enumalph}{\setcounter{kirsha}{1}
\begin{list}{(\alph{kirsha})}{\usecounter{kirsha}} }{\end{list}}
%
\newenvironment{enumarab}{\setcounter{kirshb}{1}
\begin{list}{(\arabic{kirshb})}{\usecounter{kirshb}} }{\end{list}}
%
\newenvironment{mysect}[1]{\par\noindent\setcounter{item}{1}
{\bf\S\arabic{section}.  #1 }\nopagebreak\par\nopagebreak }
{\stepcounter{section}\upshape\par\vskip20pt minus8pt}
%
\newenvironment{athm}[1]{\vskip3mm\par\noindent%\stepcounter{item}
{\bf #1 }. \slshape }
{\upshape\par\vskip10pt minus3pt}
%
\newenvironment{rem}[1]{\vskip3mm\par\noindent
{\bf #1 }.  }
{\par\vskip10pt minus3pt}
%
\newtheorem{theorem}{Theorem}[section]
\newtheorem{fact}[theorem]{Fact}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
%\renewcommand{\thetheorem}{\arabic{section}.\arabic{theorem}}
\newenvironment{demo}[1]{\noindent{\bf #1.}\upshape\mdseries}
{\nopagebreak{\hfill\rule{2mm}{2mm}\nopagebreak}\par\normalfont}
%
\theoremstyle{definition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{problem}[theorem]{Problem}
\newtheorem{example}[theorem]{Example}
\newtheorem{definition}[theorem]{Definition}
\def\R{\mathbb{R}}
%\def\N{\mathbb{N}}
\def\Q{\mathbb{Q}}
\def\C{{\mathfrak{C}}}
\def\Fm{{\mathfrak{Fm}}}
\def\Ws{{\bf Ws}}
\def\At{{\bf At}}
\def\Nr{{\mathfrak{Nr}}}
\def\Fr{{\mathfrak{Fr}}}
\def\Sg{{\mathfrak{Sg}}}
\def\Fm{{\mathfrak{Fm}}}
\def\A{{\mathfrak{A}}}
\def\B{{\mathfrak{B}}}
\def\C{{\mathfrak{C}}}
\def\D{{\mathfrak{D}}}
\def\M{{\mathfrak{M}}}
\def\N{{\mathfrak{N}}}
\def\Sn{{\mathfrak{Sn}}}
\def\CA{{\bf CA}}
\def\QA{{\bf QA}}
\def\SC{{\bf SC}}
\def\QEA{{\bf QEA}}
\def\RSC{{\bf RSC}}
\def\Df{{\bf Df}}
\def\SA{{\bf SA}}
\def\RQA{{\bf RQA}}
\def\RDf{{\bf RDf}}
\def\Lf{{\bf Lf}}
\def\Dc{{\bf Dc}}
\def\PA{{\bf PA}}
\def\PEA{{\bf PEA}}
\def\Crs{{\bf Crs}}
\def\K{{\bf K}}
\def\K{{\bf K}}
\def\RK{{\bf RK}}
\def\RCA{{\bf RCA}}
\def\SA{{\bf SA}}
\def\Rd{{\ Rd}}
\def\(R)RA{{\bf (R)RA}}
\def\RA{{\bf RA}}
\def\RRA{{\bf RRA}}
\def\Dc{{\bf Dc}}
\def\R{\mathbb{R}}
%\def\N{\mathbb{N}}
\def\Q{\mathbb{Q}}
%\def\C{\mathbb{C}}
\def\Dc{{\bf Dc}}
\def\Mn{{\bf Mn}}
\def\Crs{{\bf Crs}}
\def\Di{{\bf Di}}
\def\Sc{{\bf Sc}}
\def\RC{{\bf RC}}
\def\Mn{{\bf Mn}}

\def\c #1{{\cal #1}}
 \def\CA{{\sf CA}}
%\def\R{{\sf R}}
\def\B{{\sf B}}
\def\G{{\sf G}}
\def\w{{\sf w}}
\def\y{{\sf y}}
\def\g{{\sf g}}
\def\b{{\sf b}}
\def\r{{\sf r}}
\def\K{{\sf K}}
\def\tp{{\sf tp}}

 \def\Cm{{\mathfrak{Cm}}}
\def\Nr{{\mathfrak{Nr}}}
\def\SNr{{\bf S}{\mathfrak{Nr}}}

\def\restr #1{{\restriction_{#1}}}
\def\cyl#1{{\sf c}_{#1}}
\def\diag#1#2{{\sf d}_{#1#2}}
\def\sub#1#2{{\sf s}^{#1}_{#2}}

\def\Ra{{\mathfrak{Ra}}}
\def\Ca{{\mathfrak{Ca}}}
\def\set#1{\{#1\} }
\def\Ra{{\mathfrak{Ra}}}
\def\Nr{{\mathfrak{Nr}}}
\def\Tm{{\mathfrak{Tm}}}
\def\A{{\mathfrak{A}}}
\def\B{{\mathfrak{B}}}
\def\C{{\mathfrak{C}}}
\def\D{{\mathfrak{D}}}
\def\E{{\mathfrak{E}}}
\def\A{{\mathfrak{A}}}
\def\B{{\mathfrak{B}}}
\def\C{{\mathfrak{C}}}
\def\D{{\mathfrak{D}}}
\def\E{{\mathfrak{E}}}
\def\U{{\mathfrak{U}}}
\def\GG{{\mathfrak{GG}}}
\def\Alg{{\mathfrak{Alg}}}
\def\Bb{{\mathfrak{Bb}}}
\def\L{{\mathfrak{L}}}
\def\Rd{{\mathfrak{Rd}}}
\def\Bb{{\mathfrak{Bb}}}
\def\At{{\mathfrak{At}}}
\def\L{{\mathfrak{L}}}
\def\Bl{{\mathfrak{Bl}}}
\def\CA{{\bf CA}}
\def\RA{{\bf RA}}
\def\RRA{{\bf RRA}}
\def\RCA{{\bf RCA}}
\def\X{{\bf X}}
%\def\D{{\bf D}}
\def\G{{\bf G}}
\def\RB{{\bf RB}}
\def\CRA{{\bf CRA}}
\def\Diag{{\bf Diag}}
\def\Fl{{\mathfrak{Fl}}}
\def\F{{\mathfrak{F}}}
\def\At{{\sf{At}}}
\def\N{\mathbb{N}}
\def\R{\mathfrak{R}}
\def\CM{{\bf CM}}
\def\CRA{{\bf CRA}}
\def\CCA{{\bf CCA}}
\def\Cs{{\bf Cs}}
\def\RPEA{{\bf RPEA}}
\def\conv{{\smile}}

\def\lc{Lyndon condition}
\def\sub#1#2{{\sf s}^{#1}_{#2}}
\def\cyl#1{{\sf c}_{#1}}
\def\diag#1#2{{\sf d}_{#1#2}}
\def\ca{cylindric algebra}
\def\wa{weakly associative algebra}
\def\c #1{{\cal #1}}

\def\rb{representable}

\def\uf{ultrafilter}
\def\pa{$\forall$}
\def\pe{$\exists$}

\def\ef{Ehren\-feucht--Fra\"\i ss\'e}

\def\hb{hyperbasis}
\def\hnk{hyper\nk}


\def\nodes{{\sf nodes}}

\def\restr #1{{\restriction_{#1}}}

\def\Ra{{\mathfrak{Ra}}}
\def\Nr{{\mathfrak{Nr}}}
\def\Z{{\cal Z}}
\def\CA{{\bf CA}}
\def\RCA{{\bf RCA}}
\def\c#1{{\mathcal #1}}
\def\SS{{\bf S}}
\def\sra#1{{\\S\Ra\CA_{#1}}}
\def\set#1{ \{#1\}}
\def\rrep{relativized representation}
\def\Ca{{\mathfrak Ca}}
\def\b#1{{\bar{ #1}}}
\def\pe{$\exists$}
\def\pa{$\forall$}
\def\Cm{{\mathfrak Cm}}
\def\Sg{{\mathfrak Sg}}
\def\P{{\mathfrak P}}
\def\Rl{{\mathfrak Rl}}
\def\N{{\cal N}}
\def\Los{\L{o}\'{s}}
\def\ls { L\"owenheim--Skolem}
\def\At{{\sf At}}

\def\rg{{\sf rng}}
\def\rng{{\sf rng}}
\def\dom{{\sf dom}}


\def\w{{\sf w}}
\def\g{{\sf g}}
\def\y{{\sf y}}
\def\r{{\sf r}}
\def\bb{{\sf b}} %\b means bold, I think
\def\tp{{\sf tp}}
\def\cyl#1{{\sf c}_{#1}}
\def\sub#1#2{{\sf s}^{#1}_{#2}}
\def\diag#1#2{{\sf d}_{#1#2}}

\def\ws{winning strategy}
\def\ef{Ehren\-feucht--Fra\"\i ss\'e}
%\newenvironment{proof}{
%PROOF:
%\begin{quotation}}{
%$\Box$ \end{quotation}}
 
 


 
 
\def\lr#1{\langle#1\rangle}\def\y{{\sf y}}
\def\g{{\sf g}}
\def\bb{{\sf b}}%not to be confused with ``bold''
\def\r{{\sf r}}
\def\w{{\sf w}}

 



\title{On Completions, neat atom structures, and omitting types}
\author{Tarek Sayed Ahmed \\
Department of Mathematics, Faculty of Science,\\ 
Cairo University, Giza, Egypt.
  }
%%% ----------------------------------------------------------------------
%
\begin{document}
\maketitle
\begin{myabstract} This paper has a survey character, but it also contains several new results.
The paper tries to give a panoramic picture of the recent developments in algebraic logic.
We take a long magical tour in algebraic logic starting from classical notions due to Henkin Monk and Tarski
like neat embeddings, culminating in presenting sophisticated model theoretic constructions based on graphs, to solve problems on neat reducts.

We investigate several algebraic notions that apply to varieties of 
boolean algebras with operators in general $BAOs$, like canonicity, atom-canonicity and completions. We also show that in certain significant
special cases, when we have a Stone-like representabilty notion, like in cylindric, relation and polyadic algebras
such abtract notions turn out intimately related 
to more concrete notions, like complete representability, 
and the existence of weakly but not srongly representable atom structures.

In connection to the multi-dimensional  corresponding modal logic, we 
prove several omitting types theorem for the finite $n$ variable fragments of first order logic, 
the multi-dimensional modal logic corresponding to $CA_n$; the class of cylindric algebras of dimension $n$.

A novelty that occurs here is that
theories could be uncountable. Our constructions depend on deep model-theoretic results of Shelah. 

Several results mentioned in \cite{OTT} without proofs are proved fully here, one such result is:
There exists an uncountable atomic algebra in $\Nr_n\CA_{\omega}$ that is not completely representable. 
Another result: If $T$ is an $L_n$ theory (possibly uncountable), where $|T|=\lambda$, $\lambda$ is a regular cardinal, and $T$ admits elimination of 
quantifiers, then
$<2^{\lambda}$ non principal types {\it maximal} can be omitted. 

A central notion, that connects, representations, completions, complete representations for cylindric algebras is that of neat embedding,
which is an algebraic counterpart of Henkin constructions, and is a nut cracker in cylindric-like 
algebras for proving representation results and related results concerning various forms of the amalagmation property
for clases of representable algebras. 
For example, representable algebras are those algebras that have the neat embeding property, 
completey representable countable ones, are the atomic algebras that have the complete neat embedding 
property. We show that countability cannot be omitted which is sharp in view to our omitting types theorem mentiond above.
We show that the class $\Nr_n\CA_{\omega}$ is psuedo-elementary, not elementary, and its elementary closure is not finitely axiomatizable
for $n\geq 3$, We  characterize this class by games.

We give two constructions for weakly representable atom structures that is not 
strongly representable, that are simple variations on existing themes, and using fairly straightforward modificatons 
of constructons of Hirsch and Hodkinson, we show that
the latter class is not elementary for any reduct of polyadic algebras contaiting all cylindrifiers. 

We introduce the new notions of strongly neat, weakly neat and very weakly neat atom structures.
An $\alpha$ dimensional  atom strucure is  very weakly neat, $\alpha$
an ordinal,  if no algebra based on it in 
$\Nr_{\alpha}\CA_{\alpha+\omega};$ weakly neat if it has at least one algebra based on it that 
is in $\Nr_{\alpha}\CA_{\alpha+\omega}$, and finally  strongly neat if every algebra based on it is in 
$\Nr_{\alpha}\CA_{\alpha+\omega}.$ We give examples of the first two, show that they are distinct, 
and further show that the class of weakly 
neat atom structures is not elementary.
This is done for all dimensions $>2$, infinite included. For the third, we show that finite atom structures are strongly neat (in finite dimensions).


Modifying several constructions in the literature, as well as providing new ones, several 
results on complete representations and completions are presented, 
answering several questions posed by Robin Hirsch, and Ian Hodkinson, concerning relation algebras, and complete representabiliy
of reducts of plyadic algebras.
%In the second case, obtain a representable atomic polyadic algebra, 
%of dimension $n$, such that the completion of its $CA$ reduct
%is not in $S\Nr_nCA_{n+2};$ in particular, it is not representable. We readily infer that the class 
%$S\Nr_n\CA_{n+2}$ is not atom-canonical and is not closed under completions. Our approach is algebraic. 
%Finally a variant of Vaught's conjecture is investigated in the context of G\"odel's intuitionistic logic.
\footnote{Mathematics Subject Classification. 03G15; 06E25}
 
%Key words: multimodal logic, substitution algebras, interpolation}
\end{myabstract}


\section{Introduction}

Atom canonicity, completions, complete representations and omitting types are  four notions that could appear at first glimpse unrelated. 
The first three, are algebraic notions that apply 
to varieties of Boolean algebras with operators $BAO$s. Omitting types is a metalogical notion that applies to the corresponding multi-modal logic. 

Canonicity is one of the most important concepts of modal 
completeness theory. From an algebraic perspective, canonical models are not abstract oddities, 
they are precisey the structure one is led to 
by underlying the idea in Stones representabilty theory for Boolean algebras.

The canonical extension of an algebra  has universe the power set algebra of 
the set of ultrafilters; that is its Stone space, and  the extra non-Boolean operations induced naturally from the original ones.
A variety is canonical if it is closed under taking canonical
extensions. 

This is typically a {\it persistence property}. Persistence properties refer to closure of a variety $V$  under 
passage from a given member in $V,$ to some 'larger' algebra .

The other  persistence property, namely, atom-canonicity, concerns the atom structure $\At\A$ of an atomic algebra $\A$. 
As the name suggests, $\At\A$ is a certain relational structure based on the set of atoms of $\A$. 
A variety $V$ is atom-canonical if it contains the complex algebra $\Cm\At(\A)$ whenever it contains $\A$.
If $\At V$ is the class of all atom structures of atomic algebras in $V$, 
then atom-canonicity amounts to the requirement that $\Cm\At V\subseteq V$.%

The (canonical) models of a multi-modal logic $\L_V$, corresponding to a canonical variety $V$,  are Kripke frames; 
these are the ultrafilter frames. The  atom structures, are special cases, we call these atomic 
models.

The canonical extension of an algebra is a complete atomic algebra that the original algebra embeds into, however it only preserves finite meets. 
Another completion is the Dedekind MacNeille completion, also known as the 
minimal completion. Every completely  additive $BAO$ has such a completion. It is uniquely 
determined by the fact that it is complete and the original algebra is dense in it; hence it preserves all meets existing in the original 
algebra.
The completion is atomic if and only if the original algebra is. The completion and canonical extension of an algebra 
only coincide when the algebra is finite.

Complete representations has to do with algebras that have a notion of representations involving - using jargon of modal logic - complex algebras 
of square frames or, using algebraic logic jargon, full set algebras having square units, and also having the 
Boolean operations of (concrete) intersections and complementation,
like relation algebras and cylindric algebras.  Unlike atom-canonicity, this notion is semantical.
Such  representations is a representation that carries existing (possibly infinite) meets to set theoretic 
intersections.


Atomic representability is also related to the Orey-Henkin omitting types theorem. Let $V$ be a variety of $BAO$'s which has a notion of representation,
like for example $CA_n$. The variety $CA_n$ corresponds to the syntactical part of $L_n$ 
first order logic restricted to the first $n$ variables, while $RCA_n$ corresponds to the semantical models 

Indeed given an $L_n$ theory $T$ and a model $\M$ of $T$, let $\phi^M$ denote the set of$n$-ary assignments satisfying
$\phi$ in $\M$, notationaly $\phi^{\M}=\{s\in {}^nM: \M\models \phi[s]\}$.
Then $\{\phi^M:\phi\in L\}$ is the universe of a cylindric set algebra $Cs_n$ with the operations of cyindrifiers corresponding to 
the semantics of existential quabtifiers
and diagonal elements to equality. The class of subdirect products of algebras in 
$Cs_n$ is the class $RCA_n$.

A set $\Gamma$ in the language of $T$ is said to be omitted by the model $\M$ of $T$, 
if $\bigcap \phi^{\M}=\emptyset$. This can be formulated algebraically as follows:
Let $\A\in \CA_n$ and  let there be given a family $(X_i: i\in I)$ of subsets of $\A$,  then there exists an injective homomorphism 
$f:\A\to \wp(V)$, $V$ a disjoint union of cartesian squares, that  omits the $X_i$'s, that is $\bigcap_{x\in X_i} f(x)=0$.



The Orey-Henkin omitting types theorem says that this always happens if $\A$ is countable and locally finite,
and when $I$ is countable, and  the $X_i'$ contain only finitely many 
dimensions (free variables). 
But it is clear that the above algebraic formulation lends itself to other contexts.


Note that if omitting types theorem holds for arbitrary cardinalities, then the 
atomic representable algebras are completely representable, by finding a representation that omits the 
non principal types $\{-x: x\in \At\A\}$. The converse is false. There are easy examples. Some will be provided below.


Now let us try to find a connection between such notions. 
Consider a variety$V$ of $BAO$s that is {\it not} atom-canonical. Ths means that there is an $\A\in V$, such that $\Cm\At\A\notin V$. 
If $V$ is completely aditive, then $\Cm\At\A$, is the completion of $\A$. So $V$ is not closed 
under completions. 

There are significant varieties that are not atom-canonical, like the variety of representable cylindric algebras 
$\RCA_n$ for $n\geq 3$ and representable relation algebras $\RRA$.
These have a notion of representability involving square units. 
%Consider $\RCA_{n}\subset \CA_{n}.$

Let $\A$ be an atomic representable such that $\Cm\At\A\notin \RCA_{n}$, for $n\geq 3$. Such algebras exist, the first of its kind was constructed
by Hodkinson. The term algera $\Tm\At\A$, which is the subalgebra of the complex algebra,
is contained in $\A$, because $\RCA_n$ is completely additiv; furthermore, it is 
representable but it {\it cannot} have a complete representation, for such a representation
would necessarily induce a representation of $\Cm\At\A$. That is to say, $\A$ is an example of an atomic theory in the cylindric modal logic of dimension
$n$, but it has no atomic model. 
In such a context, $\At\A$ is also an example of a {\it weakly representable} atom structure that is not {\it strongly representable.}


A weakly representable atom structure is an atom structure such that there is at least one algebra based on this atom structure that is 
representable \cite{w}. It is strongly representable, if every algebra based on it is representable.
Hodkinson, the first to construct a weakly representable cylindric atom structure that is not strongly representable, 
used a somewhat complicated construction depending on so called Rainbow 
constructions. His proof is model-theoretic.
In \cite{w} we use the same method to construct such an atom structure for both relation and cylindric algebras.
On the one hand, the graph we used substantially simplifies Hodkinson's construction, 
and, furthermore, we get our result for relation and cylindric algebras in one go.

%Such a result have a wide range of repercussions, for example it shows that the class of representable algebras 
%is not closed under completions, and it is not
%atom canonical, there are atomic algebras with no complete representations, 
%and that this class cannot be axiomatized by Sahlqvist equations.


Hirsch and Hodkinson show that the class of strongly representable atom structures of relation algebras 
(and cylindric algebras) is not elementary \cite{strong}.
The construction makes use of the pobabilistic method of  Erd\"os to show that there are finite graphs with arbitrarily large 
chromatic number and girth.   
In his pioneering paper of 1959, Erdos took a radically new approach to construct such graphs: for each $n$ he 
defined a probability space on the set of graphs with $n$ vertices, and showed that, for some carefully chosen probability measures, 
the probability that an $n$ vertex graph has these properties is positive for all large enough $n$.
This approach, now called the {\it probabilistic method} has since unfolded into a 
sophisticated and versatile proof technique, in graph theory and in other 
branches of discrete mathematics.
This method was used first in algebraic logic by Hirsch and Hodkinson to show that the class of strongly representable atom structures of cylindric 
and relation algebras is not elementary  
and that varieties of representable relation algebras are barely canonical \cite{hv}.
But yet again using these methods of Erd\"os in  \cite{ghv} it is shown 
that there exist continuum-many canonical equational classes of Boolean algebras 
with operators that are not generated by the complex algebras of any first-order 
definable class of relational structures. Using a variant of this construction  the authors resolve the long-standing question of Fine, 
by exhibiting a bimodal logic that is valid in its canonical frames, but is not sound and complete for any first-order definable class of 
Kripke frames. 

There is an ongoing interplay between algebraic logic on the one hand, and model theory and finite combinatorics particularly graph theory, 
on the other. 
Monk was the first to use Ramsey's theorems to construct what is known an Monk's algebras.
witnessing non finite axiomatizability for 
the class of representable cylindric algebras.
The key idea of the construction of a Monk's algebra is not so hard. 
Such algebras are finite, hence atomic, more precisely their Boolean reducts are atomic.
The atoms are given colours, and 
cylindrifications and diagonals are defined by stating that monochromatic triangles 
are inconsistent. If a Monk's algebra has many more atoms than colours, 
it follows from Ramsey's Theorem that any representation
of the algebra must contain a monochromatic triangle, so the algebra
is not representable. 

Later Monk-like constructions were substantially generalized by Andr\'eka N\'emeti \cite{ANT}, 
Maddux \cite{m}, and finally Hirsch and Hodkinson \cite{HHM}.
Constructing algebras from Erdos graphs have 
proved extremely rewarding \cite{strong}, \cite{hv}, \cite{ghv}. 
Another construction invented by Robin Hirsch and Ian Hodkinson is the so-called rainbow construction, which  is an ingenious technique 
that has been used to show that several classes are not elementary \cite{HH}, \cite{HHbook}, \cite{r}, and was used together with a 
lifting argument of Hodkinson of construction polyadic algebras from relation algebras to show that it is undecidable whether a finite relation or 
cylindric algebra is representable.
This shows that  ceratin important products modal logics are undecidable.
We will use the rainbow construction below to prove the $CA$ analogue of a deep result in \cite{r}.

Constructing cylindric algebras based on certain models satisfyig certain properties like homogeniouty, saturation, 
elimination of quantifiers, using model theory like in \cite {MLQ}, will be generalized below to answer a question of Hirsch \cite{r},
on relation algebra reducts of cylindric algebras.

Another model theoretic construction of Hodkinson, 
based on rainbow graphs, considerably simplified in \cite{w} will be further simplified here 
to prove  that several varieties approximating the class of representable cylindric algebras are not closed under completions.

%something stronger.
\section*{ The main new results}

\begin{enumarab}

\item  Answering a question of Robin Hirsch in \cite{R} on complete representations for both relation and cylindric algebras 
using an example in the same paper. This example shows that in the characterization of countable completely representable algebras,
both relation and cylindric algebras, the condition of countability is necessary, it {\it cannot} be omitted.


\item Using an example by Andr\'eka et all, to show that unlike cylindric and polyadic equality algebras, 
atomic polyadic algebras, and Pinters substitution algebras, even without cylindrfiers, of 
dimension $2$ may  not be completely representable. However, the class of completely representable algebras 
is not so hard to characterize; it is finitely axiomatizable in first order logic.
This is contrary to a current belief for polyadic algebras, and is an answer to a result of Hodkinson for cylindifier free
Pinter's algebras.

\item Using the construction in Andr\'eka et all \cite{ANT}, 
showing that the omitting types theorem fails for finite first 
order definable extension of first order logic as defined by Tarski and Givant, and further pursued by others, like 
Maddux and Biro,  a result mentioned in the above cited paper without a proof.

\item Characterizing the class $\Nr_n\CA_{\omega}$ by games, and 
showing that the class $\Nr_n\CA_{\omega}$ is pseudo elementary, and its elementary closure is not finitely axiomatizable.

\item Characterizing the class of countable completely representable algebras of infinitely countable dimensions using weak representations 
(the question remains whether this class is elementary, 
the Hirsch Hodkinson example depending on a cardinality argument does not works when our units are weak spaces.)

%\item Showing that the varieties $S\Nr_n\CA_{n+k}$ $k\geq 2$ are not atom canonical, and not closed under completions.
%(This answers a question by Hirsch and Hodkinson and a question in \cite{1}.)

\item Giving full proofs to three results mentioned in \cite{OTT} without proofs, referring to a pre-print, 
concerning omitting types in uncountable theories using finitely
many variables. This is the pre print, expanded, modified and polished containing proofs of these results and much more.
The results concerning omitting types depend on deep model-theoretic constructions of Shelah's.

\item We show that the class of weakly neat atom structures, as defined in the abstract, is not elementary for every dimension.

\item Unlike the cylindric case, we show that atomic polyadic algebras of infinite dimensions are completely representable.

\end{enumarab}


This paper also simplifies existing proofs in the literature, like the proof in \cite{ANT}, concerning complete representations of relation 
atom structures, having cylindric basis. Some classical results, like Monk's non-finitizability results for relation and cylindric algebras
are also re-proved.
%Four semingly unrelated notions. Atom canonicity, completions, complete representations, are algebraic notions that apply 
%to varieties, that are $BAO$'s. To every variety $V$ of $BAO$ there corresponds a multi dimensional modal logic $\L_V.$
%The models of $\L_V$ are Kripke frames, and of importance are the atom structures which correspond to the atomic models.



%This paper also contains two results in \cite{OTT} that are mentioned without proofs, referring  instead to a preprint. 
%This is this preprint, expanded, modified and polished containing all proofs.


\section{Omitting types}

In model theory we are typically encountered by questions not only of the form find a model  (prove consistency), but rather 
of the form find a model of a theory that 
does so and so. A typical example, in this regard, is finding a model 
omitting a given set of non-isolated types. In first order logic with $\omega$ variables, this problem is completely settled 
for the countable case. Another, which a variation on Vaught's conjecture is, {\it count} the number of such models.
Vaught's conjecture is regarded as one of  the most important open problem in model theory, if not the most.

In the first case, algebraically we find ourselves looking for non-trivial homomorphisms from locally finite cylindric algebras to set algebras 
that preserve infinitary meets (or types) carrying them to set theoretic intersection, and  by the Henkin-Orey classical 
omitting types theorem we can always find one if the meets considered are not isolated.
When we truncate the dimension to be finite, it is not always the case  algebra in $\RCA_n$ have this property, 
but indeed it is worthwhile characterizing those that does. This is one of the tasks carried out in this paper. 

In the second case, we  find ourselves counting special ultrafilters in locally finite cylindric algebras, that form a dense set of its 
Stone space (This will be discussed in some detail below).
Investigating, using the well developed machinery of algebraic logic, the number of types omitted in theories, 
and the number of pairwise non-isomorphic countable models
for a countable theory that omit a given set of types, 
this paper is interdisciplinary between model theory, set theory and algebraic logic. 


There two typical types of investigations in set theory. Both of those will be adressed in this paper. 
The first type consists of theorems demonstrating the independence of mathematical statements. 
This type requires a thorough understanding of mathematics surrounding the 
source of problem in question, reducing the ambient mathematical constructions to combinatorial statements about sets, 
and finally using some method 
(primarily forcing) to show  that such combinatorial statements
are independent.
A second type involves delineating the edges of the independence proofs, giving proof in $ZFC$ of statements 
that on first sight would be suspected of being independent. 
Proofs of this kind is often extremely subtle and surprising; very similar 
statements are independent and it is hard to detect the underlying difference.

We present here results illustrating both types of investigations. 
We show that when we consider countable algebras, omitting $< {}^{\omega}2$ types turns 
out independent, whereas, if the types are maximal the statement is provable. 
Such results will be presented in the context of omitting types in $\L_n$ theories. We find this discrepancy interesting from the purely set theoretic point of 
view when maximality shifts us from the realm of independence to that of 
provability from $ZFC$.



Let us start with the classical omitting types theorem.
One direct consequence of the classical Henkin-Orey omitting types theorem states that if 
$T$ is a consistent theory in a countable language $\L$ and $\Gamma(x_1\ldots x_n)$ is 
realized in every model of $T$, then there is a formula $\phi\in \L$ such that
$\phi$ isolates $\Gamma$ in $T$. $\phi$ is called a witness of $\Gamma$. This follows directly from the contrapositive
of the following theorem, see also \cite{OTT}, \cite{OTT2}:

\begin{theorem}\label{infinite} Let $T$ be a countable consistent theory. Assume that 
$\kappa<covK$, where $covK$ is the least cardinal $\kappa$ such that the real line can be covered by
pairwise disjoint $\kappa$ nowhere-dense sets. Let $(\Gamma_i:i\in \kappa)$ be a set of non-principal types in $T$. 
Then there is a model countable $\M$ of $T$ that omits
all the $\Gamma_i$'s.
\end{theorem}
\begin{demo}{Proof} Standard Baire Category argument. Let $\A=\Fm_T$, then $\A\in \Lf_{\omega}$.
Let $X_i=\{\phi/T: \phi\in \Gamma_i\}$. Then we have $\prod X_i=0$ in $\A$. We have by \cite[1.11.6]{HMT1} that 
\begin{equation}\label{t1}
\begin{split} (\forall j<\alpha)(\forall x\in A)({\sf c}_jx=\sum_{i\in \alpha\smallsetminus \Delta x}
{\sf s}_i^jx.)
\end{split}
\end{equation}
Now let $V$ be the weak space $^{\omega}\omega^{(Id)}=\{s\in {}^{\omega}\omega: |\{i\in \omega: s_i\neq i\}|<\omega\}$.
For each $\tau\in V$ for each $i\in \kappa$, let
$$X_{i,\tau}=\{{\sf s}_{\tau}x: x\in X_i\}.$$
Here ${\sf s}_{\tau}$ 
is the unary operation as defined in  \cite[1.11.9]{HMT1}.
For each $\tau\in V,$ ${\sf s}_{\tau}$ is a complete
boolean endomorphism on $\B$ by \cite[1.11.12(iii)]{HMT1}. 
It thus follows that 
\begin{equation}\label{t2}\begin{split}
(\forall\tau\in V)(\forall  i\in \kappa)\prod{}^{\A}X_{i,\tau}=0
\end{split}
\end{equation}
Let $S$ be the Stone space of the Boolean part of $\A$, and for $x\in \A$, let $N_x$ 
denote the clopen set consisting of all
boolean ultrafilters that contain $x$.
Then from \ref{t1}, \ref{t2}, it follows that for $x\in \A,$ $j<\beta$, $i<\kappa$ and 
$\tau\in V$, the sets 
$$\bold G_{j,x}=N_{{\sf c}_jx}\setminus \bigcup_{i\notin \Delta x} N_{{\sf s}_i^jx}
\text { and } \bold H_{i,\tau}=\bigcap_{x\in X_i} N_{{\sf s}_{\bar{\tau}}x}$$
are closed nowhere dense sets in $S$.
Also each $\bold H_{i,\tau}$ is closed and nowhere 
dense.
Let $$\bold G=\bigcup_{j\in \beta}\bigcup_{x\in B}\bold G_{j,x}
\text { and }\bold H=\bigcup_{i\in \kappa}\bigcup_{\tau\in V}\bold H_{i,\tau.}$$
By properties of $covK$, it can be shown $\bold H$ is a countable collection of nowhere dense sets.
By the Baire Category theorem  for compact Hausdorff spaces, we get that $H(A)=S\sim \bold H\cup \bold G$ is dense in $S$.
Accordingly let $F$ be an ultrafilter in $N_a\cap X$.
By the very choice of $F$, it follows that $a\in F$ and  we have the following 
\begin{equation}
\begin{split}
(\forall j<\beta)(\forall x\in B)({\sf c}_jx\in F\implies
(\exists j\notin \Delta x){\sf s}_j^ix\in F.)
\end{split}
\end{equation}
and 
\begin{equation}
\begin{split}
(\forall i<\kappa)(\forall \tau\in V)(\exists x\in X_i){\sf s}_{\tau}x\notin F. 
\end{split}
\end{equation}
Next we form the canonical representation corresponding to $F$
in which satisfaction coincides with genericity. 
To handle equality. we define
$$E=\{(i,j)\in {}^2{\alpha}: {\sf d}_{ij}\in F\}.$$
$E$ is an equivalence relation on $\alpha$.   
$E$ is reflexive because ${\sf d}_{ii}=1$ and symmetric 
because ${\sf d}_{ij}={\sf d}_{ji}.$
$E$ is transitive because $F$ is a filter and for all $k,l,u<\alpha$, with $l\notin \{k,u\}$, 
we have 
$${\sf d}_{kl}\cdot {\sf d}_{lu}\leq {\sf c}_l({\sf d}_{kl}\cdot {\sf d}_{lu})={\sf d}_{ku}.$$
Let $M= \alpha/E$ and for $i\in \omega$, let $q(i)=i/E$. 
Let $W$ be the weak space $^{\alpha}M^{(q)}.$
For $h\in W,$ we write $h=\bar{\tau}$ if $\tau\in V$ is such that
$\tau(i)/E=h(i)$ for all $i\in \omega$. $\tau$ of course may
not be unique.
Define $f$ from $\B$ to the full weak set algebra with unit $W$ as follows:
$$f(x)=\{ \tau\in {}^{\omega}\omega:  {\sf s}_{\tau}x\in F\}, \text { for } x\in \A.$$ 
Then it can be checked that $f$
is a homomorphism 
such that $f(a)\neq 0$ and 
$\bigcap f(X_i)=\emptyset$ for all $i\in \kappa$, hence the desired conclusion.

\end{demo}

%If $\A$ is locally finite, then we refer to $F$ satisfying (1) and (2) above as Henkin ultrafilier. 
The above theorem can be proved using games. Indeed, the connection of games to the Baire category approach 
can be established using the famous Banach-Mazur theorem. An ultrafilter in the above proof is called a Henkn ultrfailter. 
Henkin ultrafilters give rise to representations (models) and carefully chosen one gives models that omit 
a given family of types

\section{Counting Henkin ultrafilers}


This is equivalent to countng models, and it can be used to count non isomorphic countable models for the locally finite
cylindric algebras representing a countable first order theory. This is a non-trivial topic in Model theory. This is Vaughts conjecture.


Let $\A$ be any Boolean algebra. The set of ultrafilters of $\A$ is denoted by $\mathcal{U}(\A)$. 
The Stone topology  makes $\mathcal{U}(\A)$ a compact Hausdorff space; we denote this space by $\A^*$. Recall that the Stone topology has as its basic open sets the sets $\{N_x:x\in A\}$, where
$$N_x=\{\mathcal F\in\mathcal{U}(\A):x\in\mathcal F\}.$$
It is easy to see that if $A$ is countable, then $\A^*$ is \emph{Polish}, (i.e., separable and completely metrizable).

Now, suppose $\A$ is a locally finite cylindric or quasi-polyadic $\omega$-dimensional algebra with a countable universe.
Note that if $T$ is a theory in a countable language with (without) equality, then $CA(T)$, (respectively $QPA(T)$), satisfies these requirements.
Let $$\mathcal{H}(\A)=\bigcap_{i<\omega,x\in A}(N_{-c_ix}\cup\bigcup_{j<\omega}N_{s^i_jx})$$ and, in the cylindric algebraic case, let

$$\mathcal{H}'(\A)=\mathcal{H}(\A)\cap\bigcap_{i\neq j\in\omega}N_{-d_{ij}}.$$
Note, for later use, that $\mathcal{H}(\A)$ and $\mathcal{H}'(\A)$ are $G_\delta$ subsets of $\A^*$, and are nonempty,  as a matter of face it is dense-- this latter fact can be seen,
for example, from Theorem \ref{th1} below -- and are therefore Polish spaces; (see \cite{Kechris}).
Assume $\mathcal F\in \mathcal{H}(\A).$ For any $x\in A$, define the function $\mathrm{rep}_{\mathcal F}$ to be $$\mathrm{rep}_{\mathcal F}(x)=\{\tau\in{}^\omega\omega:s^+_\tau x\in \mathcal F\}.$$
We have the following results due to G. S\'agi and D. Szir\'aki; (see \cite{Sagi}).
\begin{theorem}\label{th1}
If $\mathcal F\in \mathcal{H}'(\A)$, (respectively $\mathcal{H}(\A)$), then $\mathrm{rep}_{\mathcal F}$ is a homomorphism from $\A$ onto an element of $Lf_\omega\cap Cs_\omega^{reg}$, (respectively $LfQPA_\omega\cap Qs_\omega^{reg}$), with base $\omega.$ Conversely, if $h$ is a homomorphism from $\A$ onto an element of $Lf_\omega\cap Cs_\omega^{reg}$, (respectively $LfQPA_\omega\cap Qs_\omega^{reg}$), with base $\omega$, then there is a unique $\mathcal F\in \mathcal{H}'(\A)$, (respectively $\mathcal{H}(\A)$), such that $h=\mathrm{rep}_{\mathcal F}.$
\end{theorem}
\begin{theorem}
Let $T$ be a consistent first order theory in a countable language with (without) equality. Let $\mathcal{M}_0$ and $\mathcal{M}_1$ be two models of $T$ whose universe is $\omega$. Suppose $\mathcal{F}_0,\mathcal{F}_1\in\mathcal{H}'(CA(T))$, (respectively $\mathcal{H}(QPA(T))$), are such that $\mathrm{rep}_{\mathcal{F}_i}$ are homomorphisms from $CA(T)$, (respectively $QPA(T)$), onto $Cs(\mathcal{M}_i)$, (respectively $Qs(\mathcal{M}_i)$), for $i=0,1$. If %$\rho\in S_\infty$,
$\rho:\omega\longrightarrow\omega$ is a bijection,
then the following are equivalent:
 \begin{enumerate}
 \item $\rho:\mathcal{M}_0\longrightarrow\mathcal{M}_1$ is an isomorphism.
 \item $\mathcal{F}_1=s^+_\rho \mathcal{F}_0=\{s^+_\rho x:x\in \mathcal{F}_0\}.$
\end{enumerate}
\end{theorem}

These last two theorems allow us to study models and count them via corresponding ultrafilters. This approach was initiated by S\'agi.
The main advantage of such an approach is that results proved
for locally finite cylindric algebras transfer {\it mutatis mutandis} to quasi-polyadic algebras (without diagonal elements).
So from the algebraic point of view we do the difficult task only once,
but from the model theoretic point of view we
obtain deep theorems for first order logic {\it without} equality, as well, which are more often than not, not obvious to prove without
the process of algebraisation. There are ceratin results that have an easy metalogical proof when we have equality.
but this proof does not work  in the absence of equality.
However, a purely algebraic proof works.

Now we define an equivalence relation on the Henkin utrafilters of a locally finite cylindric algebra
and  show that it is Borel subset of the product of the Stone space.
This implies that it satisfies the Glimm-Effros dichotomy, and so has either countably many or else continuum many equivalence classes.
The equivalence relation we introduce  corresponds to a non-trivial equivalence relation between models which is weaker than isomorphism.
\begin{definition}[Notation]
Let $\mathcal{F}$ be an ultrafilter of a locally finite (cylindric or quasi-polyadic) algebra $\A$.
For $a\in A$ define $$Sat_\mathcal{F}(a)=\{t|_{\Delta a}: t\in {}^{\omega}\omega,\;s^+_t a\in\mathcal{F}\}.$$
\end{definition}

Throughout, $\A$ is countable. We define an equivalence relation $\mathcal{E}$ 
on the space $\mathcal{H}'(\A)$ (or $\mathcal{H}(\A))$  that turns out to be Borel.

\begin{definition}
Let $\mathcal{E}$ be the following equivalence relation on $\mathcal{H}'(\A)$ (or $\mathcal{H}(\A))$ :\small
$$\mathcal{E}=\{(\mathcal{F}_0,\mathcal{F}_1): (\forall a\in A) (|Sat_{\mathcal{F}_0}(a)|=|Sat_{\mathcal{F}_1}(a)|)\}.$$
\end{definition}

We say that $\mathcal{F}_0,\mathcal{F}_1\in\mathcal{H}'(\A)$(or $\mathcal{H}(\A))$ are 
\emph{distinguishable } if $(\mathcal{F}_0,\mathcal{F}_1)\notin\mathcal{E}.$ 
We also say that two models of a theory $T$ are distinguishable if their corresponding 
ultrafilters in $\mathcal{H}'(CA(T))$(or $\mathcal{H}(QPA(T)))$ are distinguishable. 
That is, two models are distinguishable if they disagree in the number of realizations they have for some formula.
$\mathcal{E}$ is Borel. 
\begin{proof}
Here is an argument. Suppose we have a language $L$ with equality.
First note that if  $L^*=L_0\cup L_1$ where $L_0$ and $L_1$ are disjoint copies of $L$,
then $X_{L^*}\cong X_{L_0}\times X_{L_1}$ (where the spaces $X_L$'s are defined as in \cite{BeckerKechris} page 22).

For each formula $\varphi,$ let $\varphi^*$ be the
sentence $\bigwedge_{n\in\omega}(\exists^n \bar{x})\varphi_0(\bar{x})\leftrightarrow (\exists^n \bar{x})\varphi_1(\bar{x})$
where $\varphi_0,$ $\varphi_1$ are the copies of $\varphi$ in $L_0,L_1$ respectively, and $\exists^n$ is a shorthand for
``there exists at least $n$ tuples such that ..."

It is then immediate that two models $M_0,M_1$ of $L$ are not distinguishable iff the model $M$ of $L^*$ such
that $M|_{L_0}=M_0$ and $M|_{L_1}=M_1$ satisfies $\bigwedge_{\varphi\in L}\varphi^*.$
This means that our equivalence relation between models corresponds to the subset of $X_{L^*}$ of models of
the formula $\bigwedge_{\varphi\in L}\varphi^*.$ Such a subset is Borel by Theorem 16.8 in \cite{Kechris}.
\end{proof}

Vaught's conjecture has been confirmed when we restrict the action on certain subgroups of $G$.
But in this  case there might be isomorphic models that the group $G$ does not 'see' (the isomorphism witnessing this
can be outside $G$) so the equivalence relation is drastically different.

\begin{theorem} Let $G\subseteq S_{\infty}$ be a cli group. Then $|{\cal H}(\A)/E_G|\leq \omega$ or $|{\cal H}(\A)/E_G|=2^{\omega}$
\end{theorem}
\begin{demo}{Proof}  It is known that the number of orbits of $E_G$  satisfies the so-called Glimm Effros Dichotomy.
By known results in the literature on the topological version of
Vaught's conjecture, we have ${\cal H}(\A)/E_G$
is either at most countable or ${\cal H}(\A)/E_G$ contains continuum many non equivalent elements
(i.e non-isomorphic models).
\end{demo}
It is known that the number of orbits of $E=E_{S_{\infty}}$ {\it does not} satisfy the Glimm Effros Dichotomy.
We note that cli groups cover all natural extensions of abelian groups, like nilpotent and solvable groups.
Now we give a topological condition that implies Vaught's conjecture.
Let everything be as above with $G$ denoting a Polish subgroup of $S_{\infty}$.
Give ${\cal H}(\A)/E_G$ the qoutient topology and let $\pi: {\cal H}(\A)\to {\cal H}(\A)/E_G$ be the projection map.
$\pi$ of course depends on $G$, we somtimes denote it by $\pi_G$ to emphasize the dependence.

%{\bf In the next theorem I do not remember the exact condition on $\pi$. Is it always open?
%This also had to do with proving that the $H(A)/E_G$ is Hausdorff?  Please adjust}.

\begin{lemma} $\pi$ is open.
\end{lemma}
\begin{proof}
To show that $\pi$ is open it is enough to show for arbitrary $a\in \A$ that $\pi^{-1}(\pi(N_a))$ is open. For,
\begin{align*}
\pi^{-1}(\pi(N_a))&=\{F\in\A^*:(\exists F'\in N_a)(F,F')\in E\}\\
&=\{F\in\A^*:(\exists F'\in N_a)(\exists \rho\in G)s^+_\rho F'=F\}\\
&=\{F\in\A^*:(\exists F'\in N_a)(\exists \rho\in G)F'=s^+_{\rho^{-1} }F\}\\
&=\{F\in\A^*:(\exists \rho\in G)s^+_{\rho^{-1} }F\in N_a\}\\
&=\{F\in\A^*:(\exists \rho\in G)a\in s^+_{\rho^{-1} }F\}\\
&=\{F\in\A^*:(\exists \rho\in G)s^+_{\rho }a\in F\}\\
&=\bigcup_{\rho\in G}N_{s^+_\rho a}
\end{align*}
\end{proof}

\begin{theorem} If $\pi$ is closed, then Vaught's conjecture holds.
\end{theorem}
\begin{demo}{Proof} We have ${\cal H}(\A)$ is Borel subset of $\A^*$, the Stone space of $\A$ and ${\cal H}(\A)/E_G$ is a continuous image of
${\cal H}(\A)$.
Because $\pi$ is open, $H(\A)/E$  is second countable.
Now, since $H(\A)$ is metrizable and second countable, it is normal.
But $\pi$ is closed, and so $H(\A)/E$ is also normal, hence regular. Thus ${\cal H}(\A)/E_G$
can also be embedded as an open set in $\mathbb{R}^{\omega},$ hence it is Polish.
If ${\cal H}(\A)/E_G$ is uncountable, then being the continuous image under a map between two Polish spaces of a Borel set, it is analytic.
Then it has the power of the continuum.
\end{demo}

\subsection{ Number of models omitting a given family of types}


%\subsection{Insert the proof of omitting types}
The way we counted the ultrafilters (corresponding to distinguishable models)
above gives a completely analogous  result when we count ultrafilters corresponding  to models {\it omitting} a countable set of non-isolated types.

Given a countable locally finite algebra $\A$, a non-zero $a\in A$ and a non-principal type
$X\subseteq \Nr_n\A$, so that $\prod X=0$, one constructs a model omitting $X$, by finding a Henkin ultrafilter preserving
the following set of infinitary joins and meets where $x\in A$, $i,j\in \omega$ and $\tau$ is a finite transformation:
$c_ix=\sum s_j^ix,$
and $\prod s_{\tau}X=0.$
Working in the Stone space, one finds an ultrafilter in $N_a$ outside the
nowhere dense sets
$N_{i,x}=S\sim \bigcup N_{s_j^i}$ and
$H_{\tau}=\bigcap_{x\in X} N_{\tau}x.$
Now suppose we want to count the number of distinguishable models omitting a
family $\Gamma=\{\Gamma_i:i<\lambda\}$ ($\lambda<covK)$
of non-isolated types of $T$.

Then $$\mathbb{H}=
\mathcal{H}(CA(T))\mbox{(or }\mathcal{H}'(QPA(T)))\cap\sim  \bigcup_{i\in\lambda,\tau\in W}\bigcap_{\varphi\in \Gamma_i}N_{s_\tau (\varphi/\equiv_T)}$$
(where $W=\{\tau\in{}^\omega\omega:|i:\tau(i)\neq i|<\omega\}$) is clearly (by the above discussion)
the space of ultrafilters corresponding to models of $T$ omitting $\Gamma.$

But then by properties of $covK$ union  $\bigcup_{i\in\lambda}$
can be reduced to a countable union.
We then have $\mathbb{H}$ a $G_\delta$ subset of a Polish space. So $\mathbb{H}$ is
Polish and moreover, $\mathcal{E}'=\mathcal{E}\cap (\mathbb{H}\times \mathbb{H})$
is a Borel equivalence relation on $\mathbb{H}.$ It follows then that the number of distinguishable models omitting $\Gamma$
is either countable or else $2^\omega.$
We readily obtain:
\begin{corollary}
Let $T$ be a first order theory  in a countable language (with or without equality).
If  $T$ has an uncountable set of countable models that omit $< covK$ many non principal types that
are pairwise distinguishable, then actually it has such a set of size $2^{\aleph_0}$.
\end{corollary}
Using the same reasoning as above conjuncted with Morleys theorem, we get
\begin{theorem}
The number of countable models of a countable theory that omits $< covK$ many types is either $\leq \omega$ or
$\omega_1$ or $^{\omega}2$.
\end{theorem}
We can also formulate a version of Vaughts conjecture for s called rich languages.
\begin{theorem}\label{2} Let $\A\in Dc_{\alpha}$ be countable simple and finitely generated.
Then the number of non-base isomorphic representations of $\A$ is $2^{\omega}$.
\end{theorem}
\begin{proof} Let $V={}^{\alpha}\alpha^{(Id)}$ and let $\A$ be as in the hypothesis. Then $\A$ cannot be atomic \cite{HMT1} corollary 2.3.33,
least hereditary atomic. Then it has $2^{\omega}$ ultrafilters.
For an ultrafilter $F$, let $h_F(a)=\{\tau \in V: s_{\tau}a\in F\}$, $a\in \A$.
Then $h_F\neq 0$, indeed $Id\in h_F(a)$ for any $a\in \A$, hence $h_F$ is an injection, by simplicity of $\A$.
Now $h_F:\A\to \wp(V)$; all the $h_F$'s have the same target algebra.
Then  $h_F(\A)$ is base isomorphic to $h_G(\A)$ iff there exists a finite bijection $\sigma\in V$ such that
$s_{\sigma}F=G$.
Define the equivalence relation $\sim $ on the set of ultrafilters by $F\sim G$, if there exists a finite permutation $\sigma$
such that $F=s_{\sigma}G$. Then any equivalence class is countable, and so we have $^{\omega}2$ many orbits, which correspond to
the non base isomorphic representations of $\A$.
\end{proof}
\begin{theorem}
Let $T$ be a countable theory in a rich language, with only finitely many relation symbols,
and $\Gamma =\{\Gamma_i: i\in covK\}$ be non isolated types.
Then $T$ has $2^{\omega}$ weak models that omit $\Gamma$.
\end{theorem}

Notice that this theorem substantially generalized the main theorem in \ref{OTT}, 
the latter shows that there exists at least one model omiytting non principal types, this theorem says that there are continuum many of 
them.
%When the Polish group is uncountable, finding the number of orbits is still an open question, of which Vaught's
%conjecture is an instance (when the group is the symmetric g

\subsection{Omitting types for finite variable fragments}

For finite variable fragments $\L_n$ for $n\geq 3$, the situation turns out to be drastically different.
But first a definition.
\begin{definition}
Assume that $T\subseteq \L_n$. We say that $T$ is $n$ complete iff for all sentences $\phi\in \L_n$ 
we have either $T\models \phi$ or $T\models \neg \phi$. We say that $T$
is $n$ atomic iff for all $\phi\in \L_n$, there is $\psi\in \L_n$ such that $T\models \psi\to \phi$ and for all $\eta\in \L_n$ either $T\models \psi\to \eta$
or $T\models \psi\to \neg \eta.$
\end{definition}


The next theorem \ref{finite} is proved using algebraic logic in \cite{ANT}, 
using combinatorial techniques depending on Ramsey's theorem. 
%The proof will be sketched in the appendix, since it connects apparently 
%remote areas in (algebraic) logic.
%It shows that the condition of strong subalgebra is essential.


\begin{theorem}\label{finite} Assume that $\L$ is a countable first order language containing a binary relation symbol. For $n>2$ and 
$k\geq 0$,  there are a consistent $n$ complete
and $n$ atomic theory $T$ using only $n$ variables, and a set $\Gamma(x_1)$ using only $3$ variables
(and only one free variable) such that $\Gamma$ is realized in all models of $T$ but each $T$-witness for $T$ uses
more that $n+k$ variables
\end{theorem}
\begin{demo}{Proof} \cite{ANT} for a full proof and \cite{OTT} for an easier sketch. 
Below we give a simplified version of the proof, but as an expense we obtain a somewaht 
weaker result.
\end{demo}

Algebraisations of finite variable fragments of first order logic with $n$ variables is 
obtained from locally finite algebras by truncating the dimensions at $n$.
Expressed, formally this corresponds to the operation of forming $n$ neat reducts.

\begin{definition}\cite{HMT1} Let $\A\in \CA_{\beta}$ and $\alpha<\beta$, then the {\it $\alpha$ neat reduct of $\A$} is the algebra obtained 
from $\A$ by discarding operations in $\beta\sim \alpha$ and restricting the remaining operations 
to the set consisting only of $\alpha$ dimensional elements. An element is $\alpha$ dimensional if its dimension set, $\Delta x=\{i\in \beta: c_ix\neq x\}$ 
is contained in $\alpha$. Such an algebra is denoted by $\Nr_{\alpha}\A$.
\end{definition}
For a class $K\subseteq \CA_{\beta}$, $\Nr_{\alpha}K=\{\Nr_{\alpha}\A: A\in K\}$. It is easy to verify that $\Nr_{\alpha}K\subseteq \CA_{\alpha}$.

A class of particular importance, is the class $S\Nr_{\alpha}\CA_{\alpha+\omega}$ where $\alpha$ is an arbitrary ordinal; 
here $S$ stands for the operation of forming subalgebras. 
This class turns out to be a variety which coincides with
the class of {\it representable} algebras of dimension $\alpha$.

Another class that is of significance is the class $S_c\Nr_{\alpha}CA_{\alpha+\omega}$. Here $S_c$ 
is the operation of forming {\it complete} subalgebras. 
(A Boolean algebra $\A$ is a complete subalgebra of $\B$, if for all $X\subseteq \A$, whenever $\sum X=1$ in $A$, then 
$\sum X=1$ in $\B)$. This class is important because a countable cylindric algebra of dimension $n$ is completey representable if and only if
$\A\in S_c\Nr_n\CA_{\omega}$, for any $\alpha\geq \omega$ and $\A$ is atomic. 
%We will extend this result to countable infinite dimensions where we change the notion of  representabilty 
This characterization even works for countable dimensions, 
by modifying the notion of complete representation' relativizing the unit to so-called weak units.


And it turns out for finite  variable fragments, that for a theory $T$ to omit types, 
whether countably or uncountably many, a sufficient condition is that 
the cylindric algebra of formulas $\Fm_T$ is in the class $S_c\Nr_n\CA_{\omega}$. 
Furthermore, the condition of  {\it complete} subalgebras, cannot be omitted.




\subsection{Omitting types for finite first order definable extensions of $\L_n$}

A question raised by Jonsson is whether non finite axiomatizability-results for the class $\RCA_n$ $n\geq 3$, can be circumvented by 
adding finitely many {\it first order definable} operations. This has to do with seeking a completeness theorem for $\L_n$.
The was answered negatively by Biro. Now we ask the same question for omitting types, 
and we also get a negative answer. But first we recall what we mean by first order definable.

\begin{definition} 
Let $\Lambda$ be a first order language with countably many 
relation symbols, $R_0, \ldots R_i,\ldots : i\in \omega$ 
each of arity $n$. 
Let $\Cs_{n,t}$ denote the following class of similarity type $t$:
\begin{enumroman}
\item $t$ is an expansion of $t_{\CA_n}.$
\item  $S\Rd_{ca}Cs_{n,t}=\Cs_n.$ In particular, every algebra in $\Cs_{n,t}$ is a boolean 
field of sets with unit $^nU$ say, 
that is closed under cylindrifications and contains diagonal elements.   
\item For any $m$-ary operation $f$ of $t$, there exists a first order formula
$\phi$ with free variables among $\{x_0,\ldots, x_n\}$ 
and having exactly $m,$ $n$-ary relation symbols
$R_0, \ldots R_{m-1}$ such that,
for any set algebra ${\cal A}\in Cs_{n,t}$ 
with base $U$, and $X_0, \ldots X_{m-1}\in {\cal A}$,
we have:
$$\langle a_0,\ldots a_{n-1}\rangle\in f(X_0,\ldots X_{m-1})$$
if and only if  
$${\cal M}=\langle U, X_0,\ldots X_{n-1}\rangle\models \phi[a_0,\ldots a_{n-1}].$$
Here $\cal M$ is the first order structure in which for each $i\leq m$, 
$R_i$ is interpreted as $X_i,$ and $\models$ is the usual satisfiability relation.
Note that cylindrifications and diagonal elements are so definable.
Indeed for $i,j<n$,  $\exists x_iR_0(x_0\ldots x_{n-1})$ 
defines $C_i$ and $x_i=x_j$ defines $D_{ij}.$ 
\item With $f$ and $\phi$ as above,   
$f$ is said to be a first order definable operation with $\phi$ defining $f$, 
or simply a first order definable 
operation, and $\Cs_{n,t}$ is said to be a first order definable
expansion of $\Cs_n.$
\item $\RCA_{n,t}$ denotes the class $SP\Cs_{n,t}$, i.e. the class of all subdirect products 
of algebras
in $\Cs_{n,t}.$ We also refer to 
$\RCA_{n,t}$ as a first order definable expansion of $\RCA_n.$
\end{enumroman}
\end{definition}

\begin{theorem} Let $t$ be a finite expansion of the $CA$ type.
Then there are atomic algebras in $\RCA_{n,t}$ that are not completely representable
\end{theorem}
\begin{proof} Let $k\in \omega$ such that all first order definable operations are built up using at most $n+k$ varialbes. 
This $k$ exists, because we have only finitely many of those operations. Let 
$\A\in RCA_n\cap \Nr_nCA_{n+k}$ be an atomic cylindric algebra that
is not completely representable. 
Then $\A$ is closed under all the first order definable operations. But the cylindric reduct of $\A$, is not completely representable, then
$\A$ itself is not.
\end{proof}
 
$\RCA_{n,t}$ is a variety, this can be proved exactly like the $\RCA_n$ case, and in fact it is a 
discriminator variety with discrimintor term $c_0\ldots c_{n-1}$.
(So it is enough to show that it is closed under ultraproducts.)
A result of Biro says that is not finitely axiomatizable. 

The logic corresponding to $\RCA_{n,t}$ has $n$ variables, and it has the usual first order connectives;
(here we view cylindrifiers as unary connectives) and it has one connective for each first order definable operation.
In atomic formulas the variables only appear in their natural order; so that they are {\it restricted}, in the sense of \cite{HMT1} sec 4.3.

Semantics are defined as follows: For simplicity we assume that we have an at most  countable collection of $n$-ary relation symbols
$\{R_i: i\in \omega\}$. 
(We will be considering countable languages anyway, to violate the omitting types theorem).

The inductive definition for the first order (usual) connectives is the usual.
Now given a model $\M=(M, R_i)_{i\in I}$ and a formula $\psi$ of $L_n^+$; with a corresponding connective 
$f_{\phi}$ which we assume is unary to simplify 
matters, and $s\in {}^nM$, then for any formula $\psi$: 
$$\M\models f_{\phi}(\psi)[s]\Longleftrightarrow (\M, \phi^{\M})\models \psi[s].$$

$_rL_n$ is the logic correponding to $\CA_n$, which has only restricted formulas, 
and $L_n$ is that correponding to $\RPEA_n$; the latter of course is a first order extension of the former 
because the substitutions corresponding to transpositions
are first order definable.

\begin{corollary}  
No first order finite extension of $L_n$  enjoys an omitting types theorem. 
\end{corollary}

\begin{proof} Let $\A$ be as above, and let $\Fm_T$ be the corresponding Tarski-Lindenbaum algebra. 
We assume that $\A=\Fm_T$ (not just isomorphic) Let $X=\At\A$ and let $\Gamma=\{\phi: \phi/\equiv\in X\}.$
Then $\Gamma$ cannot be omitted.
\end{proof}
%\begin{demo}{Proof} \cite{M71}.
%\end{demo}

\subsection{Omitting uncountably many types}

Here we give an independence result, concerning omitting a number of non-isolated types.
The we delineating the edges of the independence proofs by considering {\it complete} types,
giving a proof in $ZFC$ of the resulting statement 
that on first sight would be suspected of being independent. 
Proofs of this kind is often extremely subtle and surprising.

Inthe proof of the second item in our next theorem, we will be constructing countable models that omit
possibly uncountable types in $\L_n$, $n\in \omega$. This is theorem is formulated in \cite{OTT} without proof.
Now we provide one:
\begin{definition} 
\begin{enumroman}
\item Let $\kappa$ be a cardinal. Let $OTT(\kappa)$ be the following statement. 
If $\A\in S_c\Nr_n\CA_{\omega}$ is countable and for $i\in \kappa$, $X_i\subseteq A$ are such that $\prod X_i=0$, then for all $a\neq 0$, there exists
a set algebra $\C$ with countable base, $f:\A\to C$ such that $f(a)\neq 0$ and for all $i\in \kappa$, $\bigcap_{x\in X_i} f(x)=0.$
\item Let $OTT$ be the statement that 
$$(\forall k<{}^{\omega}2)(OTT(\kappa))$$
\item Let $OTT_m(\kappa)$ be the statement obtained from $OTT(\kappa)$ by replacing $X_i$ with ``nonprincipal ultrafilter $F_i$"
and $OTT_m$ be the statement 
$$(\forall k<{}^{\omega}2)(OTT_m(\kappa))$$
\end{enumroman}
\end{definition}
Now we formulate and prove our main theorem (this theorem is stated in \cite{OTT}):
\begin{theorem}\label{OT} 
\begin{enumroman}
\item $OTT$ is independent from $ZFC+\neg CH$. 
\item  $OTT_m$ is provable in $ZFC$.
\end{enumroman}
\end{theorem}
\begin{proof}
\begin{enumroman} 
\item Let $covK$ be the least cardinal $\kappa$ such that the real line can be covered by $\kappa$ many closed disjoint 
nowhere dense sets. It is known that 
$\omega<covK\leq 2^{\omega}$  By theorem \ref{infinite} $< covK$ types can be omitted.
But Martin's axiom implies that $covK =2^{\omega}$. 
So we have consistency. 
The independence is proved using standard iterated forcing to show that it is consistent that $covK$ could be literally anything greater 
than  $\omega$ and $\leq {}^{\omega}2$, and then show that $OTT(covK)$ is false, see example \ref{cov}. 
%(Those two statements will be proved below, by direct modifications of known results
%and methods

\item  The idea is that one can build several models such that they overlap only on isolated types.
One can build {\it two} models so that every maximal type which is realized in both is isolated. Using the jargon of Robinson's finite forcing
implemented via games, the idea
is that one distributes this job of building the two models among experts, each has a role to play, and that all have 
winning strategies. There is no difficulty in stretching the above idea to make the experts build three, four or any finite number of models 
which overlap only at principal types. 
With a pinch of diagonalisation we can extend the number to $\omega$. 

To push it still further to $^{\omega}2$ needs an entirely new idea (due to Shelah), which we will implement.
%This is one of those many places in model theory where we get continuum many models for the same price as two.

Algebraically, we first construct two representations of $\B$ such that if $F$ is an ultrafilter in $B$ that is realized in both representations, then $F$ is
necessarily principal, that is $\prod F$ is an atom generating $F$, then we sketch the idea of how to obtain $^{\omega}2$ many.
We construct two ultrafilters $T$ and $S$ of $\B$ such that (*)
$\forall \tau_1, \tau_2\in {}^{\omega}\omega( G_1=\{a\in \B: {\sf s}_{\tau_1}a\in T\},
G_2=\{a\in \B: s_{\tau_1}a\in S\})\\\implies G_1\neq G_2 \text { or $G_1$ is principal.}$
Note that $G_1$ and $G_2$ are indeed ultrafilters.
We construct $S$ and $T$ as a union of a chain. We carry out various tasks as we build the chains.
The tasks are as in (*), as well as

(**) for all $a\in A$, if $c_ka\in T$, then $s_l^kx\in T$ for $l\notin \Delta x$.

(***) for all $a\in A$ either $a\in T$ or $-a\in T$, and same for $S$.

We let $S_0=T_0=\{1\}$.
There are countably many tasks. Metaphorically we hire countably many experts and give them one task each.
We partition $\omega$ into infinitely many sets and we assign one of these tasks to each expert.
When $T_{i-1}$ and $S_{i-1}$ have been chosen and $i$ is in the set assigned to some expert $E$, then $E$ will construct
$T_i$ and $S_i$.

Let us start with  task $(**)$. The expert waits until she is given a set $T_{i-1}$ which contains $c_ka$ for some $k<\omega$.
 Every time this happen she look for a {\it witness} $l$ which is outside elements in $T_{i-1}$; 
this is possible since the latter is finite, then she sets $T_i=T_{i-1}\cup \{s_k^la\}$. 
Otherwise, she does nothing. This strategy works because her subset
of $\omega$ is infinite, hence contains arbitrarily large numbers. Same for $S_i$.

Now consider the expert who handles task (**). Let $X$ be her subset of $\omega$. Let her list as $(a_i: i\in X)$ all elements of $X$.
When $T_{i-1}$ has been chosen with $i\in X$, she should consider whether $T_{i-1}\cup \{a_i\}$ is consistent. If it is she puts
$T_i=T_{i-1}\cup \{a_i\}$. If not she puts $T_i=T_{i-1}\cup \{-a_i\}$. Same for $S_i$.
%Next consider the expert who deals with the tasks in \ref{t5}. She waits until she is gets a set $T_{i-1}$ which contains ${\sf c}_ka$.
%Every time this happens she chooses $l\notin \Delta a$
%which is not used in $T_{i-1}$, and she puts $T_i=T_{i-1}\cup \{{\sf s}_k^la\}$. Same for $S_i$.

Now finally consider the crucial tasks in (*). Suppose that $X$ contains $i,$ and $S_{i-1}$ and $T_{i-1}$ have been chosen.
Let $e=\bigwedge S_{i-1}$ and $f=\bigwedge T_{i-1}$. We have two cases.
If $e$ is an atom in $B$ then the ultrafilter $F$ containing $e$ is principal so our expert can put $S_i=S_{i-1}$ and $T_i=T_{i-1}$.
If not, then let $F_1$ , $F_2$ be distinct ultrafilters containing $e$. Let $G$ be an ultrafilter containing $e$, and assume that $F_1$ 
is different from $G$.
Let $\theta$ be in $F_1-G$. Then put $S_i=S_{i-1}\cup \{\theta\}$ and $T_i=T_{i-1}\cup \{-\theta\}.$
It is not hard to check that the canonical models, defined the usual way,  corresponding to $S$ and $T$ are as required.

To extend the idea, we allow experts at any stage to introduce a new chain of theories which is a duplicate copy of one of the chains being
constructed. The construction takes the form of a tree where each branch separately will give a chain of conditions.
By splitting the tree often enough the experts can guarantee that there are continuum many branches and hence continuum many
representations. This is a well know method, in model theory, when one gets $^{\omega}2$ many models for the price of two.
There is one expert whose job is to make sure that this property 
is enforcable for each pair of branches.
But she can do this task, because at each step the number of branches is still finite.


\item Assume not. Let ${\bold F}$ be the given set of non principal ultrafilters. Then for all $i<{}^{\omega}2$,
there exists $F$ such that $F$ is realized in $\B_i$. Let $\psi:{}^{\omega}2\to \wp(\bold F)$, be defined by
$\psi(i)=\{F: F \text { is realized in  }\B_i\}$.  Then for all $i<{}^{\omega}2$, $\psi(i)\neq \emptyset$.
Furthermore, for $i\neq j$, $\psi(i)\cap \psi(j)=\emptyset,$ for if $F\in \psi(i)\cap \psi(j)$ then it will be realized in
$\B_i$ and $\B_j$, and so it will be principal.  This implies that $|\bold F|={}^{\omega}2$ which is impossible.

\end{enumroman}
%\end{proof}
\end{proof}



\begin{example}\label{cov} Let $n>1$. To show that $OTT(cov K)$ could be false, we adapt 
an example in \cite{CF} p.242. Another example can be found in \cite{N} but also needs a bit of a modification, 
since we have only finitely many 
variables. Fix $n\in \omega$. There the example is constructed for $L_{\omega,\omega}$ to adapt 
to $L_n$ some care is required. 
Let $T$ be a theory such that for this given $n$, in $S^n(T)$, the stone space of $n$ types,  the isolated points are not dense.
(In \cite{CF}, a theory $T$ is chosen which does not have a prime model. This implies that there is an $n$ such that the isolated types in 
$S_n(T)$ are not dense; in here we need a fixed $n$, so not any theory without a 
prime model will do). 
It is easy to construct such theories, for any fixed $n$.(For example the theory of random graphs the isolated types are not dense for any $n$). 

Let $X$ be the space $S^0(T)$ of all complete $0$ types which are consistent with $T$. For an ordinal $\alpha$, let $X^{(\alpha)}$ 
be the $\alpha$-iterated Cantor-Bendixon derivative of $X$. Recall that for ordinal numbers $\alpha$ the $\alpha$ 
Cantor-Bendixon derivative of a topological space is defined by 
transfinite induction 
\begin{itemize}
\item $X^0=X$
\item $X^{\alpha+1}=[X^{\alpha}]'$
\item $X^{\beta}=\bigcap_{\alpha\in \beta} X_{\alpha}$. 
\end{itemize}
The transfinite sequence of Cantor-Bendixson derivatives of $X$ must eventually stop. 
The smallest ordinal $\alpha$ such that $X^{\alpha+1}=X^{\alpha}$
is called the Cantor-Bendixon  rank of $X$.The language is countable, there is some $\alpha<\omega_1$ 
such that $X^{(\alpha)}=X^{(\alpha+1)}$
and $X\setminus X^{\alpha}$ is countable. $X^{\alpha}$ is a perfect set and 
therefore it is homeomorphic to the Cantor space $^{\omega}2$ or it is empty.
We associate a set $P_{\infty}$ of $\leq covK$ many types with $X^{\alpha}$. Assume that $X^{\alpha}$ is non-empty, since it is a 
closed set in $X$, 
there is some extension $T_{\infty}$ of $T$
such that in $X$ 
$$X^{\alpha}=\bigcap_{\sigma\in T^{\infty}}[\sigma].$$
Hence the space $S^0(T_{\infty})$ is homeomorphic to $X^{\alpha}$ and to $^{\omega}2$. Then there are $Y_{\beta} (\beta<covK)$ 
closed nowhere dense sets in $S^0(T_{\infty})$
such that
$$S^0(T_{\infty})=\bigcup_{\beta<covK}Y_{\beta}.$$type $p_{\beta}$ such that in $S^{0}(T_{\infty})$
$$Y_{\beta}=\bigcap_{\sigma\in p_{\beta}}[\sigma].$$
As $Y_{\beta}$ is nowhere dense $p_{\beta}$ is non principal in $T_{\infty}$. Assuming, without loss, that $T_{\infty}\subseteq p_{\beta}$ we get that 
$p_{\beta}$ is non 
principal in $T$.
Set
$$P_{\infty}=\{p_{\beta}:\beta<covK\}.$$
Let us consider the $0$ types in $X\setminus X^{\alpha}$. These are complete consistent extensions of $T$.
For every $T'\in X\setminus X^{\alpha}$ we shall define a set $P_{T'}$ of $\leq covK$ many $n$ types that are not omitted in $T'$.
If $T'$ is not a finite extension of $T$, set $P_{T'}=\{T'\}$. Otherwise, in $S^n(T')$ the isolated types are not dense. 
Hence there is some non-empty $Y\subseteq
S^n(T')$ clopen and perfect. Now we can cover $Y$ with a family of $covK$ many closed nowhere dense sets of $n$ types. 
Since $Y$ is clo-open in $S^{n}(T')$, these sets are closed nowhere dense sets in $S^{n}(T')$, 
so we obtain a family of $covK$ many non principal $n$ types that cannot be omitted. We may assume that 
$T'\subseteq p$ for every $p\in P_T'$ and therefore every type in $P_T'$ is non principal in $T$.
Define
$$P=P_{\infty}\cup\bigcup\{P_{T'}: T'\in X\setminus X^{\alpha}\}.$$  
Now $P$ is a family of non-principal types $|P|=covK$ that cannot be omitted.

Let $\A=\Fm/T$ and for $p\in P$ let $X_p=\{\phi/T:\phi\in p\}$. Then $X_p\subseteq \Nr_n\A$, and $\prod X_p=0$.
However for any $0\neq a$, there is no set algebra $\C$ with countable base 
$M$ and $g:\A\to \C$ such that $g(a)\neq 0$ and $\bigcap_{x\in X_i}f(x)=\emptyset$.
But in principle, if we take the $n$ neat reduct, representations preserving meets can exist. We exclude this possibility by showing that 
such representations necessarily lift to 
all $\omega$ dimensions. Let $\B=\Nr_n\A$. Let $a\neq 0$. Assume, seeking a contradiction, that there exists 
$f:\B\to \D'$ such that $f(a)\neq 0$ and $\bigcap_{x\in X_i} f(x)=\emptyset$. We can assume that
$B$ generates $\A$ and that $\D'=\Nr_n\D$ where $\D\in \Lf_{\omega}$. Let $g=\Sg^{\A\times \D}f$. We will show  
that $g$ is a one to one function with domain $\A$ that preserves the $X_i$'s which is impossible (Note that by definition $g$ is a homomorphism). 
We have
$$Dom g=Dom\Sg^{\A\times \D}f=\Sg^{\A}Domf=\Sg^{\A}\Nr_{n}\A=\A.$$
By symmetry it is enough to show that $g$ is a function.  We first prove the following (*)
 $$ \text { If } (a,b)\in g\text { and }  {\sf c}_k(a,b)=(a,b)\text { for all } k\in \omega\sim n, \text { then } f(a)=b.$$
Indeed,
$$(a,b)\in \Nr_{n}\Sg^{\A\times \D}f=\Sg^{\Nr_{n}(\A\times \D)}f=\Sg^{\Nr_{n}\A\times \Nr_{n}\D}f=f.$$
Here we are using that $\A\times \D\in \Lf_{\omega}$, so that  $\Nr_{n}\Sg^{\A\times \D}f=\Sg^{\Nr_{n}(\A\times \D)}f.$
Now suppose that $(x,y), (x,z)\in g$.
Let $k\in \omega\sim n.$ Let $\Delta$ denote symmetric difference. Then
$$(0, {\sf c}_k(y\Delta z))=({\sf c}_k0, {\sf c}_k(y\Delta z))={\sf c}_k(0,y\Delta z)={\sf c}_k((x,y)\Delta(x,z))\in g.$$

Also,
$${\sf c}_k(0, {\sf c}_k(y\Delta z))=(0,{\sf c}_k(y\Delta z)).$$ 
Thus by (*) we have  $$f(0)={\sf c}_k(y\Delta z) \text { for any } k\in \omega\sim n.$$
Hence ${\sf c}_k(y\Delta z)=0$ and so $y=z$.
We conclude that there exists a countable $\B\in \Nr_n\CA_{\omega}$ and $(X_i:i<covK)$ such that $\prod X_i=0$ but there is no representation that
preserves the $X_i$'s. In more detail. Give any $a\in \B$, if $a$ is non zero, $\C$ is a set algebra with countable base 
and $f:\B\to \C$ is a homomorphism such 
that $f(a)\neq 0$, then there exists $i<covK$, such that $\bigcap_{x\in X_i} f(x)\neq \emptyset.$
Therefore $OTT$ is false in a model of $ZFC+\neg CH$.
\footnote{Another model in which $covK=\omega_2$ and $^{\omega}2=\omega_3$ is due to Bukovski \cite{Miller} 
who does it by starting with a model of $2^{\omega}=\omega_3$, and then doing an $\omega_2$ iteration.
At each step of the iteration he does an $\omega_3$ iteration making $MA$ true. 
Alternatively we could start with a model of $2^{\omega}=\omega_3$ and then do an $\omega_2$ iteration with $\bold D$ 
where $\bold D=\{(n,f): n<\omega, f\in {}^{\omega}\omega\}$
the order for forcing an eventually dominant real.}
\end{example}


%Assume that $\A\s
%ubseteq \Nr_n\B$ where $\B\in \Lf_{\omega}$. Then, since $A$ is countable, we can assume that $\B$ is countable.
%If not, just take $\B'=\Sg^{\B}\A$, so that $\B'$ is countable and $\A\subseteq \Nr_n\B'$, and replace $\B$ by $\B'$.

%\end{theorem}
Now we omit uncountably many types in uncountable languages.
We use the following result of Shelah's. But first a piece of notation:
For a model $M$, and $\bar{a}\in {}^{<\omega}M$, $\tp(\bar{a})$ is the set of formulas satified by $\bar{a}$ 
In our theorem, addressing $L_n$, all types considered will have of cource a uniform upper bound of free variables, namely $n$.

\begin{lemma} Suppose that $T$ is a theory, 
$|T|=\lambda$, $\lambda$ regular, then there exist models $M_i: i<\chi={}^{\lambda}2$, each of cardinality $\lambda$, 
such that if $i(1)\neq i(2)< \chi$, $\bar{a}_{i(l)}\in |M_{i(l)}|$, $l=1,2,$, $\tp(\bar{a_{l(1)}})=\tp(\bar{a_{l(2)}})$, 
then there are $p_i\subseteq \tp(\bar{a_{l(i)}}),$ $|p_i|<\lambda$ and $p_i\vdash \tp(\bar{a_ {l(i)}}).$
\end{lemma}
\begin{proof} \cite{Shelah} Theorem 5.16
\end{proof}

\begin{theorem}\label{uncountable} Let $\A=S_c\Nr_n\CA_{\omega}$. Assume that $|A|=\lambda$, where $\lambda$ is an ucountable 
cardinal. assume that  $\kappa< {}^{\lambda}2$,
and that $(F_i: i<\kappa)$ is a system of non principal ultrafilters.
Then there exists
a set algebra $\C$ with base $U$ such that $|U|\leq \lambda$, $f:\A\to \C$ such that $f(a)\neq 0$ and for all $i\in \kappa$, $\bigcap_{x\in X_i} f(x)=0.$

\end{theorem}

\begin{proof} Let $\A\subseteq_c \Nr_n\B$, where $\B$ is $\omega$ dimensional, locally finite and has the same cardinality as $\A$. 
This is possible by taking $\B$ to be the subalgebra of which $\A$ is a strong neat reduct generated by  $A$, and noting that we gave countably many 
operations. 
The $F_i$'s correspond to maximal $n$ types in the theory $T$ corresponding to $\B$, that is, the first order theory $T$ such that $\Fm_T\cong \B$. 
Applying Shelah result of the existence of $^{\lambda}2$ representations of $\B$ and restricting to ultrafilters (maximal types)  in $\Nr_n\B$, 
together  with argument (ii) above, gives a a representation with base $\M$, of the big algebra $\B$, via an injective homomorphism $g$,
omitting the given maximal types. For a sequence $s$ with finite length let $s^+=s\cup id$.
Define $f:\A\to \wp({}^{n}\M)$ via $a\mapsto \{s\in {}^nM: s\cup Id\in f(a)\},$ then clearly $f$ is as desired.

%model that omits the types.From Shelah's proof 
%\cite{Shelah} using the same ideas as above.
\end{proof}
Since $\Nr_n\CA_{\omega}\subseteq S_c\Nr_n\CA_{\omega}$, we immediately get:
\begin{corollary}
Let $\A\in \Nr_n\CA_{\omega}$ be infinite such that $|A|=\lambda$, $\lambda$ is  a regular cardinal.
Let $\kappa<2^{\lambda}$. Let $(X_i:i\in \kappa)$ be a family of non-principal ultrafilters of $\A$.
Then there exists a representation $f:\A\to \wp(^nX)$ such that $\bigcap_{x\in X_i}f(x)=\emptyset$
for all $i\in \kappa$.
\end{corollary}
%The last theorem is a new  omitting types theorem adressing the uncountable case for $\L_n$. Before we give a logical counterpart
%of the above theorems we review briefly the notion of quantifier elimination. Quantifier elimination is a concept that occurs
%in mathematical logic model theory, and 
%theoretical computer science. 
One way of classifying  formulas is by the amount of quantification. 
Formulae with less depth of quantifier alternation are thought of as simpler and the quantifier free formulae as the simplest. 
Recall that a  theory  has quantifier elimination if for every formula $\alpha$  there exists a formula $\alpha_{QF}$
without quantifiers which is equivalent to it (modulo the theory). 

%Quantifier elimination is particularly useful in proving that a given theory is decidable.
%Examples of theories that have been shown decidable using quantifier elimination are 
%Presburger arithmetic, real closed fields, atomless Boolean algebras, term algebras, dense linear orders, and random graphs.
 
Now one metalogical reading of the last two theorems is 

\begin{theorem} Let $T$ be an $\L_n$ consistent theory that admits elimination of quantifiers. 
Assume that $|T|=\lambda$ is a regular cardinal.
Let $\kappa<2^{\lambda}$. Let $(\Gamma_i:i\in \kappa)$ be a set of non-principal maximal types in $T$. Then there is a model $\M$ of $T$ that omits
all the $\Gamma_i$'s
\end{theorem}
\begin{demo}{Proof} If $\A=\Fm_T$ denotes the cylindric algebra corresponding to $T$, then since $T$ admits elimination of quantifiers, then
$\A\in \Nr_n\CA_{\omega}$. This follows from the following reasoning. Let $\B=\Fm_{T_{\omega}}$ be the locally finite cylindric algebra
based on $T$ but now allowing $\omega$ many variables. Consider the map $\phi/T\mapsto \phi/T_{\omega}$. 
Then this map is from $\A$ into $\Nr_n\B$. But since $T$ admits elimination of quantifiers the map is onto.
The Theorem now follows.
\end{demo}

We now give another natural omitting types theorem for certain uncountable languages.
Let $L$ be an ordinary first order language with 
a list $\langle c_k\rangle$ of individual constants
of order type $\alpha$. $L$ has no operation symbols, but as usual, the
list of variables is of order type $\omega$. Denote by $Sn^{L_{\alpha}}$ the set
of all $L$ sentences, the subscrpt $\alpha$ indicating that we have $\alpha$ many constants Let $\alpha=n\in \omega$. 
Let $T\subseteq Sn^{L_0}$ be consistent. Let $\M$  be an $\L_0$  model of $T$.  Then any $s:n\to M$ defines an expansion of $\M$ to $L_n$ which we denote by $\M{[s]}$.  
For $\phi\in L_n$ let $\phi^{\M}=\{s\in M^n: \M[s]\models \phi\}$. Let 
$\Gamma\subseteq Sn^{L_{n}}$. 
The question we adress is: Is there a model $\M$ of $T$ such that for no expansion $s:n\to M$ we have 
$s\in  \bigcap_{\phi\in \Gamma}\phi^{M}$. 
Such an $\M$ omits $\Gamma$. Call $\Gamma$ principal over $T$ if there exists $\psi\in L_n$ consistent with $T$ such that $T\models \psi\to \Gamma.$ 
Other wise $\Gamma$ is non principal over T.

\begin{theorem}  Let $T\subseteq Sn^{L_0}$ be consistent and assume that $\lambda$ is a regular cardinal, and $|T|=\lambda$. 
Let $\kappa<2^{\lambda}$. Let $(\Gamma_i:i\in \kappa)$ be a set of non-principal maximal types in $T$. 
Then there is a model $\M$ of $T$ that omits
all the $\Gamma_i$'s
That is, there exists a model $\M\models T$ such that is no $s:n\to \M$
such that   $s\in \bigcap_{\phi\in \Gamma_i}\phi^{\M}$. 
\end{theorem}
\begin{demo}{Proof}
Let $T\subseteq Sn^{L_0}$ be consistent. Let $\M$ be an $\L_0$  model of $T$. 
For $\phi\in Sn^L$ and $k<\alpha$
let $\exists_k\phi:=\exists x\phi(c_k|x)$ where $x$ is the first variable
not occuring in $\phi$. Here $\phi(c_k|x)$ is the formula obtained from $\phi$ by 
replacing all occurences of $c_k$, if any, in $\phi$ by $x$.
Let $T$ be as indicated above, i.e $T$ is a set of sentences in which no constants occur. Define the
equivalence relation $\equiv_{T}$ on $^2Sn^L$ as follows
$$\phi\equiv_{T}\psi \text { iff } T\models \phi\equiv \psi.$$
Then, as easily checked $\equiv_{T}$ is a 
congruence relation on the algebra 
$$\Sn=\langle Sn,\land,\lor,\neg,T,F,\exists_k, c_k=c_l\rangle_{k,l<n}$$  
We let $\Sn^L/T$ denote the quotient algebra.
In this case, it is easy to see that $\Sn^L/T$ is a $\CA_n$, in fact is an $\RCA_n$.
Let $L$ be as described in  above. But now we denote it $L_n$, 
the subsript $n$ indicating that we have $n$-many 
individual constants. Now enrich $L_{n}$ 
with countably many constants (and nothing else) obtaining
$L_{\omega}$. 
Recall that both languages, now, have a list of $\omega$ variables. 
For $\kappa\in \{n, \omega\}$ 
let $\A_{\kappa}=\Sn^{L_{k}}/{T}$. 
For $\phi\in Sn^{L_n}$, let $f(\phi/T)=\phi/{T}$. 
Then, as easily checked $f$ is an embedding  of $\A_{n}$ 
into $\A_{\omega}$. Moreover $f$ has the additional property that
it maps $\A_{n}$, into (and onto) the neat $n$ reduct of $\A_{\beta}$,
(i.e. the set of $\alpha$ dimensional elements of $A_{\beta}$). 
In short, $\A_{n}\cong \Nr_{n}\A_{\omega}$. Now agin putting $X_i=\{\phi/T: \phi\in \Gamma_i\}$ and using that the 
$\Gamma_i$'s are maximal non isolated, it follows that the 
$X_i's$ are non-principal ultrafilters
Since $\Nr_n\CA_{\omega}\subseteq S_c\Nr_n\CA_{\omega}$, then our result follows, also from Theorem 1.    
\end{demo}


%\begin{corollary} Let $T$ be an $L_n$ theory that admits elimination of quantifiers. Then


\section{Complete representability}

The notion of complete representation has turned out to be an intensely interesting topic in algebraic logic \cite{OTT}, \cite{HH}, for an overview,
as well as for undefined notions in what follows. It is known that $L_2$ enjoys a Vaughts theorem, but if we delete equality, 
and introduce substitutions, we lose this property; there are atomic theories, with no atomic models.
This is, to the best of our knowledge a new result. We start with providing a proof of algebraic version of 
the positive result  inspired by modal logic, the proof is due to Hirsch and Hodkinson:

\begin{theorem} Any atomic $\RCA_2$ is completely representable
\end{theorem}

\begin{proof} This is a nice proof inspired by modal logic taken from Hirsch and Hodkinson. 
Let $\A\in RCA_2$ be atomic. As $\RCA_2$ is conjugated and can be defined by Sahlqvist equations it is closed under completions. So
the equations are valid in $At\A$. But $At\A$ is a bounded morphic image of a disjoint union of square frames $F_i$ $(i\in I)$. 
Each $\Cm F_i$ is a full 2 dimension cylindric set
algebra. The inverse of the bounded morphism is an embedding from $\A$ into
$\prod \Cm F_i$ that preserves all meets
\end{proof}
The same can be proved for polyadic {\it equality} algebras of dimension $2$. 
Furthermore, there is a common belief that the same results holds for quasi polyadic algebras without equality.
Here we show that it is {\it not} the case; for indeed such a variety is not aditive, least conjugated.
We show that in the absence of diagonal elements, atomic algebras of dimension $2$ atomic algebras might not be completely representable.
However, we first give a simpler example showing that the omiting types theorem fails in the corresponding logic, namely, 
single non-principal types in countable languages may not be 
omitted. The proof works for every dimension, infinite included.
%We give the example for $n=2$, and then we show how it extends to higher dimensions.

\begin{example}

It suffices to show that there is an algebra $\A$, and a set $S\subseteq A$, such that $s_0^1$ does not preserves $\sum S$.
For if $\A$ had a representation as stated in the theorem, this would mean that $s_0^1$ is completely additive in $\A$.

For the latter statement, it clearly suffices to show that if $X\subseteq A$, and $\sum X=1$,
and there exists  an injection $f:\A\to \wp(V)$, such that $\bigcup_{x\in X}f(x)=V$,
then for any $\tau\in {}^nn$, we have $\sum s_{\tau}X=1$. So fix $\tau \in V$ and assume that this does not happen.
Then there is a $y\in \A$, $y<1$, and
$s_{\tau}x\leq y$ for all $x\in X$.
(Notice that  we are not imposing any conditions on cardinality of $\A$ in this part of the proof).
Now
$$1=s_{\tau}(\bigcup_{x\in X} f(x))=\bigcup_{x\in X} s_{\tau}f(x)=\bigcup_{x\in X} f(s_{\tau}x).$$
(Here we are using that $s_{\tau}$ distributes over union.)
Let $z\in X$, then $s_{\tau}z\leq y<1$, and so $f(s_{\tau}z)\leq f(y)<1$, since $f$ is injective, it cannot be the case that $f(y)=1$.
Hence, we have
$$1=\bigcup_{x\in X} f(s_{\tau}x)\leq f(y) <1$$
which is a contradiction, and we are done.
Now we turn to constructing the required  counterexample, which is an easy adaptation of a 
construction due to Andr\'eka et all in \cite{AGMNS} to our present situation.
We give the detailed construction for the reader's conveniance.

Let $\B$ be an atomless Boolean set algebra with unit $U$, that has the following property:

For any distinct $u,v\in U$, there is $X\in B$ such that $u\in X$ and $v\in {}\sim X$.
For example $\B$ can be taken to be the Stone representation of some atomless Boolean algebra.
The cardinality of our constructed algebra will be the same as $|B|$.
Let $$R=\{X\times Y: X,Y\in \B\}$$
and
$$A=\{\bigcup S: S\subseteq R: |S|<\omega\}.$$
Then indeed we have $|R|=|A|=|B|$. 

We claim that $\A$ is a subalgebra of $\wp(^2U)$.

Closure under union is obvious. To check intersections, we have:
$$(X_1\times Y_1)\cap (X_2\times Y_2)=(X_1\cap X_2) \times (Y_1\cap Y_2).$$
Hence, if $S_1$ and $S_2$ are finite subsets of $R$, then
$$S_3=\{W\cap Z: W\in S_1, Z\in S_2\}$$
is also a finite subset of $R$ and we have
$$(\bigcup S_1)\cap (\bigcup S_2)=\bigcup S_3$$
For complementation:
$$\sim (X\times Y)=[\sim X\times U]\cup [U\times \sim Y].$$
If $S\subseteq R$ is finite, then
$$\sim \bigcup S=\bigcap \{\sim Z: Z\in S\}.$$
Since each $\sim Z$ is in $A$, and $A$ is closed under intersections, we conclude that
$\sim \bigcup S$ is in $A$.
We now show that it is closed under substitutions:
$$S_0^1(X\times Y)=(X\cap Y)\times U, \\ \ S_1^0(X\times Y)=U\times (X\cap Y)$$
$$S_{01}(X\times Y)=Y\times X.$$
Let
$$D_{01}=\{s\in U\times U: s_0=s_1\}.$$
We claim that the only subset of $D_{01}$ in $\A$ is the empty set.

Towards proving this claim, assume that $X\times Y$ is a non-empty subset of $D_{01}$.
Then for any $u\in X$ and $v\in Y$ we have $u=v$. Thus $X=Y=\{u\}$ for some $u\in U$.
But then $X$ and $Y$ cannot be in $\B$ since the latter is atomless and $X$ and $Y$ are atoms.
Let
$$S=\{X\times \sim X: X\in B\}.$$
Then
$$\bigcup S={}\sim D_{01}.$$
Now we show that
$$\sum{}^{\A}S=U\times U.$$
Suppose that $Z$ is an upper bound different from $U\times U$. Then $\bigcup S\subseteq Z.$ Hence
$\sim D_{01}\subseteq Z$, hence $\sim Z=\emptyset$, so $Z=U\times U$.
Now $$S_{0}^1(U\times U) =(U\cap U)\times U=U\times U.$$
But
$$S_0^1(X\times \sim X)=(X\cap \sim X)\times U=\emptyset.$$
for every $X\in B$.
Thus $$S_0^1(\sum S)=U\times U$$
and
$$\sum \{S_{0}^1(Z): Z\in S\}=\emptyset.$$

For $n>2$, one takes $R=\{X_1\times\ldots\times X_n: X_i\in \B\}$ and the definition of $\A$ is the same. Then,
in this case, one takes $S$ to be
$X\times \sim X\times U\times\ldots\times U$
such that $X\in B$. The proof survives verbatim.
%\end{proof}

By taking $\B$ to be countable, then $\A$ can be countable, and so it violates the omitting types theorem.
\end{example}

Now let $SA_n$ be the cylindrifier free reducts of polyadic algebras. Then:

\begin{theorem}\label{counter2} For every finite $n\geq 2$, there exists an atomic algebra in 
$SA_n$ that is not completely representable.
\end{theorem}
%\begin{theorem}
%We start with $n=2$.
\begin{proof} It is enough, in view of the previous theorem,  to construct an atomic algebra such that $\Sigma_{x\in \At\A} s_0^1x\neq 1$.
In what follows we produce such an algebra. 
(This algebra will be uncountable, due to the fact that it is infinite and complete, so it cannot be countable.
In particular, it cannot be used to violate the omitting types theorem, the most it can say is that the omitting types theorem fails 
for uncountable languages, which is not too much of a surprise).

Let $\mathbb{Z}^+$ denote the set of positive integers.
Let $U$ be an infinite set. Let $Q_n$, $n\in \omega$, be a family of $n$-ary relations that form  partition of $^nU$ 
such that $Q_0=D_{01}=\{s\in {}^nU: s_0=s_1\}$. And assume also that each $Q_n$ is symmetric; for any $i,j\in n$, $S_{ij}Q_n=Q_n$. 
For example one can $U=\omega$, and for $n\geq 1$, one sets 
$$Q_n=\{s\in {}^n\omega: s_0\neq s_1\text { and }\sum s_i=n\}.$$
%For every $k\in \omega$, define the following $n$-ary relation on $\mathbb{Z}$: 
%$$Q_k=\{(x-k,x\ldots,x)\circ q:x\in\mathbb{Z},\; q\in S_n\}.$$ It is clear that $Q_0$ is the 
%identity relation on $\mathbb{Z}$ and for every $k\neq l,$ $Q_k\cap Q_l=\phi$. Also it is clear that all $Q_k$ are symmetric.
These even works for weak set algebras of infinite dimension. In this case, one takes
Let $(Q_n: n\in \omega)$ be a sequence $\alpha$-ary relations such that

\begin{enumroman}

\item $(Q_n: n\in \omega)$ is a partition of
$$V={}^{\alpha}\alpha^{(\bold 0)}=\{s\in {}^{\alpha}\alpha: |\{i: s_i\neq 0\}|<\omega\}.$$

%\item $Q_0$ is the identity relation on $V$ that is $Q_0=\{s\in V: \forall k,l\in \{i\in \alpha: s_i\neq 0\}, s_k=s_l\}.$

%\item Each $Q_n$ has $U$ as its domain and range

\item Each $Q_n$ is symmetric.
\end{enumroman}
%Such a sequence exits. For example, take $U=\Z$, the set of integers, let $\N$ be the set of positive numbers,
%and let $P$ be the predecessor function. For $n\in \N$, take $Q_n=P\circ..P$ n times.
Take $Q_0=\{s\in V: s_0=s_1\}$, and for each $n\in \omega\sim 0$, take $Q_n=\{s\in {}^{\alpha}\omega^{({\bold 0})}: s_0\neq s_1, \sum s_i=n\}.$
(Note that this is a finite sum).
Clearly for $n\neq m$, we have $Q_n\cap Q_m=\emptyset$, and $\bigcup Q_n=V.$
Furthermore, obviously each $Q_n$ is symmetric, that is  $S_{[i,j]}Q_n=Q_n$ for all $i,j\in \alpha$
Now fix $F$ a non-principal ultrafilter on $\mathcal{P}(\mathbb{Z}^+)$. For each $X\subseteq \mathbb{Z}^+$, define
\[
 R_X =
  \begin{cases}
   \bigcup \{Q_k: k\in X\} & \text { if }X\notin F, \\
   \bigcup \{Q_k: k\in X\cup \{0\}\}      &  \text { if } X\in F
  \end{cases}
\]

Let $$\A=\{R_X: X\subseteq \mathbb{Z}^+\}.$$
Notice that $\A$ is uncountable. Then $\A$ is an atomic set algebra with unit $R_{\mathbb{Z}^+}$, and its atoms are $R_{\{k\}}=Q_k$ for $k\in \mathbb{Z}^+$.
(Since $F$ is non-principal, so $\{k\}\notin F$ for every $k$).
We check that $\A$ is indeed closed under the operations.
Let $X, Y$ be subsets of $\mathbb{Z}^+$. If either $X$ or $Y$ is in $F$, then so is $X\cup Y$, because $F$ is a filter.
Hence
$$R_X\cup R_Y=\bigcup\{Q_k: k\in X\}\cup\bigcup \{Q_k: k\in Y\}\cup Q_0=R_{X\cup Y}$$
If neither $X$ nor $Y$ is in $F$, then $X\cup Y$ is not in $F$, because $F$ is an ultrafilter.
$$R_X\cup R_Y=\bigcup\{Q_k: k\in X\}\cup\bigcup \{Q_k: k\in Y\}=R_{X\cup Y}$$
Thus $A$ is closed under finite unions. Now suppose that $X$ is the complement of $Y$ in $\mathbb{Z}^+$.
Since $F$ is an ultrafilter exactly one of them, say $X$ is in $F$.
Hence,
$$\sim R_X=\sim{}\bigcup \{Q_k: k\in X\cup \{0\}\}=\bigcup\{Q_k: k\in Y\}=R_Y$$
so that  $\A$ is closed under complementation (w.r.t $R_{\mathbb{Z}^+}$).
We check substitutions. Transpositions are clear, so we check only replacements. It is not too hard to show that
\[
 S_0^1(R_X)=
  \begin{cases}
   \emptyset & \text { if }X\notin F, \\
   R_{\mathbb{Z}^+}      &  \text { if } X\in F
  \end{cases}
\]

Now
$$\sum \{S_0^1(R_{k}): k\in \mathbb{Z}^+\}=\emptyset.$$
and
$$S_0^1(R_{\mathbb{Z}^+})=R_{\mathbb{Z}^+}$$
$$\sum \{R_{\{k\}}: k\in \mathbb{Z}^+\}=R_{\mathbb{Z}^+}=\bigcup \{Q_k:k\in \mathbb{Z}^+\}.$$
Thus $$S_0^1(\sum\{R_{\{k\}}: k\in \mathbb{Z}^+\})\neq \sum \{S_0^1(R_{\{k\}}): k\in \mathbb{Z}^+\}.$$
Our next theorem gives a plathora of algebras that are not completely representable. Any algebra which shares the atom structure of $\A$ 
constructed above cannot have a complete representation. Formally:
\end{proof}



\begin{theorem} There are atomic $PA_2$s that are not completely representable; 
furthermore thay can be countable. However, the class of atomic $PA_2$ is elementary
\end{theorem}
\begin{proof}

We modify the above proof as follows. Take each $Q_n$ so it has both domain and range equal to $U$. 
This is possible; indeed it is easy to find such a 
partition of ${}^2U$.
Let $\At(x)$ be the first order formula expressing that $x$ is an atom. That is $\At(x)$ is the formula
$x\neq 0\land (\forall y)(y\leq x\to y=0\lor y=x)$. For distinct $i,j<n$ let $\psi_{i,j}$ be the formula:
$y\neq 0\to \exists x(\At(x)\land s_i^jx\neq 0\land s_i^jx\leq y).$ Let $\Sigma$ be obtained from $\Sigma_n$ by adding $\psi_{i,j}$
for every distinct $i,j\in n$.
We show that $CSA_n={\bf Mod}(\Sigma)$. Let $\A\in CSA_n$. Then, by theorem \ref{converse}, we have
$\sum_{x\in X} s_i^jx=1$ for all $i,j\in n$. Let $i,j\in $ be distinct. Let $a$ be non-zero, then $a.\sum_{x \in X}s_i^jx=a\neq 0$,
hence there exists $x\in X$, such that
$a.s_i^jx\neq 0$, and  so $\A\models \psi_{i,j}$.
Conversely, let $\A\models \Sigma$. Then for all $i,j\in n$, $\sum_{x\in X} s_i^jx=1$. Indeed, assume that for some distinct
$i,j\in n$, $\sum_{x\in X}s_i^jx\neq 1$.
Let $a=1-\sum_{x\in X} s_i^jx$. Then $a\neq 0$. But then there exists $x\in X$, such that $s_i^jx.\sum_{x\in X}s_i^jx\neq 0$
which is impossible. But for distinct $i, j\in n$, we have  $\sum_{x\in X}s_{[i,j]}X=1$ anyway, and so $\sum s_{\tau}X=1$, for all
$\tau\in {}^nn$, and so it readily follows that $\A\in CRA_n.$ So if $\A$ models $\Sigma$, then it is completely additive, and the above modal theoretic
proof works.
\end{proof}
When we do not have cylindrifiers the above example works for all dimension {\it infinite} included, using instead weak set algebras.
Indeed, we have

\begin{theorem}\label{counterinfinite2} There is an atomic $\A\in SA_{\alpha}$ such that $\A$ is not completely representable.
\end{theorem}

\begin{proof} First it is clear that if $V$ is any weak space, then $\wp(V)\models \Sigma$.
Let $(Q_n: n\in \omega)$ be a sequence $\alpha$-ary relations such that

\begin{enumroman}

\item $(Q_n: n\in \omega)$ is a partition of
$$V={}^{\alpha}\alpha^{(\bold 0)}=\{s\in {}^{\alpha}\alpha: |\{i: s_i\neq 0\}|<\omega\}.$$

%\item $Q_0$ is the identity relation on $V$ that is $Q_0=\{s\in V: \forall k,l\in \{i\in \alpha: s_i\neq 0\}, s_k=s_l\}.$

%\item Each $Q_n$ has $U$ as its domain and range

\item Each $Q_n$ is symmetric.
\end{enumroman}
%Such a sequence exits. For example, take $U=\Z$, the set of integers, let $\N$ be the set of positive numbers,
%and let $P$ be the predecessor function. For $n\in \N$, take $Q_n=P\circ..P$ n times.
Take $Q_0=\{s\in V: s_0=s_1\}$, and for each $n\in \omega\sim 0$, take $Q_n=\{s\in {}^{\alpha}\omega^{({\bold 0})}: s_0\neq s_1, \sum s_i=n\}.$
(Note that this is a finite sum).
Clearly for $n\neq m$, we have $Q_n\cap Q_m=\emptyset$, and $\bigcup Q_n=V.$
Furthermore, obviously each $Q_n$ is symmetric, that is  $S_{[i,j]}Q_n=Q_n$ for all $i,j\in \alpha$.

Now fix $F$ a non-principal ultrafilter on $\mathcal{P}(\mathbb{Z}^+)$. For each $X\subseteq \mathbb{Z}^+$, define
\[
 R_X =
  \begin{cases}
   \bigcup \{Q_n: n\in X\} & \text { if }X\notin F, \\
   \bigcup \{Q_n: n\in X\cup \{0\}\}      &  \text { if } X\in F
  \end{cases}
\]
Let $$\A=\{R_X: X\subseteq \mathbb{Z}^+\}.$$
Then $\A$ is an atomic set algebra, and its atoms are $R_{\{n\}}=Q_n$ for $n\in \mathbb{Z}^+$.
(Since $F$ is non-principal, so $\{n\}\notin F$ for every $n$.
Then one proceeds exactly as in the finite dimensional case, theorem \ref{counter2}.
\end{proof}
The latter answers an implicit question of Hodkinson's.
Another result, where weak models offer solace is:
This is a new theorem, giving a characterization of completely representable 
algebras on weak unit, and raises a natural question, namely, is the class of infinite dimensional representable 
on weak units elementary.

\begin{theorem} Let $\A\in S_c\Nr_{\alpha}\CA_{\alpha+\omega}$ be countable. 
Then $\A$ is atomic iff $\A$ has a complete representation on weak models
\end{theorem}
\begin{proof} Let $\A\subseteq_c\Nr_{\omega}\B'$, where $\B'\in CA_{\omega+\omega}$
Let $X=\At\A$. Then $\sum ^{\B'}X=1$. Let $\B=\Sg^{\B'}\A$, then $\B\in Dc_{\omega}$, and
$\sum^{\B}X=1$. Let $V={}^{\omega}\omega^{(Id)}=\{\tau \in {}^{\omega}\omega: |\{i\in \omega: \tau(i)\neq i\}|<\omega\}$.
Then for every $\tau\in V$, $s_{\tau}^{\B}X=1$. Also, in $\B$  we have $c_kx=\sum s_k^lx$. Let $F$ be an ultrafilter that preserves
these joins, and define
$$f:\A\to \wp(V)$$ via 
$$a\mapsto \{\tau \in V: s_{\tau}a\in F\}.$$
Clearly $f$ is a complete representation.
\end{proof} 



%\section{The solution of two problems of Robin Hirsch}

















We will show in example \ref{countable} that maximality cannot be omitted by showing that 
there exists an uncountable atomic simple algebra $\A\in \Nr_n\CA_{\omega}$ which is not completely representable. 
This example is sketched in a remark 31, p. 688, 
by Robin Hirsch; here we give the detailed proof, and also 
show that this construction answers a question raised by Hirsch in the same paper, see  p.674, we quote 
' Whether this characterization works for uncountable algebras remain unknown.'(End of quote) 
Here we show it does not. This theorem was also mentioned in \cite{OTT} without proof. 
We shall need the following form of the Erdos-Rado cominatorial principle (Extending Ramsey's theorem to uncountable sets)

\begin{theorem}(Erdos-Rado)
If $r\geq 2$ is finite, $k$  an infinite cardinal, then
$$exp_r(k)^+\to (k^+)_k^{r+1}$$

where $exp_0(k)=k$ and inductively $exp_{r+1}(k)=2^{exp_r(k)}$.
\end{theorem}
The above partition symbol describes the following statement. If $f$ is a coloring of the $r+1$ 
element subsets of a set of cardinality $exp_r(k)^+$
in $k$ many colors, then there is a homogeneous set of cardinality $k^+$+ 
(a set, all whose $r+1$ element subsets get the same $f$-value).

%Example \ref{countable} also show that Vaughts theorem on exsitence of atomic models fail for $\L_n$ when $n\geq 3$
%when theories have cardinality $\omega_1$, i.e just at the first uncountable cardinal; 
%this consequence will be explained in the second item of the example.
Our cylindric algebra will be based on the atom structure of a relation algebra that has an $\omega$-dimensional cylindric base.
So we review some notions from relation and cylindric algebras, concerning atom structures and networks (which are graphs labelled by atoms).

%We also assume familiarity with the notion of networks over a relation algebra atom structure, and that of 
%$n$ - dimensional cylindric bases.

A \textit{relation atom structure} $\Upsilon=(A, Id, \breve{}, C)$
consists of a non-empty set $A$ of atoms, a unary function
$\breve{}:A\to A$ and a ternary relation $C$ satisfying
\begin{itemize}
\item $a=b$ $\longleftrightarrow  \exists e\in A$ such that $
 e \in Id~ \land~ (a,e,b) \in C$,
\item $(a,e,b) \in C \implies (\breve{a}, c, b) \in C$ and $(c, \breve{b}, a) \in
C$,
\item $(a,b,c) \in C\land (c,d,g) \in C\implies \exists f\in A$ such that $(a,f,g) \in C\land
(b,d,f) \in C$.
\end{itemize}
Let $\A=(A, +, ., -, 0,1, Id, \breve{}, ;)$ be an atomic  relation
algebra. Let $At \A$ denote the set of atoms. Then this determines
an atom structure
$$\At \A=(At \A, Id, \breve{} ,C)$$
whose domain is the set of atoms of $A$, the identity $Id=\{e\in At
\A: e\leq 1'\}$, converse is the restriction of $~\breve{}~$ on the
atoms of $\A$, and the ternary relation $C$ is defined by
$$ (a,b,c) \in C\longleftrightarrow a;b\geq c, \text { for all atoms }a,b,c.$$
A triple in $C$ is said to be a \textit{consistent triple}.
\\

Conversely from an atom structure $\Upsilon=(A, Id, \breve{}, C)$ it
is possible to define an \textit{atomic relation algebra} $\Cm
\Upsilon$ whose domain is the power set of $A$, the boolean
operations $\cup$, complement in $A$, and whose non-boolean
operations are defined by
$$1'=Id,$$
$$\breve{r}=\{\breve{a}: a\in r\},$$
$$r;s=\{c: \exists a\in r, \exists b\in s, (a,b,c) \in C\}.$$
$\Cm\Upsilon$ is called the complex algebra. The set of forbidden
triples of $\Upsilon$ is $A\times A\times A\smallsetminus C$, i.e.,
it is the complement of the consistent triples. Listing the
forbidden triples specifies the composition part of the atom
structure.

\begin{definition}
Let $\A$ be a relation algebra. An $\A$-labelled graph $N=(N_1,
N_2)$ consists of a set $N_1$ of nodes, and a map $N_2:N_1\times
N_1\to \A.$ $N$ is called an $\A$ network, if for all nodes
$x,y,z\in N_1$:
\begin{itemize}
\item $N_2(x,x)\leq 1'$,
\item $N_2(x,y)\cdot (N_2(x,z);N_2(z,y))\neq 0$.
\end{itemize}
\end{definition}


\begin{definition} { \ }

\begin{enumerate}

\item An $\A$ network is atomic if $N(x,y)$ is an atom for all $x,y\in N.$
For an atomic $\A$ network it is not difficult to show that the
following hold for $x,y,z\in N$:

\begin{itemize}

\item $N(x,y)=N(y,x)$,

\item $N(x,y)\leq N(x,z);N(z,y)$,

\item $N(x,x)=1'.(N(x,y);N(y,x))$.

\end{itemize}
For networks $N$ and $M$, we write $M\equiv_{x_1\ldots x_m}N$ if
$M(\bar{a})=N(\bar{a})$ for all $\bar{a}\in {}^2(n-\{x_1,\ldots,
x_m\}).$

\item Let $n\leq \omega$. An $n$-dimensional cylindric basis for $\A$ is a set $\cal M$
of atomic networks $N$ with nodes $(N)=n$ such that

(a) For $a\in At \A$ there is $N\in \cal M$ with $N(0,1)=a.$

(b) If $N\in \cal M$ and $x,y<n$, and $a,b\in At \A$, $N(x,y)\leq
a;b$ and $x,y\neq z<n$, then there is some $M\in \cal M$ such that
$M\equiv_zN$, $M(x,z)=a$ and $M(z,y)=b$.

(c) If $M, N\in \cal M$ $x,y<n$, $x\neq y$ and $M\equiv_{xy}N$, then
there is some $L\in \cal M$ such that $M\equiv_x L\equiv_yN.$
%Here $M\equiv_xN$  means that $M(\bar{a})=N(\bar{b})$
%for all $\bar{a}, \bar{b}\in {}^2(n-\{x\}).$
%and $M\equiv_{xy}N$  means that $M(\bar{a})=N(\bar{b})$ for all
%$\bar{a}, \bar{b}\in {}^2(n-\{x,y\}).$
\end{enumerate}
\end{definition}
Maddux calls networks basic matrices. 
A relation algebra can be specified either by specifying the consistent triples of atoms or the forbbiden triples, a forbidden triple is one that
is not consistent.

%This exaple is sketced in a rematk in robin. here we give a complete prrof. Also this constructions answers a question in the same paper! we will also indicate this.
\begin{example}\label{countable}

\begin{enumarab}

\item We define an atomic relation algebra $\A$ with uncountably many
atoms. Let $\kappa$ be an infinite cardinal. This algebra will be used to construct cylindric algebras of dimension 
$n$ showing that countability is essential in the above characterization.  

The atoms are $1', \; a_0^i:i<2^{\kappa}$ and $a_j:1\leq j<
\kappa$, all symmetric.  The forbidden triples of atoms are all
permutations of $(1',x, y)$ for $x \neq y$, \/$(a_j, a_j, a_j)$ for
$1\leq j<\kappa$ and $(a_0^i, a_0^{i'}, a_0^{i^*})$ for $i, i',
i^*<2^{\kappa}.$  In other words, we forbid all the monochromatic
triangles.  

Write $a_0$ for $\set{a_0^i:i<2^{\kappa}}$ and $a_+$ for 
$\set{a_j:1\leq j<\kappa}$. Call this atom
structure $\alpha$.  

Let $\A$ be the term algebra on this atom
structure; the subalgebra of $\Cm\alpha$ generated by the atoms.  $\A$ is a dense subalgebra of the complex algebra
$\Cm\alpha$. We claim that $\A$, as a relation algebra,  has no complete representation.

Indeed, suppose $\A$ has a complete representation $M$.  Let $x, y$ be points in the 
representation with $M \models a_1(x, y)$.  For each $i<\omega_1$ there is a 
point $z_i \in M$ such that $M \models a_0^i(x, z_i) \wedge a_1(z_i, y)$.  

Let $Z = \set{z_i:i<2^{\kappa}}$.  Within $Z$ there can be no edges labelled by 
$a_0$ so each edge is labelled by one of the $\kappa$ atoms in 
$a_+$.  The Rado Erdos theorem forces the existence of three points 
$z^1, z^2, z^3 \in Z$ such that $M \models a_j(a^1, z^2) \wedge a_j(z^2, z^3) 
\wedge a_j(z^3, z_1)$, for some single $j<\kappa$.  This contradicts the 
definition of composition in $\A$.

Let $S$ be the set of all atomic $\A$-networks $N$ with nodes
 $\omega$ such that\\ $\set{a_i: 1\leq i<\omega,\; a_i \mbox{ is the label 
of an edge in }
 N}$ is finite.
Then it is straightforward to show $S$ is an amalgamation class, that is for all $M, N 
\in S$ if $M \equiv_{ij} N$ then there is $L \in S$ with 
$M \equiv_i L \equiv_j N.$  
Hence the complex cylindric algebra $\Ca(S)\in \CA_\omega$.

Now let $X$ be the set of finite $\A$-networks $N$ with nodes
$\subseteq\omega$ such that 
\begin{enumerate}
\item each edge of $N$ is either (a) an atom of
$\c A$ or (b) a cofinite subset of $a_+=\set{a_j:1\leq j<\kappa}$ or (c)
a cofinite subset of $a_0=\set{a_0^i:i<2^{\kappa}}$ and
\item $N$ is `triangle-closed', i.e. for all $l, m, n \in nodes(N)$ we
have $N(l, n) \leq N(l,m);N(m,n)$.  That means if an edge $(l,m)$ is
labelled by $1'$ then $N(l,n)= N(m,n)$ and if $N(l,m), N(m,n) \leq
a_0$ then $N(l,n).a_0 = 0$ and if $N(l,m)=N(m,n) =
a_j$ (some $1\leq j<\omega$) then $N(l,n).a_j = 0$.
\end{enumerate}
For $N\in X$ let $N'\in\Ca(S)$ be defined by 
\[\set{L\in S: L(m,n)\leq
N(m,n) \mbox{ for } m,n\in nodes(N)}\]
For $i,\omega$, let $N\restr{-i}$ be the subgraph of $N$ obtained by deleting the node $i$.
Then if $N\in X, \; i<\omega$ then $\cyl i N' =
(N\restr{-i})'$.
The inclusion $\cyl i N' \subseteq (N\restr{-i})'$ is clear.

Conversely, let $L \in (N\restr{-i})'$.  We seek $M \equiv_i L$ with
$M\in N'$.  This will prove that $L \in \cyl i N'$, as required.
Since $L\in S$ the set $X = \set{a_i \notin L}$ is infinite.  Let $X$
be the disjoint union of two infinite sets $Y \cup Y'$, say.  To
define the $\omega$-network $M$ we must define the labels of all edges
involving the node $i$ (other labels are given by $M\equiv_i L$).  We
define these labels by enumerating the edges and labelling them one at
a time.  So let $j \neq i < \omega$.  Suppose $j\in nodes(N)$.  We
must choose $M(i,j) \leq N(i,j)$.  If $N(i,j)$ is an atom then of
course $M(i,j)=N(i,j)$.  Since $N$ is finite, this defines only
finitely many labels of $M$.  If $N(i,j)$ is a cofinite subset of
$a_0$ then we let $M(i,j)$ be an arbitrary atom in $N(i,j)$.  And if
$N(i,j)$ is a cofinite subset of $a_+$ then let $M(i,j)$ be an element
of $N(i,j)\cap Y$ which has not been used as the label of any edge of
$M$ which has already been chosen (possible, since at each stage only
finitely many have been chosen so far).  If $j\notin nodes(N)$ then we
can let $M(i,j)= a_k \in Y$ some $1\leq k < \omega$ such that no edge of $M$
has already been labelled by $a_k$.  It is not hard to check that each
triangle of $M$ is consistent (we have avoided all monochromatic
triangles) and clearly $M\in N'$ and $M\equiv_i L$.  The labelling avoided all 
but finitely many elements of $Y'$, so $M\in S$. So
$(N\restr{-i})' \subseteq \cyl i N'$.

Now let $X' = \set{N':N\in X} \subseteq \Ca(S)$.
Then the subalgebra of $\Ca(S)$ generated by $X'$ is obtained from 
$X'$ by closing under finite unions.
Clearly all these finite unions are generated by $X'$.  We must show
that the set of finite unions of $X'$ is closed under all cylindric
operations.  Closure under unions is given.  For $N'\in X$ we have
$-N' = \bigcup_{m,n\in nodes(N)}N_{mn}'$ where $N_{mn}$ is a network
with nodes $\set{m,n}$ and labelling $N_{mn}(m,n) = -N(m,n)$. $N_{mn}$
may not belong to $X$ but it is equivalent to a union of at most finitely many 
members of $X$.  The diagonal $\diag ij \in\Ca(S)$ is equal to $N'$
where $N$ is a network with nodes $\set{i,j}$ and labelling
$N(i,j)=1'$.  Closure under cylindrification is given.
Let $\c C$ be the subalgebra of $\Ca(S)$ generated by $X'$.
Then $\A = \Ra(\c C)$.
Each element of $\A$ is a union of a finite number of atoms and
possibly a co-finite subset of $a_0$ and possibly a co-finite subset
of $a_+$.  Clearly $\A\subseteq\Ra(\c C)$.  Conversely, each element
$z \in \Ra(\c C)$ is a finite union $\bigcup_{N\in F}N'$, for some
finite subset $F$ of $X$, satisfying $\cyl i z = z$, for $i > 1$. Let $i_0,
\ldots, i_k$ be an enumeration of all the nodes, other than $0$ and
$1$, that occur as nodes of networks in $F$.  Then, $\cyl 
{i_0} \ldots
\cyl {i_k}z = \bigcup_{N\in F} \cyl {i_0} \ldots
\cyl {i_k}N' = \bigcup_{N\in F} (N\restr{\set{0,1}})' \in \A$.  So $\Ra(\c C)
\subseteq \A$.
$\A$ is relation algebra reduct of $\c C\in\CA_\omega$ but has no
complete representation.
Let $n>2$. Let $\B=\Nr_n \c C$. Then
$\B\in \Nr_n\CA_{\omega}$, is atomic, but has no complete representation.

\item We have $\A\in \Nr_n\CA_{\omega}$ such that $|A|=\omega_1$, $\A$ is atomic, $A$ is simple and $\A$ has no complete representation.
Let $X=At\A$, the set of atoms of $\A$. Then $\sum X=1$. But $\A$ is not completely representable, 
this means that for any isomorphism $f:\A\to \B$, where $\B$ is a  set algebra with base $U$, we have 
$\bigcup_{x\in X} f(x)\neq {}^nU$. Let $T$ be an $L_n$ theory such that $\A\cong \Fm_T$. Then $T$ is an atomic theory, but it has not atomic model $\M$.
$T$ is an atomic theory because for evey $\psi$ consistent with $T$, there is an $\psi,$ such that $\psi/T$ is a an atom and
$\vdash \psi\to \phi$. On the other hand, if $\M$ were an atomic model of $T$, then 
the map $\Psi:\Fm_T\to Cs_{n}^{\M}$ defined by $\phi_T\mapsto \phi^{\M}$ would establish a complete
representation, which is a contradiction.
\end{enumarab} 
\end{example}



A theorem of Johnson says that if $\A$ is a cylindric algebra of dimension $n$  generated by a set $X$, such that $\Delta x\neq n$, for all $x\in X$,
has a representable diagonal free reduct, then the algebra itself is representable. Our next theorem shows that a similar situation holds 
when we replace representable by completely
representable. For $\A\in CA_n$, $\Rd_{df}\A$ denotes its diagonal free reduct.

For an $n$ dimensional cylindric algebra $c_{(n)}$ stands for the term $c_0\ldots c_{n-1}$.
The next theorem is due to Ian Hodkinson, based on ideas of Johnson. 

\begin{theorem} Let $\D$ be a cylindric algebra of dimension $n\geq 3$, that is generated by the set $\{x\in D: \Delta x\neq n\}.$
Then if $\Rd_{df}\D$ is completely representable, then so is $\D$.
\end{theorem}
\begin{proof} First suppose that $\D$ is simple, and let $h: \D\to \wp(V)$ be a complete representation, where $V=\prod_{i<n}U_i$ for sets $U_i$.
We can assume that $U_i=U_j$ for all $i,j<n$, and if $s\in V$, $i,j<n$ and $a_i=a_j$ then $a\in h(d_{ij}$. Indeed, let $\delta =\prod d_{ij}\in \D$. 
As $C$ is a cylndric algebra, we have $c_{(n)}\delta =1$, so for each $u\in U_i$ there is an $s\in h(\delta)$ with $a_i=u$. 
So there exists a function $s_i: U_i\to h(\delta)$ such that $(s_i(u))_i=u$ for each $u\in U_i$.

Let $U$ be the disjoint union of the $U_i$s. Let $t_i:U\to U_i$ be the surjection defined by $t_i(u)=(s_j(u))_i$.
Let $g: \D\to \wp(^nU)$ be defined via 
$$d\mapsto \{s\in {}^nU: (t_0(a_0),\ldots, t_{n-1}(a_{n-1}))\in h(d)\}.$$
Then $g$ is a complete representation of $\D.$ Now suppose $s\in {}^nU$, satisfies $s_i=s_j$ with $a_i\in U_k$, say, where $k<n$. 
Let $\bar{b}=s_k(a_i)=s_k(a_j)\in h(\delta).$ Then $t_i(a_i)=b_i$ and $t_j(a_j)=b_j$, so $(t_i(a_i): i<n)$ 
agrees with $\bar{b}$ on coordinates $i,j$. Since $\bar{b}\in h(\delta)$ and 
$\Delta d_{ij}=\{i,j\}$, then $(t_i(a_i): i<n)\in h(d_{ij}$ and so $s\in g(d_{ij}),$ as required.

Now define $\sim_{ij}=\{(a_i, a_j): \bar{a}\in h(d_{ij})$. Then it easy to 
check that $\sim_{01}=\sim_{i,j}$ is an equivalence relation on $U$. For 
$s,t\in {}^nU$, define $s\sim t$, if $s_i\sim t_i$ for each $i<n$, then $\sim$ is an equivalence relation
on $^nU$. Let
$$E=\{d\in D: h(d)\text { is a union of $\sim$ classes }\}.$$
Then $$\{d\in D: \Delta d\neq n\}\subseteq E.$$

Furthermore, $E$ is the domain of a complete subalgebra of $\C$. 
Let us check this. We have $\{0,1, d_{ij}: i,j<n\}\subseteq E$, since $\Delta 0=\Delta 1=\emptyset$ 
and $\Delta d_{ij}=\{i,j\}\neq n$ (as $n\geq 3$).
If $h(d)$ is a union of $\sim$ classes, then so
is $^nU\setminus h(d)= h(-d)$. If $S\subseteq E$ and $\sum S$ exists in $\D$, then because $h$ is complete representation 
we have $h(\sum^{\D}S)=\bigcup h[S]$, a union of $\sim$ classes
so $\sum S\in E$. Hence $E=C$. Now define $V=U/\sim_{01}$, and
define $g:\C\to \wp(^nV)$ via
$$c\mapsto \{(\bar{a}/\sim_{01}): \bar{a}\in h(c).$$
Then $g$ is a complete representation.

Now we drop the assumption that $\D$ is simple. Suppose that $h:\D\to \prod_{k\in K} Q_k$ is a complete representation. 
Fix $k\in K$, let $\pi_k: Q\to Q_k$ be the canonical projection, and let 
$\D_k=rng(\pi_k\circ h)$. We define diagonal elements in $\D_k$ by $d_{ij}=\pi_k(h^{\C}(d_{ij}))$. This expands $\D_k$ 
to a cylindric-type algebra $\C_k$ that is a homomorphic image of $\C$, and hence
is a cylindric algebra with diagonal free reduct $\D_k$. Then the inclusion map $i_k:\D_k\to Q_k$ is a complete
representation of $D_k$. 
Since obviously
$$\pi_k[h[\{c\in C: \Delta c\neq n\}]\subseteq \{c\in C_k: \Delta c\neq n\}$$ 
and $\pi_k, h$ preserve arbitrary sums, then $C_k$ is completely generated by $\{c\in C_k: \Delta c\neq n\}$. 
Now $c_{(n)}x$ is a discriminator term in $Q_k$, so $D_k$ is simple.
So by the above $\C_k$ has complete represenation $g_k:\C_k\to Q_k'$. Define
$g: \C\to \prod_{k\in K}Q_k'$ via
$$g(c)_k=g_k(\pi_k(h(c))).$$
Then $g$ defines a complete representation.
\end{proof}

Let $\B$ be the algebra constructed in example \ref{countable}; that is $\B\in \Nr_n\CA_{\omega}$ and $\B$ has no complete representation.
Since $\B$ is generated by its $\Ra$ reduct, that is its $2$ dimensional elements, then it follows that the diagonal free reduct of $\A$ is not 
completely representable, see theorem below.




A classical theorem of Vaught for first order logic says that countable atomic theories have countable atomic models,
such models are necessarily prime, and a prime model omits all non principal types.
We have a similar situation here:

\begin{theorem} Let $f:\A\to \wp(V)$ be an atomic representation of $\A\in CA_n$.
Then for any given family $(Y_i:i\in I)$ of subsets of $\A$,  if $\prod Y_i=0$ for all $i\in I$, then we have
$\bigcap_{y\in Y_i} f(y)=\emptyset$ for all $i\in I$.
\end{theorem}
\begin{proof} Let $i\in I$. Let $Z_i=\{-y: y\in Y_i\}$. Then $\sum Z_i=1$. Let $x$ be an atom. Then $x.\sum Z_i=x\neq 0$.
Hence there exists $z\in Z_i$, such that
$x.z\neq 0$. But $x$ is an atom, hence $x.z=x$ and so $x\leq z$. We have shown that for every atom $x$, there exists $z\in Z_i$ such that $x\leq z$.
It follows immediately that
$V=\bigcup_{x\in \At\A}f(x)\leq \bigcup_{z\in Z_i} f(z)$, and so $\bigcap_{y\in Y_i} f(y)=\emptyset,$
and we are done.
\end{proof}










Our last theorem shows that the omitting types fails in countable complete $L_n$ theories in a very strong sense.
theories chosen can be complete, atomic, having no atomic models.

\begin{corollary} For each finite $n\geq 3$, there exists a simple countable atomic representable polyadic equality 
algebra of dimension $n$ whose diagonal free reduct  not completely representable
\end{corollary}
\begin{demo}{Proof} Let $\A$ be a countable atomic representable polyadic algebra, that is not necessarily simple, satisfying that its $Df$ reduct is not 
completely representable. Such algebras exist, see the next section. 
Consider the elements $\{{\sf c}_na: a\in At\A'\}$. Then every simple component $S_a$ of $\A$ can be
obtained by relativizing to ${\sf c}_na$ for an atom $a$. 
Then one of the $S_a$'s should have no complete representation . Else for each atom $a$ $S_a$ has a complete representation $h_a$.
From those one constructs a complete representation for $\A$. 
The domain of the representation will be the disjoint union of the domains of $h_a$, and now represent $\A$ by
$$h(\alpha)=\bigcup_{a\in AtA}\{h(\alpha\cdot {\sf c}_na)\}.$$

\end{demo}  





\section{Complete representability of polyadic algebras of infinite dimension}


\begin{theorem} Let $\alpha$ be an infinite ordinal. Let $\A\in \PA_{\alpha}$ be atomic. 
Then $\A$ has a complete representation.
\end{theorem}

\begin{demo}{Proof} Let $c\in A$ be non-zero. We will find a set $U$ and a homomorphism 
from $\A$ into the set algebra with universe $\wp(^{\alpha}U)$ that preserves arbitrary suprema
whenever they exist and also satisfies that  $f(c)\neq 0$. $U$ is called the base of the set algebra. 
Let $\mathfrak{m}$ be the local degree of $\A$, $\mathfrak{c}$ its effective cardinality and 
$\mathfrak{n}$ be any cardinal such that $\mathfrak{n}\geq \mathfrak{c}$ 
and $\sum_{s<\mathfrak{m}}\mathfrak{n}^s=\mathfrak{n}$. The cardinal 
$\mathfrak{n}$ will be the base of our desired representation.

Now there exists $\B\in \PA_{\mathfrak{n}}$ 
such that $\A\subseteq \Nr_{\alpha}\B$ and $A$ generates $\B$. 
The local degree of $\B$ is the same as that of $\A$, 
in particular each $x\in \B$ admits a support of cardinality $<\mathfrak{m}$. Furthermore, $|\mathfrak{n}\sim \alpha|=|\mathfrak{n}|$ and
for all $Y\subseteq A$, we have $\Sg^{\A}Y=\Nr_{\alpha}\Sg^{\B}Y.$ 
All this can be found in \cite{DM}, see the proof of theorem 1.6.1 therein; in such a proof, 
$\B$ is called a minimal dilation of $\A$. 
Without loss of generality, we assume that $\A=\Nr_{\alpha}\B$,  since $\Sg^{\A}A=\Nr_{\alpha}\Sg^{\B}A=\Nr_{\alpha}\B.$ (In the last equality 
we are using that $A$ generates $\B$).
Hence $\A$ is first order interpretable in $\B$. In particular, any first order sentence (e.g. the one expressing that $\A$ is atomic)
of the language of $\PA_{\alpha}$ translates effectively to a sentence $\hat{\sigma}$ of the language of $\PA_{\beta}$ such that for all 
$\C\in \PA_{\beta}$, we have
$\Nr_{\alpha}\C\models \sigma\longleftrightarrow \C\models \hat{\sigma}$. 
Here the languages are uncountable, even if $\A$ is countable and has countable dimension,  so we use effective in a loose sense, 
but it roughly means that there is an effective 
procedure or algorithm that does 
this translation. Since $\A=\Nr_{\alpha}\B$ and $\A$ is atomic, it 
follows that $\B$ is also atomic.
Let $\Gamma\subseteq \alpha$ and $p\in \A$. Then in $\B$ we have, see \cite{DM} the proof of theorem 1.6.1,  
\begin{equation}\label{tarek1}
\begin{split}
{\sf c}_{(\Gamma)}p=\sum\{{\sf s}_{\bar{\tau}}p: \tau\in {}^{\alpha}\mathfrak{n},\ \  \tau\upharpoonright \alpha\sim\Gamma=Id\}.
\end{split}
\end{equation}
Here, and elsewhere throughout the paper,  for a transformation $\tau$ wth domain $\alpha$ and range included in $\mathfrak{n}$,
$\bar{\tau}=\tau\cup Id_{\mathfrak{n}\sim \alpha}$. 
Let $X$ be the set of atoms of $\A$. Since $\A$ is atomic, then  $\sum^{\A} X=1$. By $\A=\Nr_{\alpha}\B$, we also have $\sum^{\B}X=1$.
We will further show that
for all $\tau\in {}^{\alpha}\mathfrak{n}$ we have,
\begin{equation}\label{tarek2}
\begin{split}
\sum {\sf s}_{\bar{\tau}}^{\B}X=1.
\end{split}
\end{equation}
It suffices to show that for every $\tau\in {}^{\alpha}\alpha$ and $a\neq 0\in \A$, there exists $x\in X$, such that ${\sf s}_{\tau}x\leq a.$
This will show that for any $\tau\in {}^{\alpha}\alpha$, the sum  $\sum{\sf s}_{\tau}^{\A}X$ 
is equal to the top element in $\A$, which is the same as that of $\B$. 
The required will then follow since for $Y\subseteq A$, we have
$\sum^{\A}Y=\sum ^{\B}Y$ by $\A=\Nr_{\alpha}\B$, and for all $x\in A$ and $\tau\in {}^{\alpha}\alpha$, we have 
${\sf s}_{\tau}^{\A}x={\sf s}_{\bar{\tau}}^{\B}x$, since $\B$ is a dilation of $\A$.

Our proof proceeds by certain non-trivial manipulations of substitutions. 
Assume that $\tau\in {}^{\alpha}\alpha$ and non-zero $a\in A$ are given. Suppose for the time being that $\tau$ is onto. 
We define a right inverse $\sigma$ of $\tau$ the usual way. That is for $i\in \alpha$ choose $j\in \tau^{-1}(i)$ and set $\sigma(i)=j$. 
Then $\sigma\in {}^{\alpha}\alpha$ and in fact $\sigma$ is one to one. 
Let $a'={\sf s}_{\sigma}a$. Then $a'\in A$ and  $a'\neq 0$, for if it did, then by polyadic axioms (7) and (8), we would get 
$0={\sf s}_{\tau}a'={\sf s}_{\tau\circ \sigma}a=a$ 
which is not the case, since $a\neq 0$.
Since $\A$ is atomic, then there exists an atom $x\in X$ such that $x\leq a'$.
Hence using that substitutions preserve the natural order on the boolean algebras in question, 
since they are boolean endomorphisms, and axioms (7) and (8)
in the polyadic axioms, we obtain
$${\sf s}_{\tau}x\leq {\sf s}_{\tau}a'={\sf s}_{\tau}{\sf s}_{\sigma}a={\sf s}_{\tau\circ \sigma}a={\sf s}_{Id}a=a.$$
Here all substitutions are evaluated in the algebra $\A$, and we are done in this case.
Now assume that $\tau$ is not onto. Here we use the spare dimensions of $\B$. 
By a simple cardinality argument, baring in mind that $|\mathfrak{n}\sim \alpha|=|\mathfrak{n}|$, 
we can find $\bar{\tau}\in {}^{\mathfrak{n}}\mathfrak{n}$, such that 
$\bar{\tau}\upharpoonright \alpha=\tau$, and 
$\bar{\tau}$ is onto. We can also assume that $\bar{\tau}\upharpoonright (\mathfrak{n}\sim \alpha)$ is one to one since 
$|\mathfrak{n}\sim \alpha|=|\mathfrak{n}\sim Range(\tau)|=\mathfrak{n}.$
Notice also that we have $\bar{\tau}^{-1}(\mathfrak{n}\sim \alpha)\cap \alpha=\emptyset$.  
Indeed if $x\in \bar{\tau}^{-1}(\mathfrak{n}\sim \alpha)\cap \alpha$, then $\bar{\tau}(x)\notin \alpha$, while $x\in \alpha$, 
which is impossible, since $\bar{\tau}\upharpoonright \alpha=\tau$, 
so that $\bar{\tau}(x)=\tau(x)\in \alpha$.
Now let $\sigma\in {}^{\mathfrak{n}}\mathfrak{n}$ be a right inverse of $\tau$ (on $\mathfrak{n}$), so that
for chosen $j\in \bar{\tau}^{-1}(i)$, we have $\sigma(i)=j.$
We distinguish between two cases:

(i) $\sigma(\alpha)\subseteq \alpha$.
Let $a'={\sf s}_{\sigma}^{\B}a={\sf s}_{\sigma\upharpoonright \alpha}^{\A}a.$ The last equation holds because $\B$ is a dilation of $\A$.
Then $a'\in A$, since $\sigma(\alpha)\subseteq \alpha$. Also $a'\neq 0$, by same reasoning as above.
Now there exists an atom $x\in X$, such that $x\leq a'$. A fairly straightforward computation, using that $\B$ is a dilation of $\A$,
together with the fact that substitutions preserve order, and axioms 
(7) and (8)
in the polyadic axioms applied to $\B$, gives
$${\sf s}_{{\tau}}^{\A}x\leq {\sf s}_{{\tau}}^{\A}a'={\sf s}_{\bar{\tau}}^{\B}a'= {\sf s}_{\bar{\tau}}^{\B}{\sf s}_{\sigma}^{\B}a=
{\sf s}_{\bar{\tau}\circ \sigma}^{\B}a={\sf s}_{Id}^{\B}a=a,$$
which finishes the proof in the first case.

(ii) $\sigma(\alpha)$ is not contained in $\alpha.$ 

We proceed as follows. Let $a'={\sf s}_{\sigma}^{\B}a\in \B$, then by the same reasoning as above, $a'\neq 0$. 
But $a'\in \B$, hence, there exists $\Gamma\subseteq \mathfrak{n}\sim \alpha$, such that 
$a''={\sf c}_{(\Gamma)}a'\in \A,$ since $\A=\Nr_{\alpha}\B$. But  $a'\leq a''$, so $a''$ is also a non-zero element in $\A$. 
Let $x\in X$ be an atom below $a''$. Then we have
$${\sf s}_{\tau}^{\A}x\leq {\sf s}_{\bar{\tau}}^{\B}a''={\sf s}_{\bar{\tau}}^{\B}{\sf c}_{(\Gamma)}^{\B}a'=
{\sf s}_{\bar{\tau}}^{\B}{\sf c}_{(\Gamma)}^{\B}{\sf s}_{\sigma}a=
{\sf c}_{(\Delta)}^{\B}{\sf s}_{\bar{\tau}}^{\B}{\sf s}_{\sigma}^{\B}a.$$
Here $\Delta=\bar{\tau}^{-1}\Gamma$, and the last equality follows from axiom (10) in the polyadic axioms noting that
$\bar{\tau}\upharpoonright \Delta$ is one to one since  
$\bar{\tau}\upharpoonright (\mathfrak{n}\sim \alpha)$ is one to one and  
$\Delta\subseteq \mathfrak{n}\sim \alpha$. The last inclusion holds by noting that $\bar{\tau}^{-1}(\mathfrak{n}\sim \alpha)\cap \alpha=\emptyset,$ 
$\Gamma\subseteq \mathfrak{n}\sim \alpha,$ and $\Delta=\bar{\tau}^{-1}\Gamma.$
But then going on, we have also by axioms (7) and (8) of the polyadic axioms
$${\sf c}_{(\Delta)}^{\B}{\sf s}_{\bar{\tau}}^{\B}{\sf s}_{\sigma}^{\B}a= 
{\sf c}_{(\Delta)}^{\B}{\sf s}_{\bar{\tau}\circ \sigma}^{\B}a={\sf c}_{(\Delta)}^{\B}{\sf s}_{Id}^{\B}a={\sf c}_{(\Delta)}^{\B}a=a.$$
The last equality follows from the fact that $\Delta\subseteq \mathfrak{n}\sim \alpha,$ 
and $a$ is $\alpha$-dimensional, that is $a\in \A=\Nr_{\alpha}\B.$  
We have proved that ${\sf s}_{\tau}^{\A}x\leq a$, and so we are done with the second slightly more difficult case, as well.

Let $S$ be the Stone space of $\B$, whose underlying set consists of all boolean ulltrafilters of 
$\B$. Let $X^*$ be the set of principal ultrafilters of $\B$ (those generated by the atoms).  
These are isolated points in the Stone topology, and they form a dense set in the Stone topology since $\B$ is atomic. 
So we have $X^*\cap T=\emptyset$ for every nowhere dense set $T$ (since principal ultrafilters, which are isolated points in the Stone topology,
lie outside nowhere dense sets). 
For $a\in \B$, let $N_a$ denote the set of all boolean ultrafilters containing $a$.
Now  for all $\Gamma\subseteq \alpha$, $p\in B$ and $\tau\in {}^{\alpha}\mathfrak{n}$, we have,
by the suprema, evaluated in (1) and (2):
\begin{equation}\label{tarek3}
\begin{split}
G_{\Gamma,p}=N_{{\sf c}_{(\Gamma)}p}\sim \bigcup_{{\tau}\in {}^{\alpha}\mathfrak{n}} N_{s_{\bar{\tau}}p}
\end{split}
\end{equation}
and
\begin{equation}\label{tarek4}
\begin{split}
G_{X, \tau}=S\sim \bigcup_{x\in X}N_{s_{\bar{\tau}}x}.
\end{split}
\end{equation}
are nowhere dense. 
Let $F$ be a principal ultrafilter of $S$ containing $c$. 
This is possible since $\B$ is atomic, so there is an atom $x$ below $c$; just take the 
ultrafilter generated by $x$.
Then $F\in X^*$, so $F\notin G_{\Gamma, p}$, $F\notin G_{X,\tau},$ 
for every $\Gamma\subseteq \alpha$, $p\in A$
and $\tau\in {}^{\alpha}\mathfrak{n}$.
Now define for $a\in A$
$$f(a)=\{\tau\in {}^{\alpha}\mathfrak{n}: {\sf s}_{\bar{\tau}}^{\B}a\in F\}.$$
Then $f$ is a homomorphism from $\A$ to the full
set algebra with unit $^{\alpha}\mathfrak{n}$, such that $f(c)\neq 0$. We have $f(c)\neq 0$ because $c\in F,$ so $Id\in f(c)$. 
The rest can be proved exactly as in \cite{super}; the preservation of the boolean operations and substitutions is fairly 
straightforward. Preservation of cylindrifications
is guaranteed by the condition that $F\notin G_{\Gamma,p}$ for all $\Gamma\subseteq \alpha$ and all $p\in A$. (Basically an elimination 
of cylindrifications, this condition is also used in \cite{DM}
to prove the main representation result for polyadic algebras.)
Moreover $f$ is an 
atomic representation since $F\notin G_{X,\tau}$ for every $\tau\in {}^{\alpha}\mathfrak{n}$, 
which means that for every $\tau\in {}^{\alpha}\mathfrak{n}$, 
there
exists $x\in X$, such that
${\sf s}_{\bar{\tau}}^{\B}x\in F$, and so $\bigcup_{x\in X}f(x)={}^{\alpha}\mathfrak{n}.$ 
We conclude that $f$ is a complete  representation by Lemma \ref{r}.
\end{demo}
Contrary to cylindric algebras, we have:
\begin{corollary} The class of completely representable polyadic algebras of infinite dimension is elementary
\end{corollary}
\begin{demo}{Proof} Atomicity can be expressed by a first order sentence.
\end{demo}















\section{ Non elementary classes related to complete representations and neat reducts}


\begin{theorem} Let $\K$ be any of cylindric algebra, polyadic algebra, with and without equality, or Pinter's substitution algebra.
We give a unified model theoretic construction, to show the following:
\begin{enumarab}
\item For $n\geq 3$ and $m\geq 3$, $\Nr_n\K_m$ is not elementary, and $S_c\Nr_n\K_{\omega}\nsubseteq \Nr_n\K_m.$
\item For any $k\geq 5$, $\Ra\CA_k$ is not elementary
and $S_c\Ra\CA_{\omega}\nsubseteq \Ra\CA_k$. 
\end{enumarab}
\end{theorem}



%\section{ On relation algebra reducts of cylindric algebras}

In the abstract $S_c$ stands for the operation of forming {\it complete} subalgebras. 
The relation algebra part formulated in the abstract reproves a result of Hirsch in \cite{r}, and answers a question of his 
posed in {\it op.cit}.
For $\CA$ and its relatives the idea is very much like that in \cite{MLQ}, the details implemented, in each separate case, 
though are significantly distinct, because we look for terms not in the clone of operations
of the algebras considered; and as much as possible, we want these to use very little spare dimensions.

The relation algebra part is more delicate. We shall construct a relation algebra $\A\in \Ra\CA_{\omega}$ with a complete subalgebra $\B$, 
such that $\B\notin \Ra\CA_k$, and $\B$ is elementary equivalent to $\A.$ (In fact, $\B$ will be an elementary subalgebra of $\A$.)

Roughly the idea is to use an uncountable cylindric algebra in $\Nr_3\CA_{\omega}$, hence representable, and a finite atom structure
of another cylindric algebra.
%The latter will also be a set algebra.
We construct a finite product of the the uncountable cylindric  algebra; the product will be indexed by the atoms of the atom structure; 
the $\Ra$ reduct of the former will be as desired; it will be a full $\Ra$ reduct of an $\omega$ dimensional algebra and it has a 
complete elementary equivalent subalgebra not in 
$\Ra\CA_k$. This is the same idea for $\CA$, but in this case, and the other cases of its relatives, one spare dimension suffices.
 
This subalgebra is obtained by replacing one of the components of the product with an elementary 
{\it countable} algebra. First order logic will not see this cardinality twist, but a suitably chosen term 
$\tau_k$ not term definable in the language of relation algebras will, witnessing that the twisted algebra is not in $\Ra\CA_k$. 
For $\CA$'s and its relatives, as mentioned in the previous paragraph, we are lucky enough to have $k$ just $n+1,$
proving the most powerful result.

We concentrate on relation algebras. Let $\tau_k$ be an $m$-ary term of $\CA_k$, with $k$ large enough, 
and let $m\geq 2$ be its rank. We assume that $\tau_k$ is not definable
of relation algebras (so that $k$ has to be $\geq 5$); such terms exist.
Let $\tau$ be a term expressible in the language of relation algebras, such that
$\CA_k\models \tau_k(x_1,\ldots x_m)\leq \tau(x_1,\ldots x_m).$ (This is an implication between two first order formulas using $k$-variables).
Assume further that  whenever $\A\in {\bf Cs}_k$ (a set algebra of dimension $k$) is uncountable, 
and $R_1,\ldots R_m\in A$  are such that at least one of them is uncountable, 
then $\tau_k^{\A}(R_1\ldots R_m)$ is uncountable as well. 
(For $\CA$'s and its relatives, $k=n+1$, for $\CA$'s and $\Sc$s, it is a unary term, 
for polyadic algebras it also uses one extra dimension, it is a generalized composition hence it is a binary term.)
%Let $\G$ be a finite group, and $\Cm\G$ be the group relation algebra on $\G$. Then $\Cm\G\cong \wp(\G\times \G)$; 
%this is a complete representation, so  $\Cm\G\in S_c\Ra\CA_{\omega}$; but even more, we have $\Cm\G\in \Ra\CA_{\omega}.$ 
%If $\G=S_n$, we readily infer, that  for any $u\in {}^nn$, $\{u\}\in \Cm S_n$.

%For $u_1\ldots u_m\in {}^33$ and $\sigma$ a relation algebra term 
%we write $\sigma(u_1\ldots u_l)$ for $\sigma^{\R}(\{u_1\},\ldots \{u_n\})$, 
%the latter evaluated in the group relation algebra $\R=\Cm S_3$.
\begin{lemma} Let $V=(\At, \equiv_i, {\sf d}_{ij})_{i,j<3}$ be a finite cylindric atom structure, 
such that $|\At|\geq {}^33.$  
Let $L$ be a signature consisting of the unary relation 
symbols $P_0,P_1,P_2$ and
uncountably many tenary predicate symbols. 
For $u\in V$, let $\chi_u$
be the formula $\bigwedge_{u\in V}  P_{u_i}(x_i)$.  
Then there exists an $L$-structure $\M$ with the following properties:
\begin{enumarab}

\item $\M$ has quantifier elimination, i.e. every $L$-formula is equivalent
in $\M$ to a boolean combination of atomic formulas.

\item The sets $P_i^{\M}$ for $i<n$ partition $M$,

\item For any permutation $\tau$ on $3,$
$\forall x_0x_1x_2[R(x_0,x_1,x_2)\longleftrightarrow R(x_{\tau(0)},x_{\tau(1)}, x_{\tau(2)}],$
\item $\M \models \forall x_0x_1(R(x_0, x_1, x_2)\longrightarrow 
\bigvee_{u\in V}\chi_u)$, 
for all $R\in L$,

\item $\M\models  \exists x_0x_1x_2 (\chi_u\land R(x_0,x_1,x_2)\land \neg S(x_0,x_1,x_2))$ 
for all distinct tenary $R,S\in L$, 
and $u\in V.$

\item For $u\in V$, $i<3,$ 
$\M\models \forall x_0x_1x_2
(\exists x_i\chi_u\longleftrightarrow \bigvee_{v\in V, v\equiv_iu}\chi_v),$

\item For $u\in V$ and any $L$-formula $\phi(x_0,x_1,x_2)$, if 
$\M\models \exists x_0x_1x_2(\chi_u\land \phi)$ then 
$\M\models 
\forall x_0x_1x_2(\exists x_i\chi_u\longleftrightarrow 
\exists x_i(\chi_u\land \phi))$ for all $i<3$
%\item For all $r_1, r_2\in G$, $\M\models \exists x_2[\exists x_0(R_{r_1}(x_0, x_1)\land x_0=x_1)\land 
%\exists x_1(R_{r_2}(x_0, x_1)\land x_0=x_2)\longleftrightarrow R_{r_1+r_2}(x_0, x_1)]$
%\item $\forall x_0x_1x_2[ \phi_{\tau_k}(\chi_{u_1},\ldots \chi_{u_m})(x_0,x_1,x_2)\longrightarrow \chi_{\tau(u_1,\ldots u_{m})}(x_0,x_1,x_2)]$

\end{enumarab}
\end{lemma} 
\begin{demo}{Proof}  
We cannot apply Frassie's theorem to our signature, because it is uncountably infinite. What we do 
instead is that we introduce a new $4$ -ary relation symbol, that will be used to code the uncountably many tenary
relation symbols. Let $\cal L$ be the relational signature containing unary relation symbols
$P_0,\ldots, P_{3}$ and a $4$-ary relation symbol $X$. 
Let $\bold K$ be the class of all finite 
$\cal L$-structures $\D$ satsfying
\begin{enumarab}
\item The $P_i$'s are disjoint : $\forall x \bigvee_{i<j<4}(P_i(x)\land \bigwedge_
{j\neq i}\neg P_j(x)).$ 
\item $\forall x_0 x_1x_2x_3(X(x_0, x_1, x_2, x_3)\longrightarrow P_{3}(x_{3})\land \bigvee_{u\in V}\chi_u).$ 
%\item $\forall x_0x_1x_2[ \phi_{\tau_k}(\chi_{u_1},\ldots \chi_{u_m})(x_0,x_1,x_2)\longrightarrow \chi_{\tau(u_1,\ldots u_{m})}(x_0,x_1,x_2)]$
\end{enumarab}
Then $\bold K$ contains countably many 
isomorphism types. Also it is easy to check that $\bold K$ is closed under
substructures and that $\bold K$ has the the amalgamation Property
From the latter it follows that it has the Joint Embedding Property.
Then there is a  countably infinite homogeneous $\cal L $-structure
$\N$ with age $\bold K$.  $\N$ has quantifier elimination, and obviously, so does 
any elementary extension of $\N$. 
$\bold K$ contains structures 
with arbitrarily large $P_{3}$-part, so $P_{3}^{\N}$ is infinite.
Let $\N^*$ be an elementary extension of $\N$ such that $|P_3|^{\N^*}|=|L|$, and
fix a bijection $*$ from the set of binary relation symbols of $L$ to $P_{3}^{\cal \N^*}.$
Define an $L$-structure $\M$ with domain 
$P_0^{\N^*}\cup P_1^{\N^*}\cup P_2^{\N^*}\cup \ldots P_{3}^{\N^*}$, by:
$P_i^{\M}=P_i^{\N^*}$ for $i<3$ and for
binary $R\in L$, and $\tau\in S_3$ 
$$\M\models R(\tau\circ \bar{a})\text { iff }{\N^*}\models X(\bar{a},R^*).$$ 
If $\phi(\bar {x})$ is any $L$-formula, let $\phi^*(\bar {x},\bar {R})$ be the 
$\cal L$-formula with parameters
$\bar {R}$ from $\N^*$ obtained from $\phi$ by replacing each atomic subformula 
$R(\bar{x})$ by $X(\bar{x},R^*)$ and relativizing quantifiers to $\neg P_n$, 
that is replacing $(\exists x)\phi(x)$ and $(\forall x)\phi(x)$ 
by $(\exists x)(\neg P_3(x)\to \phi(x))$
and $(\forall x)(\neg P_3(x)\to \phi(x)),$ respectively.
A straightforward induction on complexity of formulas
gives that for $\bar {a}\in \M$  
$$\M\models \phi(\bar {a})\text { iff } {\N^*}\models \phi^*(\bar {a},\bar {R}).$$

We show that $\M$ is as required. For quantifier elimination, if $\phi(\bar {x})$ is an
$L$-formula , then $\phi^*(\bar {x} ,\bar {R}^*)$ is equivalent in $\N^*$ 
to a quantifier free $\cal L$-formula
$\psi(\bar {x},\bar {R^*})$. 
Then replacing $\psi$'s atomic subformulas 
$X(x,y,z,R^*)$ by $R(x,y,z)$,
replacing all $X(t_0,\cdots t_3)$ not of this form by $\bot$ , replacing subformulas $P_3(x)$
by $\bot$, and $P_i(R^*)$ by $\bot$ if $i<3$ and $\top$ if $i=3$, gives a quantifier free $L$
-formula $\psi$ equivalent in $\M$ to $\phi$. 
(2) follows from the definition of satisfiability.
\par Let $$\sigma=\forall x(\neg P_3(x)\longrightarrow \bigvee_{i<3}
(P_i(x)\land \bigwedge_{j\neq i}\neg P_j(x))).$$
Then $K\models \sigma$, so $\M\models \sigma$ and ${\N^*}\models \sigma.$
It follows from the definition that $\M$ satisfies (3); (4) is similar.
\par For (5), let $u\in V$ and let $r,s\in P_3^{\M}$ be distinct. 
Take a finite $\cal L$-structure
$D$ with points $a_i\in P_{u_i}^D(i<3)$ and distinct $r',s'\in P_3^D$ with 
$$D\models X(a_0,a_1,a_2,r')\land \neg X(a_0,a_1,a_2,s').$$ 
Then $D\in K$, so $D$ embeds into $\M$. 
By homogeneity, we can assume that the embedding
takes $r'$ to $r$ and $s'$ to $s$. Therefore 
$${\M}\models \exists \bar {x}(\chi_u\land X(\bar {x},r)\land \neg X(\bar {x},s)),$$ 
where $\bar {x}=\langle x_0,x_1,x_2\rangle.$
Since $r,s$ were arbitrary and $\N^*$ is an elementary extension
of $\M$, we get that  
$${\N*}\models \forall yz
(P_3(y)\land P_3(z)\land y\neq z\longrightarrow \exists \bar {x}(\chi_u\land 
X(\bar {x},y)\land \neg (X(\bar {x},z))).$$
The result for $\M$ now follows.
\par Note that it follows from (4,5) that $P_i^{\M}\neq \emptyset$ for each $i<3$.
So it is clear that 
$$\M\models \forall x_0x_1x_2(\exists x_i\chi_u\longleftrightarrow 
\bigvee_{v\in {}V, v\equiv_i u}\chi_v);$$ giving
(6).

\par Finally consider $(7)$. Clearly, it is enough to show that for any
$\cal L$-formula $\phi(\bar {x})$ with parameters
$\bar {r}\in P_3^{\M}, u\in S_3, i<3$, we have 
$${\M}\models \exists \bar {x}(\chi_u\land \phi)\longrightarrow \forall \bar{x}
(\exists x_i(\chi_u\longrightarrow \exists x_i(\chi_u\land \phi)).$$
For simplicity of notation assume $i=2.$
Let $\bar {a} ,\bar {b}\in \M$ with 
$${\M}\models (\chi_u\land \phi)(\bar {a})\text { and } {\M}\models \exists x_2(\chi_u(\bar {b})).$$
We require 
$${\M}\models \exists x_2(\chi_u\land \phi)(\bar {b}).$$
\par It follows from the assumptions that
$${\M}\models P_{u_0}(a_0)\land P_{u_1}(a_1)\land a_0\neq a_1,\text { and }
{\M}\models P_{u_0}(b_0)\land P_{u_1}(b_1)\land b_0\neq b_1.$$
These are the only relations on $a_0a_r\bar {r} $ and on $b_0b_1\bar {r}$
(cf. property (4) of Lemma), so 
$$\theta^{-}=\{(a_0,b_0)(a_1,b_1) (r_l,r_l): l<|\bar {r}|\}$$ 
is a partial isomorphism
of $\M$. By homogeneity, it is induced by an automorphism $\theta$ of $\M$.
Let $c=\theta(\bar {a})=(b_0,b_1,\theta(a_2))$.
Then ${\M}\models (\chi_u\land \phi)(\bar {c}).$
Since $\bar {c}\equiv_2 \bar {b}$, we have 
${\M}\models \exists x_2(\chi_u\land \phi)(\bar {b})$
as required. 
%\end{demo}   
\end{demo}

\begin{lemma}\label{term}
\begin{enumarab} 

\item For $\A\in \CA_3$ or $\A\in \SC_3$, there exist
a unary term $\tau_4(x)$ in the language of $\SC_4$ and a unary term $\tau(x)$ in the language of $\CA_3$
such that $\CA_4\models \tau_4(x)\leq \tau(x),$
and for $\A$ as above, and $u\in \At={}^33$, 
$\tau^{\A}(\chi_{u})=\chi_{\tau^{\wp(^nn)}(u).}$ 

\item For $\A\in \PEA_3$ or $\A\in \PA_3$, there exist a binary
term $\tau_4(x,y)$ in the language of $\SC_4$ and another  binary term $\tau(x,y)$ in the language of $\SC_3$
such that $PEA_4\models \tau_4(x,y)\leq \tau(x,y),$
and for $\A$ as above, and $u,v\in \At={}^33$, 
$\tau^{\A}(\chi_{u}, \chi_{v})=\chi_{\tau^{\wp(^nn)}(u,v)}.$ 


%\begin{enumarab}


\item Let $k\geq 5$.
Then there exist a term $\tau_k(x_1,\ldots x_m)$ in the language of $\CA_k$ and a term $\tau(x_1,\ldots, x_m)$ in the language of $\RA$,
expressible in $\CA_3$, such that
$\CA_k\models \tau_k(x_1,\ldots x_m)\leq \tau(x_1,\ldots x_m),$
and for  $\A$ as above, and $u_1,\ldots u_m\in \At$, 
$\tau^{\A}(\chi_{u_1},\ldots \chi_{u_m})=\chi_{\tau^{\Cm\At}(u_1,\ldots u_m)}.$ 
%\item There exist $u_1,\ldots u_m, v\in {}^33$ such that
%$\tau^{\R}(\{u_1\},\ldots \{u_m\})=v$.
\end{enumarab}
\end{lemma}

\begin{proof} 

\begin{enumarab}

\item For all reducts of polyadic algebras, these terms are given in \cite{FM}, and \cite{MLQ}.
For cylindric algebras $\tau_4(x)={}_3 s(0,1)x$ and $\tau(x)=s_1^0c_1x.s_0^1c_0x$.
For polyadic algebras, it is a little bit more complicated because the former term above is definable.
In this case we have $\tau(x,y)=c_1(c_0x.s_1^0c_1y).c_1x.c_0y$, and $\tau_4(x,y)=c_3(s_3^1c_3x.s_3^0c_3y)$.

\item For relation algebras, we take the term corresponding to the following generalization of Johnson's $Q$'s. 
Given $1\leq n<\omega$ and $n^2$ tenary relations
we define
$$Q(R_{ij}: i,j<n)(x_0, x_2, x_3)\longleftrightarrow$$ 
$$\exists z_0\ldots z_{n-1}(z_0=x\land z_1=y\land 
z_2=z\land \bigwedge_{i,j,l<n} 
R_{ij}(z_i, z_j, z_l).$$
{\bf Still working out the required term $\tau$}
\end{enumarab}
\end{proof}

\begin{theorem}
\begin{enumarab} 
\item There exists $\A\in \Nr_3\QEA_{\omega}$
with an elementary equivalent cylindric  algebra, whose $\SC$ reduct is not in $\Nr_3\SC_4$.
Furthermore, the latter is a complete subalgebra of the former.
 
\item There exists a relation algebra $\A\in \Ra\CA_{\omega}$, with an elementary equivalent relation algebra not in $\Ra\CA_k$.
Furthermore, the latter is a complete subalgebra of the former.
\end{enumarab}
\end{theorem}
\begin{proof} Let $\L$ and $\M$ as above. Let 
$\A_{\omega}=\{\phi^M: \phi\in \L\}.$ 
Clearly $\A_{\omega}$ is a locally finite $\omega$-dimensional cylindric set algebra.
For the first part, we prove the theorem for $\CA$; and its relatives.
%The algebra $\A$  is a relation algebra; indeed it is closed under converse 
%because of condition (iii) and under
%composition by condition (viii), in the previous lemma.

Then $\A\cong \Nr_3\A_{\omega}$, the isomorphism is given by 
$$\phi^{\M}\mapsto \phi^{\M}.$$
Quantifier elimination in $\M$ guarantees that this map is onto, so that $\A$ is the full $\Ra$ reduct.

For $u\in {}V$, let $\A_u$ denote the relativisation of $\A$ to $\chi_u^{\M}$
i.e $$\A_u=\{x\in A: x\leq \chi_u^{\M}\}.$$ $\A_u$ is a boolean algebra.
Also  $\A_u$ is uncountable for every $u\in V$
because by property (iv) of the above lemma, 
the sets $(\chi_u\land R(x_0,x_1,x_2)^{\M})$, for $R\in L$
are distinct elements of $\A_u$.  

Define a map $f: \Bl\A\to \prod_{u\in {}V}\A_u$, by
$$f(a)=\langle a\cdot \chi_u\rangle_{u\in{}V}.$$

Here, and elsewhere, for a relation algebra $\C$, $\Bl\C$ denotes its boolean reduct.
We will expand the language of the boolean algebra $\prod_{u\in V}\A_u$ by constants in 
such a way that
the relation algebra reduct of $\A$ becomes interpretable in the expanded structure.
For this we need.

Let $\P$ denote the 
following structure for the signature of boolean algebras expanded
by constant symbols $1_{u}$ for $u\in V$ 
and ${\sf d}_{ij}$ for $i,j\in 3$: 
We now show that the relation algebra reduct of $\A$ is interpretable in $\P.$  
For this it is enough to show that 
$f$ is one to one and that $Rng(f)$ 
(Range of $f$) and the $f$-images of the graphs of the cylindric algebra functions in $\A$ 
are definable in $\P$. Since the $\chi_u^{\M}$ partition 
the unit of $\A$,  each $a\in A$ has a unique expression in the form
$\sum_{u\in {}V}(a\cdot \chi_u^{\M}),$ and it follows that 
$f$ is boolean isomorphism: $bool(\A)\to \prod_{u\in {}V}\A_u.$
So the $f$-images of the graphs of the boolean functions on
$\A$ are trivially definable. 
$f$ is bijective so $Rng(f)$ is 
definable, by $x=x$. For the diagonals, $f({\sf d}_{ij}^{\A})$ is definable by $x={\sf d}_{ij}$.

Finally we consider cylindrifications for $i<3$. Let $S\subseteq {}V$ and  $i<3$, 
let $t_S$ be the closed term
$$\sum\{1_v: v\in {}V, v\equiv_i u\text { for some } u\in S\}.$$
Let
$$\eta_i(x,y)=\bigwedge_{S\subseteq {}V}(\bigwedge_{u\in S} x.1_u\neq 0\land 
\bigwedge_{u\in {}V\smallsetminus S}x.1_u=0\longrightarrow y=t_S).$$
We claim that for all $a\in A$, $b\in P$, we have 
$$\P\models \eta_i(f(a),b)\text { iff } b=f({\sf c}_i^{\A}a).$$
To see this, let $f(a)=\langle a_u\rangle_{u\in {}V}$, say. 
So in $\A$ we have $a=\sum_ua_u.$
Let $u$ be given; $a_u$ has the form $(\chi_i\land \phi)^{\M}$ for some $\phi\in L^3$, so
${\sf c}_i^A(a_u)=(\exists x_i(\chi_u\land \phi))^{\M}.$ 
By property (vi), if  $a_u\neq 0$, this is 
$(\exists x_i\chi_u)^M$; by property $5$,
this is $(\bigvee_{v\in {}V, v\equiv_iu}\chi_v)^{\M}.$
Let $S=\{u\in {}V: a_u\neq 0\}.$
By normality and additivity of cylindrifications we have,
$${\sf c}_i^A(a)=\sum_{u\in {}V} {\sf c}_i^Aa_u=
\sum_{u\in S}{\sf c}_i^Aa_u=\sum_{u\in S}(\sum_{v\in {}V, v\equiv_i u}\chi_v^{\M})$$
$$=\sum\{\chi_v^{\M}: v\in {}V, v\equiv_i u\text { for some } u\in S\}.$$
So $\P\models f({\sf c}_i^{\A}a)=t_S$. Hence $\P\models \eta_i(f(a),f({\sf c}_i^{\A}a)).$
Conversely, if $\P\models \eta_i(f(a),b)$, we require $b=f({\sf c}_ia)$. 
Now $S$ is the unique subset of $V$ such that 
$$\P\models \bigwedge_{u\in S}f(a)\cdot 1_u\neq 0\land \bigwedge_{u\in {}V\smallsetminus S}
f(a)\cdot 1_u=0.$$  So we obtain 
$$b=t_S=f({\sf c}_i^Aa).$$
The rest is the same as in \cite{MLQ} while for other relatives, the idea implemented is also the same;
one just uses the corresponding terms as in lemma \ref{term}.




For relation algebras we proceed as follows:
Now the $\Ra$ reduct of $\A$ is a generalized reduct of $\A$, hence $\P$ is first order interpretable in $\Ra\A$, as well.
 It follows that there are closed terms $1_{u, v},d_{i,j}$ and a formula $\eta$ built out of these closed terms such that 
$$\P\models \eta(f(a), b, c)\text { iff }b= f(a\circ c),$$
where the composition is taken in $\Ra\A$.

We have proved that $\Ra\A$ is interpretable in $\P$.
Furthermore it is easy to see that the interpretation 
is two dimensional and quantifier free. 


For each $u\in V$, choose any countable boolean elementary 
complete subalgebra of $\A_{u}$, $\B_{u}$ say.
%Thus $\B_{Id}\preceq \A_{Id}$, and is also a complete subalgebra.
Le $u_i: i<m$ be elements in $V$ and let
$$Q=(\prod_{u_i: i<m}\A_{u_i}\times \B_{\tau^{\Cm\At}(u_1,\ldots u_m)}\times \prod_{u\in {}V\smallsetminus \{u_1,\ldots u_m, \tau^{\Cm\At\A}(u_1,\ldots u_m)\}} 
\A_u), 1_{u,v},d_{ij})_{u,v\in {}V,i,j<3}\equiv$$  
$$(\prod_{u\in V} \A_u, 1_{u,v}, {\sf d}_{ij})_{u\in V, i,j<3}=P.$$

Let $\B$ be the result of applying the interpretation given above to $Q$.
Then $\B\equiv \Ra\A$ as relation  algebras, furthermore $\Bl\B$ is a complete subalgebra of $\Bl\A$. 
%(Indeed $\B\preceq \A$. 
%Therefore $\B\in RCA_3$; it is easy to see that
%$B$ is simple the latter expressible by a first order formula
%so that it is actually a $Cs_3$.)
Assume for contradiction that $\B=\Ra\D$ with $\D\in \CA_k$. Let $u_1,\ldots u_m\in V$ be such that 
$\tau_k^{\D}(\chi_{u_1},\ldots \chi_{u_n})$, is uncountable in $\D$. 

Because $\B$ is a full $\RA$ reduct, 
this set is contained in $\B.$ 

For simplicity assume that $\tau^{\Cm\At}(u_1\ldots u_m)=Id.$
On the other hand for $x_i\in B$, with $x_i\leq \chi_{u_i}$, we have 
$$\tau_k^{\D}(x_1,\ldots x_m)\leq \tau(x_1\ldots x_m)\in \tau(\chi_{u_1},\ldots {\chi_{u_m}})=\chi_{\tau(u_1\ldots u_m)}=\chi_{Id}.$$

But this is a contradiction, since  $B_{Id}=\{x\in B: x\leq \chi_{Id}\}$ is  countable.


\end{proof}


\section{Weakly representable atom structures that are not strongly representable}

There are two types of constructions of atom structures as in the title, the Andreka-N\'emeti construction,  
and the Hodkinson's style (model theoretic) style construction.
In the first, for any $n\geq 3$, for any pre assigned $k\in \omega$,  
a relation algeba atom structure which has an $n$ dimensional cylindric basis, 
such that the term algebra over the
cylindric atom structure of  the basic matrices is representable, and is in $\Nr_nCA_{n+k},$ but the complex algebra is not. 
Furthermore, the relation algebra atom structure itself is only weakly representable. 

In the second construction, one move backwards.
Hodkinson constructs, by model theoretic methods that are not too difficult,
an atomic representable cylindric algebra, whose atom structure is isomorphic to the cylindric algebra consisting 
of basic matrices over a relation atom structure. 
Both structures are again only weakly representable.

In the first case, we will simply the construction. In this second case, we will also simplify the (rainbow) construction, which we did before in \cite{w},
but our real gain here , is that the complex algebra over the cylindric atom structure, consisting of basic matrices,
which is the completion of $\A$, is not only not representable, but in fact it is not in $S\Nr_n\CA_{n+k}$ for any $k\in \omega$.
From this we show that each of the varieties$S\Nr_n\CA_{n+k}$ ($k\geq 2)$ is not closed under completions. 



\subsection{First construction}

Here we give a simplified version of the construction in \cite{1}. Also several new consequences are implemented.
Basically this is the same argument used in opcit, except that is considerably simplified and hence is more streamlined, and we believe easier to 
grasp. Indeed the idea underlying some construction maybe simple, 
when they first occur to the reserach, but then trying to get the most general result, 
things become apprently much more complicated, though in esence the idea is the same.
Another thing is the first solution to a problem is usualy not the simplest.



Throughout, $H$ is a fixed  
infinite set on which we will impose necessary conditions as we proceed.  In fact,  
$H$ denotes the set of non-identity atoms in our future 
algebras  (i.e. all the algebras we construct in a while.) 
$$F=H\cup \{Id\}$$ is the universe
of our intended future atom structure $\cal F.$ 

%Let $I$, $J$ be sets, and let 
%$\langle H^P: P\in I\rangle,$ 
%and 
%$\langle E^W: W\in J\rangle$ 
%be two partitions
%of $H$. 


%Now we implement a concrete plan. 
%We provide a concrete $H$ that satisfies the desired conditions listed as $(1^*)-(6^*)$ 
%above.  
Let $I$ be a finite set with $|I|\geq 6$. Let $J$ be the set of all $2$ element
subsets of $I$, and let
$$H=\{a_i^{P,W}: i\in \omega, P\in I, W\in J, P\in W\}.$$
For $P\in I$, let 
$$H^P=\{a_i^{P,W}: i\in \omega, W\in J, P\in W\}.$$
For $W\in J,$ let
$$E^W=\{a_i^{P,W}: i\in \omega, P\in W\}.$$
%This is illustrated in figure 1 below:
This way we have defined our {\it two} partitions of $H$.
For $i,j,k\in \omega$ $e(i,j,k)$ abbreviates that $i,j,k$ are {\it evenly distributed}, i.e.
$$e(i,j,k)\text { iff } (\exists p,q,r)\{p,q,r\}=\{i,j,k\}, r-q=q-p$$
For example $3,5,7$ are evenly distributed, but $3,5,8$ are not. 

All atoms are self-converse. We define the consistent triples as follows
Let $i,j,k\in \omega$, $P,Q,R\in I$ and $S,Z,W\in J$ such that
$P\in S$, $Q\in Z$ and $R\in W$. Then the triple
$(a_i^{P,S},a_j^{Q,Z}, a_k^{R,W})$ is consistent iff 
either 
\begin{enumroman}
\item $S\cap Z\cap W=\emptyset,$
or 
\item $e(i,j,k)\&\{P,Q,R\}|\neq 1.$
\end{enumroman}
It is easy to check that $e$ hence $\bold T$ is symmetric. 
Therefore $;$ satisfies the triangle
rule. Now choose a (finite) relation algebra $\bold M$ with atoms $I\cup \{1d\}$ such that
for all $P,Q\in I$, $P\neq Q$ we have
$$P;P=\{Q\in I: Q\neq P\}\cup \{Id\}\text { and } P;Q=H$$
Such an $\bold M$ exists. Such algebras are constructed by Maddux. It is known that
$\bold M$ cannot be represented on finite sets.
\begin{enumroman}
\item $\Cm{\cal F}$ is a relation algebra that is not representable.
\item ${\R}$ the term algebra over $\cal F$ is representable.
In other words, $\cal F$ is an example of a weakly representable atom structure
that is not strongly representable.
\end{enumroman}
\begin{proof} 
\begin{enumarab}
\item Non representabiliy uses the first partition of $H$. Note that $;$ is defined on $\Cm({\cal F})$ so  that 
$$H^P; H^Q=\bigcup \{H^Z: Z\leq P; Q\in {\bold M}\}.$$
So $\bold M$ is isomorphic to a subalgebra of $Cm F$. But $\Cm F$ 
can only be represented o infinite sets, while $\M$ only on finite ones, hence we are done.

\item The representability of the term algebra uses the second partition. 
It can be checked that $R=\{X\subseteq F: X\cap E^W\in Cof(E^W), \forall W\in J\}$. 
For any $a\in F$ and $W\in J$, let
$$U^a=\{X\in R: a\in X\}$$
and
$$U^W=\{X\in R: |Z\cap E^W|\geq \omega\}$$
Let
$$Uf=\{U^a: a\in F\}\cup \{U^W: W\in J: |E^W|\geq \omega \}$$.
$Uf$ denotes the set of ultrafilters of $\cal R$, that include at least one non-principal 
ultrafilter, that is an element of the form $U^W$.  
We call $(G,l)$ a  {\it consistent coloured graph} if $G$ is a set , 
$l:G\times G\to Uf$
such that for all $x,y,z\in G$, the following hold:

\begin{enumroman}

\item $l(x,y)=U^{Id}$ iff $x=y,$

\item $l(x,y)=l(y,x)$

\item The triple $(l(x,y),l(x,z),l(y,z))$ is consistent.

\end{enumroman}

We say that a consistent coloured graph $(G,l)$ is complete if for all 
$x,y\in G$, and $F,K\in Uf$, whenever 
$(l(x,y),F,K)$ is consistent, then there is a node $z$ such that
$l(z,x)=F$ and $l(z,y)=K$.
%Consistent coloured graphs can 
%be used to buid up representations either in a step by step manner or by games \cite{HH}.  
We will build a complete consistent graph step-by-step. So assume (inductively) 
that$(G,l)$ is a consistent coloured graph and $(l(x,y), F, K)$ is a consistent triple.
We shall extend $(G,l)$ with a new point $z$ such that $(l(x,y), l(z,x), l(z,y)) =(l(x,y),G,K).$
Let $z\notin G.$ We define $l(z,p)$ for $p\in G$ as follows:
$$l(z,x)=F$$
$$l(z,y)=K, \text { and if } p\in G\smallsetminus \{x,y\}, \text { then }$$
$$l(z,p)=U^W\text { for some }W\in J' \text { such that both }$$
$$(U^W, F, l(x,p))\text { and } (U^W, K, l(y,p))\text { are consistent }.$$
Such a $W$ exists by our assumptions (i)-(iii).
Conditions (i)-(ii) guarantee that this extension is again a consistent coloured graph.



We now show that any non-empty complete coloured graph $(G,l)$ gives a representation 
for $\cal R.$ For any $X\in R$ define
$$rep(X)=\{(u,v)\in G\times G: X\in l(u,v)\}$$
We show that 
$$rep:{\cal R}\to R(G)$$ 
is an embedding. $rep$ is a boolean homomorphism
because all the labels are ultrafilters. 
$$rep(Id)=\{(u,u): u\in G\},$$  
and for all $X\in R$,
$$rep(X)^{-1}=rep(X).$$
The latter follows from the first condition in the definition of a consistent coloured graph.
From the second condition in the definition of a consistent coloured graph, we have:
$$rep(X);rep(Y)\subseteq rep(X;Y).$$
Indeed, let $(u,v)\in rep(X), (v,w)\in rep(Y)$ I.e. $X\in l(u,v), Y\in l(v,w).$
Since $(l(u,v), l(v,w), l(u,w))$ is consistent, then $X;Y\in l(u,w)$, i.e. $(u,w)\in 
rep(X;Y).$
On the other hand, since $(G,l)$ is complete and because (i)-(ii) hold, we have:
$$rep(X;Y)\subseteq rep(X);rep(Y),$$
because $(G,l)$ is complete and because (i) and (ii) hold.
Indeed, let $(u,v)\in rep(X;Y)$. Then $X;Y\in l(u,v)$. 
We show that there are $F,K\in Uf$ such that
$$X\in F, Y\in K \text { and } (l(u,v), F,K)\text { is consistent }.$$
We distinguish between two cases:

{\bf Case 1}. $l(u,v)=U^a$ for some $a\in F$. By $X;Y\in U^a$ we have $a\in X;Y.$
Then there are $b\in X$, $c\in Y$ with $a\leq b;c.$ Then $(U^a, U^b, U^c)$ is consistent.

{\bf Case 2.} 
$l(u,v)=U^W$ for some $W\in J'$. Then $|X;Y\cap E^W|\geq \omega$ 
by $X;Y\in U^W$.
Now if both $X$ and $Y$ are finite, then there are $a\in X$, $b\in Y$ with 
$|a;b\cap E^W|\geq \omega$.
Then $(U^W, U^a, U^b)$ is consistent by (i). Assume that one of $X,Y$, say $X$ is infinite.
Let $S\in J'$ such that $|X\cap E^S|\geq \omega$ and let $a\in Y$ be arbitrary. 
Then $(U^W, U^S, U^a)$ is consistent by (ii)
and $X\in U^S, Y\in U^a.$

Finally, $rep$ is one to one because $rep(a)\neq \emptyset$ for all $a\in A$. 
Indeed $(u,v)\in rep(Id)$ for any
$u\in G$. Let $a\in H$. Then $(U^{Id}, U^a, U^a)$ is consistent, so there is a $v\in G$ with 
$l(u,v)=U^a$. Then $(u,v)\in rep(a).$

\end{enumarab}

\end{proof}

%\subsubsection*{ Atom structures giving rise to cylindric atom structures}

We define the atom structure like we did before. The basic matrices of the atom structure above form a 
$3$ dimensional cylindric algebra. We want an 
$n$ dimensional one.  Our previous construction of the atom structure
satisfies $(\forall a_1\ldots a_n3_1\ldots b_3\in H)(\exists W\in J')(a_1;b_1)\cap (a_3;b_3)\in U^W.$
We strengthen this condition to 
$$(\forall a_1\ldots a_nb_1\ldots b_n\in H)(\exists W\in J')(a_1;b_1)\cap (a_n;b_n)\in U^W.$$
Accordingly instead of $H$ we write $H_n$. 
This condition will entail that the set of {\it all} $n$ by $n$ 
basic matrices of ${\R}_n$ forms a cylindric bases as we proceed
to show:
Let $n>2$ be given. Let $I$ be a finite set with $|I|\geq 2n+2$. 
Let $J$ be the set of all $2$ element
subsets of $I$, and let
$$H_n=\{a_j^{P,W}: j\in \omega, P\in I, W\in J, P\in W\}.$$
For $P\in I$, let 
$$H^P=\{a_j^{P,W}: j\in \omega, W\in J, P\in W\}.$$
For $W\in J,$ let
$$E^W=\{a_j^{P,W}: j\in \omega, P\in W\}.$$
This way we have defined our two partitions of $H_n$.
For $i,j,k\in \omega$ $e(i,j,k)$ abbreviates that $i,j,k$ are {\it evenly distributed}, i.e.
$$e(i,j,k)\text { iff } (\exists p,q,r)\{p,q,r\}=\{i,j,k\}, r-q=q-p$$
%For example $3,5,7$ are evenly distributed, but $3,5,8$ are not. 

\begin{definition}
We define the consistent triples as before
Let $i,j,k\in \omega$, $P,Q,R\in I$ and $S,Z,W\in J$ such that
$P\in S$, $Q\in Z$ and $R\in W$. Then we set
$(a_i^{P,S},a_j^{Q,Z}, a_k^{R,W})$ is consistent iff 
\begin{enumarab}
\item $S\cap Z\cap W=\emptyset$,
or
\item $e(i,j,k)\text {and }|\{P,Q,R\}|\neq 1.$
\end{enumarab}
\end{definition}

Now we have:


Let $n>2$ be finite. Let 
$H_n$ be as specified above and let ${\R}_n$ be the relation algebra 
based
on $F=H_n\cup \{Id\}$. Then the following hold:
\begin{athm}{Theorem 2}
Suppose that $n$ is a finite ordinal with $n>2$ and $k\geq 0$.
There is a countable 
symmetric integral representable 
relation algebra ${\R}$
such that
\begin{enumroman}
\item Its completion, i.e. the complex algebra of its atom structure is 
not representable, so $\R$ is representable but not completely representable 
\item $\R$ is generated by a single element.
\item The (countable) set $\B_n{\R}$ of all $n$ by $n$ basic matrices over $\R$ 
constitutes an $n$-dimensional cylindric basis. 
Thus $\B_n{\R}$ is a cylindric atom structure 
and the full complex algebra $\Cm(\B_n{\R})$ 
with universe the power set of $\B_n{\R}$
is an $n$-dimensional cylindric algebra 
\item The {\it term algebra} over the atom structure 
$\B_n{\R}$, which is the countable subalgebra of $\Cm(\B_n{\R})$ 
generated by the countable set of 
$n$ by $n$ basic matrices, $\C=\Tm(B_n \R)$ for short,
is a countable representable $\CA_n$, but $\Cm(\B_n)$ is not representable. 
\item Hence $\C$ is a simple, atomic representable but not completely representable $\CA_n$
\item $\C$ is generated by a single $2$ dimensional element $g$, the relation algebraic reduct
of $\C$ does not have a complete representation and is also generated by $g$ as a relation algebra, and 
$\C$ is a sub-neat reduct of some simple representable $\D\in \CA_{n+k}$ 
such that the relation algebraic reducts of $\C$ and $\D$
coincide.  
%If it did, then $R_n$ will also have a complete representation, which is not the case. 
\end{enumroman}
\end{athm}


%\begin{theorem} Let  $H_n$, $P$ and $Q$ be as described above.
%Recall that $H_n$ is the set of non-identity atoms. 



%\subsection{More Implementations}
Here we include more examples. 
%It is not hard to check that they satisfy $(1^*)-(6^*)$.

\begin{example}


Let $l\in \omega$, $l\geq 2$, and let $\mu$ be a non-zero cardinal. Let $I$ be a finite set,
$|I|\geq 3l.$ Let 
$$J=\{(X,n): X\subseteq I, |X|=l,n<\mu\}.$$
Let $H$ be as before, i.e.
$$H=\{a_i^{P,W}: i\in \omega, P\in I, W\in J\}.$$
Define
$(a_i^{P,S,p}, a_j^{Q,Z,q}, a_k^{R,W,r})$ is consistent ff 

$S\cap Z\cap W=\emptyset$ or 
$ e(i,j,k) \text { and } |\{P,Q,R\}|\neq 1.$

Pending on $l$ and $\mu$, let us call these atom structures ${\cal F}(l,\mu).$
Then the example in section 2 is just 
${\cal F}(2,1).$

If $\mu\geq \omega$, then $J$ as defined in section 2 would be infinite, 
and $Uf$ will be a proper subset of the ultrafilters.
It is not difficult to show that if $l\geq \omega$ 
(and we relax the condition that $I$ be finite), then
$\Cm{\cal F}(l,\mu)$ is completely representable, 
and if $l<\omega$ then $Cm{\cal F}(l,\mu)$ is non-representable.

%Hirsch and Hodkinson prove that the class of 
%completely representable $RRA$'s is not elementary.
%We now have the machinery to prove:

%\end{enumarab}
\end{example}

\begin{corollary}
\begin{enumarab}
\item  The classes $\RRA$ is not finitely axiomatizable.
\item  The elementary 
closure of the class ${\bf CRA}$ is not finitely axiomatizable.
\end{enumarab}
\end{corollary}
\begin{demo}{Proof} 
Let ${\cal D}$ be a non-
trivial ultraproduct of ${\cal F}(i,1)$, $i\in \omega$. Then $Cm{\cal D}$
is completely representable. 
Thus ${\R}(i,1)$ are $RRA$'s 
without a complete representation while their ultraproduct has a complete representation.
Also $Cm{\cal D}(i,1)$ $i\in \omega$ 
are non representable with a completely representable ultraproduct.
This yields the desired result.
\end{demo}
We note that a variation on the theme, usiraltion atom structures with $n$ dimensional cylindric bases,
yields the analogous result for cylindric algebras.
In more detail

\begin{corollary} Let $2<n<\omega$
\begin{enumarab}
\item  The classes $\RCA_n$ is not finitely axiomatizable.
\item  The elementary 
closure of the class  of completely representable $\CA_n$ 
is not finitely axiomatizable.
\end{enumarab}
\begin{demo} Let $n\geq 3$.  For $k\geq 0$, let $\B_k\in RCA_n\cap \Nr_n\CA_{n+k}$ 
such that $\B$ has no complete representation. Then $\B_k=\Nr_n\C_k$, with  $C_k\in \CA_{n+k}$. 
Let $C_n^+\in \CA_{\omega}$ 
be such that $\Rd_{n+k}\C_k^+=C_k$. Then $\B=\prod \B_k/F=\prod Nr_nC_k/F=\Nr_n\prod C_k^+/F$. 
Then $\B$ is completely representable. 
\end{demo}

\end{corollary}

%\begin{demo}{Proof} This proves theorem..above. We leave this to the reader. 
%\end{demo}

\subsection{Completions of sub neat reducts}

We use a construction of Hirsch and Hodkinson \cite{HHbook}. We follow the notation in op.cit. 
%$\RA$ stands for the class of relation algebras and
%$\CA_n$ stands for the class of cylindric algebras of dimension $n$.
$\Ra\CA_n$ stands for the class of relation algebra reducts of $\CA_n$.
%$\RCA_n$ stands for the class of representable $\CA_n$'s and for $k<n$, $\Nr_k\CA_n$ stands for 
%the class of neat $k$ reducts of algebras in $\CA_n$.
It is known that $\RCA_n={\bf S}\Nr_n\CA_{\omega}$ for any finite $n$ and that for $k\in \omega$ and $n>2$, 
$\RCA_n\subset {\bf S}\Nr_n\CA_{n+k+1}\subset 
{\bf S}\Nr_n\CA_{n+k}$ \cite{HHbook}. 

Let $\RA_n$ be the class of subalgebras of atomic relation algebras having $n$ dimensional relational basis.
Then ${\bf S}\Ra\CA_n\subseteq \RA_n$ \cite{HHbook}. 
The full complex algebra of an atom structure $S$
will be denoted by $\Cm S$, and the term algebra by $\Tm S.$
$S$ could be a relation atom structure or a cylindric atom structure.

\begin{theorem} Let $n\geq 3$. Assume that for any simple atomic relation algebra $\cal A$ with atom structure $S$, 
there is a cylindric atom structure $H$ such that:
\begin{enumarab}
\item If $\Tm S\in \RRA$, then $\Tm H\in \RCA_n$.
\item $\Cm S$ is embeddable in $\Ra$ reduct of $\Cm H$.
\end{enumarab}
Then for all $k\geq 3$, $S\Nr_n\CA_{n+k}$ is not closed under completions.

\end{theorem}
\begin{demo}{Proof} Let $S$ be a relation atom structure such that $\Tm S$ is representable while $\Cm S\notin \RA_6$.
Such an atom structure exists \cite{HHbook} Lemmas 17.34-17.36 and are finite. 
It follows that $\Cm S\notin {\bf S}\Ra\CA_n$.
Let $H$ be the $\CA_n$ atom structure provided by the hypothesis of the previous theorem.  
Then $\Tm H\in \RCA_n$. We claim that $\Cm H\notin {\bf S}\Nr_n\CA_{n+k}$, $k\geq 3$. 
For assume not, i.e. assume that $\Cm H\in {\bf S}\Nr_n\CA_{n+k}$, $k\geq 3$.
We have $\Cm S$ is embeddable in $\Ra\Cm H.$  But then the latter is in ${\bf S}\Ra\CA_6$
and so is $\Cm S$, which is not the case.
 \end{demo}
 
Monk and Maddux constructs such an $H$ for $n=3$ and Hodkinson constructs  an $H,$ but $H$ does not satisfy (2).


Now we prove, in a different way that classes of subneat reducts are not closed under completions. the condition theorem above. 
We use the model theoretic methods in \cite{Hodkinson} and \cite{w}, 
to construct strongly representable atom structures that are not weakly representable.
Our chosen graph is yet simpler than the one used in \cite{w}, and we prove an even  stronger result namely that for each $k\geq 2$
the variety $S\Nr_n\CA_{n+k}$ is not closed under completions, hence is it also not atom-canonical.
This answers a question of Hirsch and Hodkinson. When the proofs are completely analogous to profs in \cite{w} 
(the esence of the construction is the same) we refer to \cite{w}.

In this section by a graph we shall mean a pair $\G=(G,E)$ where $G$ is a non-empty set of vertices 
and $E\subseteq G\times G$ is an irreflexive and
symmetric binary relation on $G$. A pair $(x,y)\in E$ is called an edge of $\G$. Fix finite $N\geq n(n-1)/2$. 
 $\G$ will denote the graph $\G=(\N,E)$ with nodes $\N$ and $i,l$ is an edge i.e $(i,l)\in E$ if  
$0<|i-l|<N$. The graph used in \cite{w} is an countable union of cliques, denote by $N\times \omega$. 
%Another one is the one used in \cite{HHbook} based on $\mathbb \Z$.


\begin{definition}
A \textit{labelled graph} is an undirected graph $\Gamma$ such that
every edge ( \textit{unordered} pair of distinct nodes ) of $\Gamma$
is labelled by a unique label from $(\G \cup \{\rho\}) \times n$, where
$\rho \notin \G$ is a new element. The colour of $(\rho, i)$ is
defined to be $i$. The \textit{colour} of $(a, i)$ for $a \in \G$ is
$i$.
\end{definition}

We will write $ \Gamma (x, y)$ for the label of an edge $ (x, y)$ in
the labelled graph $\Gamma$. Note that these may not always be
defined: for example, $ \Gamma (x, x)$ is not.\\
If $\Gamma$ is a labelled graph, and $ D \subseteq \Gamma$, we write
$ \Gamma \upharpoonright D$ for the induced subgraph of $\Gamma$ on
the set $D$ (it inherits the edges and colours of $\Gamma$, on its
domain $D$). We write $\triangle \subseteq \Gamma$ if $\triangle$ is
an induced subgraph of $\Gamma$ in this sense.



\begin{definition}
Let $ \Gamma, \triangle$ be labelled graphs, and $\theta : \Gamma
\rightarrow \triangle$ be a map. $\theta$ is said to be a
\textit{labelled graph embedding}, or simple an \textit{embedding},
if it is injective and preserves all edges, and all colours, where
defined, in both directions. An \textit{isomorphism} is a bijective
embedding.
\end{definition}

Now we define a class $\GG$ of certain labelled graphs.


\begin{definition}
The class $\GG$ consists of all complete labelled graphs $\Gamma$ (possibly
the empty graph) such that for all distinct $ x, y, z \in \Gamma$,
writing $ (a, i) = \Gamma (y, x)$, $ (b, j) = \Gamma (y, z)$, $ (c,
l) = \Gamma (x, z)$, we have:\\
\begin{enumarab}
\item $| \{ i, j, l \} > 1 $, or
\item $ a, b, c \in \G$ and $ \{ a, b, c \} $ has at least one edge
of $\G$, or
\item exactly one of $a, b, c$ -- say, $a$ -- is $\rho$, and $bc$ is
an edge of $\G$, or
\item two or more of $a, b, c$ are $\rho$.
\end{enumarab}
Clearly, $\GG$ is closed under isomorphism and under induced
subgraphs.
\end{definition}


\begin{theorem}
There is a countable labelled graph $M\in \GG$ with the following
property:\\
$\bullet$ If $\triangle \subseteq \triangle' \in \GG$, $|\triangle'|
\leq n$, and $\theta : \triangle \rightarrow M$ is an embedding,
then $\theta$ extends to an embedding $\theta' : \triangle'
\rightarrow M$.
\end{theorem}
\begin{proof} \cite{Hodkinson}, \cite{w} prop. 3.5
\end{proof}
Let $L^n$ denote the first-order fragment of $L^n_{\infty \omega}$

\begin{definition}
An $n$-\textit{back-and-forth system} on $A$ is a set $\Theta$ of
one-to-one partial maps : $ A \rightarrow A$ such that:\\
\begin{enumerate}
\item if $\theta \in \Theta$ then $|\theta| \leq n$
\item if $ \theta' \subseteq \theta \in \Theta$ then $\theta' \in \Theta$
\item if $\theta \in \Theta$, $|\theta| \leq n$, and $ a \in A$,
then there is $ \theta' \supseteq \theta$ in $\Theta$ with $ a \in
dom(\theta')$ (forth)
\item if $\theta \in \Theta$, $|\theta| \leq n$, and $ a \in A$,
then there is $ \theta' \supseteq \theta$ in $\Theta$ with $ a \in
rng(\theta')$ (back).

\end{enumerate}
\end{definition}

 Recall that a \textit{partial isomorphism} of $A$ is a
partial map $ \theta : A \rightarrow A$ that preserves all
quantifier-free $L$-formulas.
The next theorem is a well known result in model theory.
\begin{theorem} 
\begin{enumarab}
\item Let $\Theta$ be an $n$-back-and-forth system
of partial isomorphism on $A$, let $\bar{a}, \bar{b} \in {}^{n}A$,
and suppose that $ \theta = ( \bar{a} \mapsto \bar{b})$ is a map in
$\Theta$. Then $ A \models \phi(\bar{a})$ iff $ A \models
\phi(\bar{b})$, for any formula $\phi$ of $L^n_{\infty \omega}$.

\item If $W$ is $L^n_{\infty \omega}$ definable, $\Theta$ is an
 $n$-\textit{back-and-forth} system
of partial isomorphisms on $A$, $\bar{a}, \bar{b} \in W$, and $
\bar{a} \mapsto \bar{b} \in \Theta$, then $ A \models \phi(\bar{a})$
iff $ A \models \phi(\bar{b})$ for any formula $\phi$ of
$L^n_{\infty \omega}$.
\end{enumarab}
\end{theorem}
\begin{proof} \cite{w} fact 4.5, corollary 4.6
\end{proof}
\begin{definition} Let $L^+$ be the signature consisting of the binary
relation symbols $(a, i)$, for each $a \in \G \cup \{ \rho \}$ and
$ i < n$. Let $L = L^+ \setminus \{ (\rho, i) : i < n \}$. From now
on, the logics $L^n, L^n_{\infty \omega}$ are taken in this
signature.
\end{definition}

We may regard any non-empty labelled graph equally as an
$L^+$-structure, in the obvious way. The $n$-homogeneity built into
$M$ by its construction would suggest that the set of all partial
isomorphisms of $M$ of cardinality at most $n$ forms an
$n$-back-and-forth system. This is indeed true, but we can go
further.

\begin{definition}
Let $\chi$ be a permutation of the set $\omega \cup \{ \rho\}$. Let
$ \Gamma, \triangle \in \GG$ have the same size, and let $ \theta :
\Gamma \rightarrow \triangle$ be a bijection. We say that $\theta$
is a $\chi$-\textit{isomorphism} from $\Gamma$ to $\triangle$ if for
each distinct $ x, y \in \Gamma$,
\begin{itemize}
\item If $\Gamma ( x, y) = (a, j)$ with $a\in \N$, then there exist unique $l\in \N$ and $r$ with $0\leq r<N$ such that
$a=Nl+r$. 
\begin{equation*}
\triangle( \theta(x),\theta(y)) =
\begin{cases}
( N\chi(i)+r, j), & \hbox{if $\chi(i) \neq \rho$} \\
(\rho, j),  & \hbox{otherwise.} \end{cases}
\end{equation*}
\end{itemize}

\begin{itemize}
\item If $\Gamma ( x, y) = ( \rho, j)$, then
\begin{equation*}
\triangle( \theta(x),\theta(y)) \in
\begin{cases}
\{( N\chi(\rho)+s, j): 0\leq s < N \}, & \hbox{if $\chi(\rho) \neq \rho$} \\
\{(\rho, j)\},  & \hbox{otherwise.} \end{cases}
\end{equation*}
\end{itemize}
\end{definition}

\begin{definition}
For any permutation $\chi$ of $\omega \cup \{\rho\}$, $\Theta^\chi$
is the set of partial one-to-one maps from $M$ to $M$ of size at
most $n$ that are $\chi$-isomorphisms on their domains. We write
$\Theta$ for $\Theta^{Id_{\omega \cup \{\rho\}}}$.
\end{definition}


\begin{lemma}
For any permutation $\chi$ of $\omega \cup \{\rho\}$, $\Theta^\chi$
is an $n$-back-and-forth system on $M$.
\end{lemma}
\begin{demo}{Proof}
The proof is similar to theorem 4.11 in \cite{w}; however, the graph used is different, so we include the proof.
Clearly, $\Theta^\chi$ is closed under restrictions. We check the
``forth" property. Let $\theta \in \Theta^\chi$ have size $t < n$.
Enumerate $ dom(\theta)$, $rng(\theta)$ respectively as $ \{ a_0,
\ldots, a_{t-1} \}$, $ \{ b_0,\ldots b_{t-1} \}$, with $\theta(a_i)
= b_i$ for $i < t$. Let $a_t \in M$ be arbitrary, let $b_t \notin M$
be a new element, and define a complete labelled graph $\triangle
\supseteq M \upharpoonright \{ b_0,\ldots, b_{t-1} \}$ with nodes
$\{ b_0,\ldots, b_{t} \}$ as follows.\\

Choose distinct "nodes"$ e_s < N$ for each $s < t$, such that no
$(e_s, j)$ labels any edge in $M \upharpoonright \{ b_0,\dots,
b_{t-1} \}$. This is possible because $N \geq n(n-1)/2$, which
bounds the number of edges in $\triangle$. We can now define the
colour of edges $(b_s, b_t)$ of $\triangle$ for $s = 0,\ldots, t-1$.

\begin{itemize}
\item If $M ( a_s, a_t) = ( Ni+r, j)$, for some $i\in \N$ and $0\leq r<N$, then
\begin{equation*}
\triangle( b_s, b_t) =
\begin{cases}
( N\chi(i)+r, j), & \hbox{if $\chi(i) \neq \rho$} \\
\{(\rho, j)\},  & \hbox{otherwise.} \end{cases}
\end{equation*}
\end{itemize}


\begin{itemize}
\item If $M ( a_s, a_t) = ( \rho, j)$, then assuming that $e_s=Ni+r,$ $i\in \N$ and $0\leq r<N$,
\begin{equation*}
\triangle( b_s, b_t) =
\begin{cases}
( N\chi(\rho)+r, j), & \hbox{if $\chi(\rho) \neq \rho$} \\
\{(\rho, j)\},  & \hbox{otherwise.} \end{cases}
\end{equation*}
\end{itemize}

This completes the definition of $\triangle$. It is easy to check 
that $\triangle \in \GG$. Hence, there is a graph embedding $ \phi : \triangle \rightarrow M$
extending the map $ Id_{\{ b_0,\ldots b_{t-1} \}}$. Note that
$\phi(b_t) \notin rng(\theta)$. So the map $\theta^+ = \theta \cup
\{(a_t, \phi(b_t))\}$ is injective, and it is easily seen to be a
$\chi$-isomorphism in $\Theta^\chi$ and defined on $a_t$.
The converse,``back" property is similarly proved ( or by symmetry,
using the fact that the inverse of maps in $\Theta$ are
$\chi^{-1}$-isomorphisms).
\end{demo}

But we can also derive a connection between classical and
relativised semantics in $M$, over the following set $W$:\\

\begin{definition}
Let $W = \{ \bar{a} \in {}^n M : M \models ( \bigwedge_{i < j < n,
l < n} \neg (\rho, l)(x_i, x_j))(\bar{a}) \}.$
\end{definition}
\begin{theorem}
$M \models_W \varphi(\bar{a})$ iff $M \models \varphi(\bar{a})$, for
all $\bar{a} \in W$ and all $L^n$-formulas $\varphi$.
\end{theorem}
\begin{demo} {Proof} \cite{w}
\end{demo}

We can now extract form the labelled graph $M$ of Theorem  a
relativised set algebra $\A$, which will turn out to be
representable atomic polyadic algebra.
\begin{definition}
\begin{enumerate}
\item For an $L^n_{\infty \omega}$-formula $\varphi $, we define
$\varphi^W$ to be the set $\{ \bar{a} \in W : M \models_W \varphi
(\bar{a}) \}$. 

\item We define $\A$ to be the relativised set algebra with domain
$$\{\varphi^W : \varphi \,\ \textrm {a first-order} \;\ L^n-
\textrm{formula} \}$$  and unit $W$, endowed with the algebraic
operations ${\sf d}_{ij}, {\sf c}_i, $ ect., in the standard way .

\end{enumerate}

Note that $\A$ is indeed closed under the operations and so is a
bona fide relativised set algebra. For, reading off from the
definitions of the standard operations and the relativised
semantics, we see that for all $L^n$-formulas $\varphi,\psi,$
\begin{itemize}
\item $-^{\A}(\varphi^W)=(\neg\varphi)^W$
\item $\varphi^W \cdot ^{\A} \psi^W = (\varphi \wedge \psi)^W$
\item ${\sf d}^{\A}_{ij}=(x_i = x_j)^W \,\ \textrm{for all} \;\ i,j<n.$
\item ${\sf c}_i^{\A}(\varphi^W)=(\exists x_i \varphi)^W \,\ \textrm{for all} \,\ i <
n.$\\ 
For a formula $\phi$ and $i,j<n$ , $\phi[x_i,x_j]$ stands for the formula obtained from $\phi$ by interchanging the free occurences of $x_i$ 
and $x_j$. Then we have:
\item ${\sf p}_{ij}^{\A}(\varphi^W)=\varphi[x_i,x_j]^W$
\end{itemize}

\end{definition}

\begin{theorem}
$\A$ is a representable (countable) atomic polyadic algebra
\end{theorem}
\begin{demo}{Proof}
\cite{w} prop. 5.2
\end{demo}

Define $\cal C$ to be the complex algebra over $At\A$, the atom structure of $\A$.
Then $\cal C$ is the completion of $\A$. The domain of $\cal C$ is $\wp(At\A)$. The diagonal ${\sf d}_{ij}$ is interpreted as the set of all $S\in At\A$ with $a_i=a_j$ for some $\bar{a}\in S$.
The cylindrification ${\sf c}_i$ is interpreted by ${\sf c}_iX=\{S\in At\A: S\subseteq c_i^{\A}(S')\text { for some } S'\in X\}$, for $X\subseteq At\A$.
Finally ${\sf p}_{ij}X=\{S\in At\A: S\subseteq {\sf p}_{ij}^{\A}(S')\text { for some } S'\in X\}.$
Let $\cal D$ be the relativized set algebra with domain $\{\phi^W: \phi\text { an $L_{\infty\omega}^n$ formula }\}$,  unit $W$
and operations defined like those of $\cal A$.
Then it is not hard to show that ${\cal C}\cong \cal D$, via the map $X\mapsto \bigcup X$.

We shall use the following form of Ramsey's theorem:
If $n<\omega$, $S$ is a finite set, $f:[\omega]^n\to S$ is a map, then there exists an infinite $H\subseteq \omega$ such that 
$f\upharpoonright [H]^n$ is constant. Here $[X]^n$ denotes the set of all subsets of $X$ of size $n$.

\begin{theorem} 
$\Rd_{ca}{\cal C}\notin S\Nr_n\CA_{n+2}.$ In particular, $\Rd_{ca}\cal C$ is not representable.
\end{theorem} 

\begin{demo}{Proof} We define a relation algebra atom structure $\alpha(\G)$ of the form
$(\{1'\}\cup (\G\times n), R_{1'}, \breve{R}, R_;)$.
The only identity atom is $1'$. All atoms are self converse, 
so $\breve{R}=\{(a, a): a \text { an atom }\}.$
The colour of an atom $(a,i)\in \G\times n$ is $i$. The identity $1'$ has no colour. A triple $(a,b,c)$ 
of atoms in $\alpha(\G)$ is consistent if
$R;(a,b,c)$ holds. Then the consistent triples are $(a,b,c)$ where

\begin{itemize}

\item one of $a,b,c$ is $1'$ and the other two are equal, or

\item none of $a,b,c$ is $1'$ and they do not all have the same colour, or

\item $a=(a', i), b=(b', i)$ and $c=(c', i)$ for some $i<n$ and 
$a',b',c'\in \G$, and there exists at least one graph edge
of $G$ in $\{a', b', c'\}$.

\end{itemize}
$\alpha(\G)$ can be checked to be a relation atom structure. 
The atom structure of $\Rd_{ca}\A$ is isomorphic (as a cylindric algebra
atom structure) to the atom structure ${\cal M}_n$ of all $n$-dimensional basic
matrices over the relation algebra atom structure $\alpha(\G)$.
Indeed, for each  $m  \in {\cal M}_n, \,\ \textrm{let} \,\ \alpha_m
= \bigwedge_{i,j<n}  \alpha_{ij}. $ Here $ \alpha_{ij}$ is $x_i =
x_j$ if $ m_{ij} = 1$' and $R(x_i, x_j)$ otherwise, where $R =
m_{ij} \in L$. Then the map $(m \mapsto
\alpha^W_m)_{m \in {\cal M}_n}$ is a well - defined isomorphism of
$n$-dimensional cylindric algebra atom structures.
We can show that thae $\Cm\alpha(\G)$ is not representable like exactly in \cite{w} and Hodkinson using Ramseys theore.
Here we show something stronger.


We shall first show that $\Cm\alpha(\G)$ is not in $S\Ra \CA_{n+2}$. The idea is to use relativized representations. Such algebras are 
localy representable, but the epresentation is global enough so that Ramseys theorem applies. 
Hence the full complex cylindric algebra over the set of $n$ by $n$ basic matrices
- which is isomorphic to $\cal C$ is not in $S\Nr_n\CA_{n+2}$ for we have a relation algebra
embedding of $\Cm\alpha(\G)$ onto $\Ra\Cm{\cal M}_n$.
Assume for contradiction that $\Cm\alpha(\G)\in S\Ra \CA_{n+2}$. Then $\Cm\alpha(\G)$ has an $n$-flat representation $M$ \cite{HH}  13.46, 
which is $n$ square \cite{HH} 13.10. 
In particular, there is a set $M$, $V\subseteq M\times M$ and $h: \Cm\alpha(\G)\to \wp(V)$ 
such that $h(a)$ ($a\in \Cm\alpha(\G)$) is a binary relation on $M$, and
$h$ respects the relation algebra operations. Here $V=\{(x,y)\in M\times M: M\models 1(x,y)\}$, where $1$ is the greatest element of 
$\Cm\alpha(\G)$.
A clique $C$ of $M$ is a subset of the domain $M$ such that for $x,y\in C$ we have $M\models 1(x,y)$, equivalently $(x,y)\in V$.
Since $M$ is $n+2$ square, then for all cliques $C$ of $M$ with $|C|<n+2$, all $x,y\in C$ and $a,b\in \Cm\alpha(\G)$, $M\models (a;b)(x,y)$ 
there exists $z\in M$ such that $C\cup \{z\}$ is a clique and $M\models a(x,z)\land b(z,y)$.
For $Y\subseteq \N$ and $s<n$, set 
$$[Y,s]=\{(l,s): l\in Y\}.$$
For $r\in \{0, \ldots N-1\},$ $N\N+r$ denotes the set $\{Nq+r: q\in \N\}.$
Let $$J=\{1', [N\N+r, s]: r<N,  s<n\}.$$
Then $\sum J=1$ in $\Cm\alpha(\G).$
As $J$ is finite, we have for any $x,y\in M$ there is a $P\in J$ with
$(x,y)\in h(P)$.
Since $\Cm\alpha(\G)$ is infinite then $M$ is infinite. 
By Ramsey's Theorem, there are distinct
$x_i\in X$ $(i<\omega)$, $J\subseteq \omega\times \omega$ infinite
and $P\in J$ such that $(x_i, x_j)\in h(P)$ for $(i, j)\in J$, $i\neq j$. Then $P\neq 1'$. 
Also $(P;P)\cdot P\neq 0$. 
This follows from $n+2$ squareness and that if $x,y, z\in M$, 
$a,b,c\in \Cm\alpha(\G)$, $(x,y)\in h(a)$, $(y, z)\in h(b)$, and 
$(x, z)\in h(c)$, then $(a;b)\cdot c\neq 0$. 
A non -zero element $a$ of $\Cm\alpha(\G)$ is monochromatic, if $a\leq 1'$,
or $a\leq [\N,s]$ for some $s<n$. 
Now  $P$ is monochromatic, it follows from the definition of $\alpha$ that
$(P;P)\cdot P=0$. This contradiction shows that 
$\Cm\alpha(\G)$ is not in $S\Ra\CA_{n+2}$. Hence $\Cm{\cal M}_n\notin S\Nr_n\CA_{n+2}$.
\end{demo}

\section{Strongly representable atom structures}


From now on we follow closely \cite{strong}, generalizing this result to other algebras, starting from diagonal free to quasipolyadic equality algebras. 
In \cite{strong} definition 3.5, 
the authors define a cylindric atom structure based on a graph $\Gamma$.
We enrich this atom structure by the relations corresponding to the polyadic operations:

\begin{definition} We define an atom structure $\eta(\Gamma)=(H, D_{ij}, \equiv_i, P_{ij})$
as follows.
\begin{enumarab}
\item $H$ is the set of all pairs $(K, \sim)$ where $K:n\to \Gamma\times n$ is a partial map and $\sim$ is an equivalence relation
on $n$ satisfying the followng conditions

(a) If $|n/\sim|=n$ then $dom(K)=n$ and $rng(K)$ is not  independent subset of $n$. 

(b) If $|n/\sim|=n-1$, then $K$ is defined only on the unique $\sim$ class $\{i,j\}$ say of size $2$
and $K(i)=K(j)$

(c) If $|n/\sim|\leq n-2$, then $K$ is nowhere defined.

\item $D_{ij}=\{(K,\sim)\in H: i\sim j\}$

\item $(K,\sim)\equiv_i (K', \sim')$ iff $K(i)=K'(i)$ and $\sim\upharpoonright (n\setminus \{i\})=\sim'\upharpoonright (n\setminus \{i\})$

\item $(K.\sim)\equiv_{ij}(K',\sim')$ iff $K(i)=K'(j)$ and $K(j)=K'(i),$ $K\upharpoonright n\sim \{i,j\}=K'\upharpoonright n\sim \{i,j\}$ 
and if $i\sim j$ then $\sim=\sim'$, if not, then $\sim'$ is related to $\sim$ as follows
For all $k\notin [i]_{\sim}\cup [j]_{\sim}$ $[k]_{\sim'}=[k]_{\sim}$ $[i]_{\sim'}=[j]_{\sim}\setminus \{j\}\cup \{i\}$ and 
$[j]_{\sim'}=[i]_{\sim}\setminus \{i\}\cup \{j\}.$ 
\end{enumarab}
\end{definition}

\begin{definition} Let $\C(\Gamma)$ be the complex algebra of polyadic type of the above atom structure.
That is $\C(\Gamma)=(\B(\eta(\Gamma)), {\sf c}_i, {\sf s}_i^j, {\sf p}_{ij}, {\sf d}_{ij})_{i,j<n}$ with extra non-Boolean operations
defined by: 
$${\sf d}_{ij}=D_{ij}$$
$${\sf c}_iX=\{c: \exists a\in X, a\equiv_ic\}.$$
$${\sf p}_{ij}X=\{c:\exists a\in X, a\equiv_{ij}c\}$$
and $${\sf s}_i^jx={\sf c}_j(x\cap d_{ij}).$$
\end{definition}
For $\A\in \CA_n$ and $x\in A$, recall that $\Delta x,$ the dimension set of $x$, is the set $\{i\in n: {\sf c}_ix\neq x\}$.

\begin{theorem} For any graph $\Gamma$, $\C(\Gamma)$ is a simple $\PEA_n$, that is generated by the set
$\{x\in C: \Delta x\neq n\}$.
\end{theorem}
\begin{demo}{Proof} $\Rd_{ca}\C(\Gamma)$ is a simple $\CA_n$ by \cite{h} lemma 5.1, 
hence if we prove that $\C(\Gamma)$ is a polyadic equality algebra, then as a polyadic equality algebra it will be simple. 
This follows from the simple observation that any polyadic ideal in $\C(\Gamma)$ is a cylindric ideal.
Furthermore for any atom $x=\{(K, \sim)\}$ of $\C$ we have $x={\sf c}_0x\cap {\sf c}_1x\ldots\cap {\sf c}_nx$.
We need to check the polyadic axioms. 
Since $\Rd_{ca}\A$ is a $\CA_n$ we need to show that 
the following hold for all $i,j,k\in n$:
\begin{enumarab}
\item ${\sf p}_{ij}$'s  are boolean endomorphisms
\item ${\sf p}_{ij}{\sf p}_{ij}x=x$
\item ${\sf p}_{ij}{\sf p}_{ik}={\sf p}_{jk}{\sf p}_{ij}x   \text { if } |\{i,j,k\}|=3$
\item ${\sf p}_{ij}{\sf s}_i^jx={\sf s}_j^ix$
\end{enumarab}
These properties, follow from the definitions, and are therefore left to the reader.
\end{demo}
Let $\Gamma=(G,E)$ be a graph. Then A set $X\subseteq G$ is independent if $E\cap (X\times X)=\emptyset$.
The chromatic number $\chi(\Gamma)$ of $\Gamma$ is the least  $k<\omega$ such that $G$ can be partitioned into $k$ independent sets, and
$\infty$ if there is no such set.

\begin{theorem} 
\begin{enumroman}
\item Suppose that $\chi(\Gamma)=\infty$. Then $\C(\Gamma)$ is representable  as a polyadic equality algebra.
\item If $\Gamma$ is infinite and $\chi(\Gamma)<\infty$ then $\Rd_{df}\C(\Gamma)$ is not representable.
\end{enumroman}
\end{theorem}
\begin{demo}{Proof} (i) We have $\Rd_{ca}\C(\Gamma)$ is representable. 
Let $J=\{x\in C: \Delta x\neq n\}$. Then $\C(\Gamma)$ is generated from $J$ using infinite intersections and complementation.
Let $f$ be an isomorphism of of $\Rd_{ca}\C(\Gamma)$ onto a cylindric set algebra with base $U$. Since the ${\sf p}_{ij}$'s  
distribute over arbitrary (unions and) intersections, 
it suffices to  show that $f{\sf p}_{kl}x={\sf p}_{kl}fx$ for all $x\in J$. Let $\mu\in n\setminus \Delta x$.
If $k=\mu$ or $l=\mu$, say $k=\mu$, then using the polyadic axioms we have
$$f{\sf p}_{kl}x=f{\sf }{\sf p}_{kl}{\sf c}_kx=f{\sf s}_l^kx={\sf s}_l^kfx={\sf s}_l^k{\sf c}_kfx= {\sf p}_{kl}fx.$$ 
If $\mu\neq k,l$ then again using the polyadic axioms we get
$$f{\sf p}_{kl}x=f{\sf s}_{\mu}^l{\sf s}_l^k {\sf s}_k^{\mu}{\sf c}_{\mu}x=
{\sf s}_{\mu}^l{\sf s}_l^k {\sf s}_k^{\mu}{\sf c}_{\mu}fx={\sf p}_{kl}f(x)$$

(ii) Note that $\Rd_{ca}\C(\Gamma)$ is generated by $\{x\in C: \Delta x\neq n\}$ using infinite intersections and complementation.
\end{demo}
Recall that an atom structure is strongly representable if the complex algebra over this atom structure is representable \cite{h}.
We now have:
\begin{theorem}\label{st} Let $t$ be any signature between 
$\Df_n$ and $\PEA_n$. Then the class of strongly representable atom structures of type $t$
is not elementary.
\end{theorem}
\begin{demo}{Proof} \cite{strong} theorem 6.1. 
By a famous theorem of Erdos, for every $k<\omega$, there is a finite graph $G_k$ with $\chi(G_k)>k$ 
and with no cycles of length $<k$.
Let $\Gamma_k$ be the disjoint union of of the $G_l$ for $l>k$. Then $\chi(\Gamma_k)=\infty$. Thus, by the previous theorem 
$\C(\Gamma_k)\in {\bf RPEA_n}$. In fact, being simple, $\C(\Gamma_k)$ is actually a  polyadic set algebra.
Let $\Gamma$ be a non principal ultraproduct $\prod_D \Gamma_k$. So $\Gamma$ has no cycles, and so $\chi(\Gamma)\leq 2$.
It follows, again from the previous theorem,  that $\Rd_{df}\C(\Gamma)$ is not representable.
From $\prod_D\C(\Gamma_k)\cong \C(\prod_D\Gamma_k)$ we are done.
\end{demo}
\begin{theorem} For any such case, the classes of strongly representable and completely representable 
atom structures are not elementary
\end{theorem} 


\section{Neat reducts, complete representations and games}

Next we characterize the class $\Nr_n\CA_{\omega}$ using games.
Our treatment in this part 
follows very closely \cite{R}. The essential difference is that we deal with $n$ 
dimensional networks and composition moves are replaced by cylindrifier moves in the games.

\begin{definition}\label{def:string} 
Let $n$ be an ordinal. An $s$ word is a finite string of substitutions $({\sf s}_i^j)$, 
a $c$ word is a finite string of cylindrifications $({\sf c}_k)$.
An $sc$ word is a finite string of substitutions and cylindrifications
Any $sc$ word $w$ induces a partial map $\hat{w}:n\to n$
by
\begin{itemize}

\item $\hat{\epsilon}=Id$

\item $\widehat{w_j^i}=\hat{w}\circ [i|j]$

\item $\widehat{w{\sf c}_i}= \hat{w}\upharpoonright(n\sim \{i\}$ 

\end{itemize}
\end{definition}
If $\bar a\in {}^{<n-1}n$, we write ${\sf s}_{\bar a}$, or more frequently 
${\sf s}_{a_0\ldots a_{k-1}}$, where $k=|\bar a|$,
for an an arbitary chosen $sc$ word $w$
such that $\hat{w}=\bar a.$ 
$w$  exists and does not 
depend on $w$ by \cite[definition~5.23 ~lemma 13.29]{HHbook}. 
We can, and will assume \cite[Lemma 13.29]{HHbook} 
that $w=s{\sf c}_{n-1}{\sf c}_n.$
%Let $n\geq 3$ be an ordinal and $i, j<n$. We define a string of
%substitutions $s_{i_0i_1\ldots i_{\mu-1}}$ 
%that `move dimensions $0, 1,\ldots \mu-1$ to $i_0,i_1\ldots i_{\mu-1}$
%asfollows.
%\[s_{ij}=\left\{\begin{array}{ll}
%s^0_is^1_j &\mbox{if }j\neq 0\\
%s^1_0s^0_i&\mbox{if }j=0, \; i\neq 1\\
%s^2_0s^0_1s^1_2&\mbox{if } j=0,\;i=1
%\end{array}\right.
%\]
[In the notation of \cite[definition~5.23,~lemma~13.29]{HHbook}, 
$\widehat{s_{ijk}}$ for example is the function $n\to n$ taking $0$ to $i,$
$1$ to $j$ and $2$ to $k$, and fixing all $l\in n\setminus\set{i, j,k}$.]
Let $\delta$ be a map. Then $\delta[i\to d]$ is defined as follows. $\delta[i\to d](x)=\delta(x)$
if $x\neq i$ and $\delta[i\to d](i)=d$. We write $\delta_i^j$ for $\delta[i\to \delta_j]$.


\begin{definition}
From now on let $2\leq n<\omega.$ Let $\C$ be an atomic $\CA_{n}$. 
An \emph{atomic  network} over $\C$ is a map
$$N: {}^{n}\Delta\to At\cal C$$ 
such that the following hold for each $i,j<n$, $\delta\in {}^{n}\Delta$
and $d\in \Delta$:
\begin{itemize}
\item $N(\delta^i_j)\leq {\sf d}_{ij}$
\item $N(\delta[i\to d])\leq {\sf c}_iN(\delta)$ 
\end{itemize}
\end{definition}
Note than $N$ can be viewed as a hypergraph with set of nodes $\Delta$ and 
each hyperedge in ${}^{\mu}\Delta$ is labelled with an atom from $\C$.
We call such hyperedges atomic hyperedges.
We write $\nodes(N)$ for $\Delta.$ But it can happen 
let $N$ stand for the set of nodes 
as well as for the function and the network itself. Context will help.

Define $x\sim y$ if there exists $\bar{z}$ such that $N(x,y,\bar{z})\leq {\sf d}_{01}$.
Define an equivalence relation
$\sim$ over the set of all finite sequences over $\nodes(N)$ by $\bar
x\sim\bar y$ iff $|\bar x|=|\bar y|$ and $x_i\sim y_i$ for all
$i<|\bar x|$.

(3) A \emph{ hypernetwork} $N=(N^a, N^h)$ over $\cal C$ 
consists of a network $N^a$
together with a labelling function for hyperlabels $N^h:\;\;^{<
\omega}\!\nodes(N)\to\Lambda$ (some arbitrary set of hyperlabels $\Lambda$)
such that for $\bar x, \bar y\in\; ^{< \omega}\!\nodes(N)$ 
\begin{enumerate}
\renewcommand{\theenumi}{\Roman{enumi}}
\setcounter{enumi}3%{\ref{net:third}}
\item\label{net:hyper} $\bar x\sim\bar y \Rightarrow N^h(\bar x)=N^h(\bar y)$. 
\end{enumerate}
If $|\bar x|=k\in nats$ and $N^h(\bar x)=\lambda$ then we say that $\lambda$ is
a $k$-ary hyperlabel. $(\bar x)$ is referred to a a $k$-ary hyperedge, or simply a hyperedge.
(Note that we have atomic hyperedges and hyperedges) 
%[It is possible, though not very desirable, that
%$\lambda$ could be a $k$-ary hyperlabel and an $l$-ary hyperlabel for
%$k\neq l$.]
When there is no risk of ambiguity we may drop the superscripts $a,
h$. 

The following notation is defined for hypernetworks, but applies
equally to networks.  

(4) If $N$ is a hypernetwork and $S$ is any set then
$N\restr S$ is the $n$-dimensional hypernetwork defined by restricting
$N$ to the set of nodes $S\cap\nodes(N)$.  For hypernetworks $M, N$ if
there is a set $S$ such that $M=N\restr S$ then we write $M\subseteq
N$.  If $N_0\subseteq N_1\subseteq \ldots $ is a nested sequence of
hypernetworks then we let the \emph{limit} $N=\bigcup_{i<\omega}N_i$  be
the hypernetwork defined by
$\nodes(N)=\bigcup_{i<\omega}\nodes(N_i)$,\/ $N^a(x_0,\ldots x_{n-1})= 
N_i^a(x_0,\ldots x_{n-1})$ if
$x_0\ldots x_{\mu-1}\in\nodes(N_i)$, and $N^h(\bar x)=N_i^h(\bar x)$ if $\rng(\bar
x)\subseteq\nodes(N_i)$.  This is well-defined since the hypernetworks
are nested and since hyperedges $\bar x\in\;^{<\omega}\nodes(N)$ are
only finitely long.  


For hypernetworks $M, N$ and any set $S$, we write $M\equiv^SN$
if $N\restr S=M\restr S$.  For hypernetworks $M, N$, 
and any set $S$, we write $M\equiv_SN$ 
if the symmetric difference $\Delta(\nodes(M), \nodes(N))\subseteq S$ and
$M\equiv^{(\nodes(M)\cup\nodes(N))\setminus S}N$. We write $M\equiv_kN$ for
$M\equiv_{\set k}N$.


Let $N$ be a network and let $\theta$ be any function.  The network
$N\theta$ is a complete labelled graph with nodes
$\theta^{-1}(\nodes(N))=\set{x\in\dom(\theta):\theta(x)\in\nodes(N)}$,
and labelling defined by 
$(N\theta)(i_0,\ldots i_{\mu-1}) = N(\theta(i_0), \theta(i_1), \theta(i_{\mu-1}))$,
for $i_0, \ldots i_{\mu-1}\in\theta^{-1}(\nodes(N))$.  Similarly, for a hypernetwork
$N=(N^a, N^h)$, we define $N\theta$ to be the hypernetwork
$(N^a\theta, N^h\theta)$ with hyperlabelling defined by
$N^h\theta(x_0, x_1, \ldots) = N^h(\theta(x_0), \theta(x_1), \ldots)$
for $(x_0, x_1,\ldots) \in \;^{<\omega}\!\theta^{-1}(\nodes(N))$.

Let $M, N$ be hypernetworks.  A \emph{partial isomorphism}
$\theta:M\to N$ is a partial map $\theta:\nodes(M)\to\nodes(N)$ such
that for any $
i_i\ldots i_{\mu-1}\in\dom(\theta)\subseteq\nodes(M)$ we have $M^a(i_1,\ldots i_{\mu-1})= 
N^a(\theta(i), \ldots\theta(i_{\mu-1}))$
and for any finite sequence $\bar x\in\;^{<\omega}\!\dom(\theta)$ we
have $M^h(\bar x) = 
N^h\theta(\bar x)$.  
If $M=N$ we may call $\theta$ a partial isomorphism of $N$.

 
\begin{definition}\label{def:games} Let $2\leq n<\omega$. For any $\CA_{n}$  
atom structure $\alpha$, and $n\leq m\leq
\omega$, we define two-player games $F_{n}^m(\alpha),$ \; and 
$H_{n}(\alpha)$,
each with $\omega$ rounds, 
and for $m<\omega$ we define $H_{m,n}(\alpha)$ with $n$ rounds.

\begin{itemize}
\item 
Let $m\leq \omega$.  
In a play of $F_{n}^m(\alpha)$ the two players construct a sequence of
networks $N_0, N_1,\ldots$ where $\nodes(N_i)$ is a finite subset of
$m=\set{j:j<m}$, for each $i$.  In the initial round of this game \pa\
picks any atom $a\in\alpha$ and \pe\ must play a finite network $N_0$ with
$\nodes(N_0)\subseteq  n$, 
such that $N_0(\bar{d}) = a$ 
for some $\bar{d}\in{}^{\mu}\nodes(N_0)$.
In a subsequent round of a play of $F_{n}^m(\alpha)$ \pa\ can pick a
previously played network $N$ an index $\l<n$, a ``face" 
$F=\langle f_0,\ldots f_{n-2} \rangle \in{}^{n-2}\nodes(N),\; k\in
m\setminus\set{f_0,\ldots f_{n-2}}$, and an atom $b\in\alpha$ such that 
$b\leq {\sf c}_lN(f_0,\ldots f_i, x,\ldots f_{n-2}).$  
(the choice of $x$ here is arbitrary, 
as the second part of the definition of an atomic network together with the fact
that $\cyl i(\cyl i x)=\cyl ix$ ensures that the right hand side does not depend on $x$).
This move is called a \emph{cylindrifier move} and is denoted
$(N, \langle f_0, \ldots f_{\mu-2}\rangle, k, b, l)$ or simply $(N, F,k, b, l)$.
In order to make a legal response, \pe\ must play a
network $M\supseteq N$ such that 
$M(f_0,\ldots f_{i-1}, k, f_i,\ldots f_{n-2}))=b$ 
and $\nodes(M)=\nodes(N)\cup\set k$.

\pe\ wins $F_{n}^m(\alpha)$ if she responds with a legal move in each of the
$\omega$ rounds.  If she fails to make a legal response in any
round then \pa\ wins.

 
\item
Fix some hyperlabel $\lambda_0$.  $H_{n}(\alpha)$ is  a 
game the play of which consists of a sequence of
$\lambda_0$-neat hypernetworks 
$N_0, N_1,\ldots$ where $\nodes(N_i)$
is a finite subset of $\omega$, for each $i<\omega$.  
%The other main
%difference is that \pa\ can play a more general kind of amalgamation
%move.  
In the initial round \pa\ picks $a\in\alpha$ and \pe\ must play
a $\lambda_0$-neat hypernetwork $N_0$ with nodes contained in
$\mu$ and $N_0(\bar d)=a$ for some nodes $\bar{d}\in {}^{\mu}N_0$.  
At a later stage
\pa\ can make any cylindrifier move $(N, F,k, b, l)$ by picking a
previously played hypernetwork $N$ and $F\in {}^{n-2}\nodes(N), \;l<n,  
k\in\omega\setminus\nodes(N)$ 
and $b\leq {\sf c}_lN(f_0, f_{l-1}, x, f_{n-2})$.  
[In $H_{n}$ we
require that \pa\ chooses $k$ as a `new node', i.e. not in
$\nodes(N)$, whereas in $F_{n}^m$ for finite $m$ it was necessary to allow
\pa\ to `reuse old nodes'. This makes the game easior as far as $\forall$ is concerned.) 
For a legal response, \pe\ must play a
$\lambda_0$-neat hypernetwork $M\equiv_k N$ where
$\nodes(M)=\nodes(N)\cup\set k$ and 
$M(f_0, f_{i-1}, k, f_{n-2})=b$.
Alternatively, \pa\ can play a \emph{transformation move} by picking a
previously played hypernetwork $N$ and a partial, finite surjection
$\theta:\omega\to\nodes(N)$, this move is denoted $(N, \theta)$.  \pe\
must respond with $N\theta$.  Finally, \pa\ can play an
\emph{amalgamation move} by picking previously played hypernetworks
$M, N$ such that $M\equiv^{\nodes(M)\cap\nodes(N)}N$ and
$\nodes(M)\cap\nodes(N)\neq \emptyset$.  
This move is denoted $(M,
N)$.  To make a legal response, \pe\ must play a $\lambda_0$-neat
hypernetwork $L$ extending $M$ and $N$, where
$\nodes(L)=\nodes(M)\cup\nodes(N)$.

Again, \pe\ wins $H_n(\alpha)$ if she responds legally in each of the
$\omega$ rounds, otherwise \pa\ wins. 

\item For $m< \omega$ the game $H_{m,n}(\alpha)$ is similar to $H_n(\alpha)$ but
play ends after $m$ rounds, so a play of $H_{m,n}(\alpha)$ could be
\[N_0, N_1, \ldots, N_m\]
If \pe\ responds legally in each of these
$m$ rounds she wins, otherwise \pa\ wins.
\end{itemize}

\end{definition}

\begin{definition}\label{def:hat}
For $m\geq 5$ and $\c C\in\CA_m$, if $\A\subseteq\Nr_n(\C)$ is an
atomic cylindric algebra and $N$ is an $\A$-network then we define
$\widehat N\in\C$ by
\[\widehat N =%\prod_{i\leq j\in\nodes(N)}s^0_is^1_jN(i, j)=
 \prod_{i_0,\ldots i_{n-1}\in\nodes(N)}{\sf s}_{i_0, \ldots i_{n-1}}N(i_0\ldots i_{n-1})\]
$\widehat N\in\C$ depends
implicitly on $\C$.
\end{definition}
We write $\A\subseteq_c \B$ if $\A\in S_c\{\B\}$. 
\begin{lemma}\label{lem:atoms2}
Let $n<m$ and let $\A$ be an atomic $\CA_n$, 
$\A\subseteq_c\Nr_n\C$
for some $\C\in\CA_m$.  For all $x\in\C\setminus\set0$ and all $i_0, \ldots i_{n-1} < m$ there is $a\in\At(\A)$ such that
${\sf s}_{i_0\ldots i_{n-1}}a\;.\; x\neq 0$.
\end{lemma}
\begin{proof}
We can assume, see definition  \ref{def:string}, 
that ${\sf s}_{i_0,\ldots i_{n-1}}$ consists only of substitutions, since ${\sf c}_{m}\ldots {\sf c}_{m-1}\ldots 
{\sf c}_nx=x$ 
for every $x\in \A$.We have ${\sf s}^i_j$ is a
completely additive operator (any $i, j$), hence ${\sf s}_{i_0,\ldots i_{\mu-1}}$ 
is too  (see definition~\ref{def:string}).
So $\sum\set{{\sf s}_{i_0\ldots i_{n-1}}a:a\in\At(\A)}={\sf s}_{i_0\ldots i_{n-1}}
\sum\At(\A)={\sf s}_{i_0\ldots i_{n-1}}1=1$,
for any $i_0,\ldots i_{n-1}<n$.  Let $x\in\C\setminus\set0$.  It is impossible
that ${\sf s}_{i_0\ldots i_{n-1}}\;.\;x=0$ for all $a\in\At(\c A)$ because this would
imply that $1-x$ was an upper bound for $\set{{\sf s}_{i_0\ldots i_{n-1}}a:
a\in\At(\A)}$, contradicting $\sum\set{{\sf s}_{i_0\ldots i_{n-1}}a :a\in\At(\c A)}=1$.
\end{proof}


We now prove two Theorems relating neat embeddings
to the games we defined:

\begin{theorem}\label{thm:n}
Let $n<m$, and let $\A$ be a $\CA_m$.  
If $\A\in{\bf S_c}\Nr_{n}\CA_m, $
then \pe\ has a \ws\ in $F^m(\At\A)$. In particular if $\A$ is $CR$ then \pe has a \ws in $F^{\omega}(\At\A)$
\end{theorem}
\begin{proof}
If $\A\subseteq\Nr_n\C$ for some $\C\in\CA_m$ then \pe\ always
plays hypernetworks $N$ with $\nodes(N)\subseteq n$ such that
$\widehat N\neq 0$. In more detail, in the initial round , let $\forall$ play $a\in \At \cal A$.
$\exists$ play a network $N$ with $N(0, \ldots n-1)=a$. Then $\widehat N=a\neq 0$.
At a later stage suppose $\forall$ plays the cylindrifier move 
$(N, \langle f_0, \ldots f_{\mu-2}\rangle, k, b, l)$ 
by picking a
previously played hypernetwork $N$ and $f_i\in \nodes(N), \;l<\mu,  k\notin \{f_i: i<n-2\}$, 
and $b\leq {\sf c}_lN(f_0,\ldots  f_{i-1}, x, f_{n-2})$.
Let $\bar a=\langle f_0\ldots f_{l-1}, k\ldots f_{n-2}\rangle.$
Then ${\sf c}_k\widehat N\cdot {\sf s}_{\bar a}b\neq 0$.
Then there is a network  $M$ such that
$\widehat{M}.\widehat{{\sf c}_kN}\cdot {\sf s}_{\bar a}b\neq 0$. Hence 
$M(f_0,\dots  k, f_{n-2})=b.$
\end{proof}

\begin{theorem}\label{thm:RaC}
Let $\alpha$ be a countable 
$\CA_n$ atom structure.  If \pe\ has a \ws\ in $H_n(\alpha),$ then
there is a representable cylindric algebra $\C$ of
dimension $\omega$ such that $\Nr_n\c C$ is atomic 
and $\At \Nr_n\C\cong\alpha$.
\end{theorem}

\begin{proof} 
We shall construct a generalized atomic weak set algebra of dimension $\omega$ such that the atom 
structure of its full neat reduct is isomorphic to
the given atom structure. Suppose \pe\ has a \ws\ in $H_n(\alpha)$. Fix some $a\in\alpha$. We can define a
nested sequence $N_0\subseteq N_1\ldots$ of hypernetworks
where $N_0$ is \pe's response to the initial \pa-move $a$, requiring that
\begin{enumerate}
\item If $N_r$ is in the sequence and 
and $b\leq {\sf c}_lN_r(\langle f_0, f_{n-2}\rangle\ldots , x, f_{n-2})$.  
then there is $s\geq r$ and $d\in\nodes(N_s)$ such 
that $N_s(f_0, f_{i-1}, d, f_{n-2})=b$.
\item If $N_r$ is in the sequence and $\theta$ is any partial
isomorphism of $N_r$ then there is $s\geq r$ and a
partial isomorphism $\theta^+$ of $N_s$ extending $\theta$ such that
$\rng(\theta^+)\supseteq\nodes(N_r)$.
\end{enumerate}
%Since $\alpha$ is countable there are countably many requirements  to extend. 
%Since the sequence of networks is nested , these requirements
%to extend
We can schedule these requirements
to extend so that eventually, every requirement gets dealt with.
If we are required to find $k$ and $N_{r+1}\supset N_r$
such that 
$N_{r+1}(f_0, k, f_{n-2})=b$ then let $k\in \omega\setminus \nodes(N_r)$
where $k$ is the least possible for definiteness, 
and let $N_{r+1}$ be \pe's response using her \ws, 
to the \pa move $N_r, (f_0,\ldots f_{n-1}), k, b, l).$
For an extension of type 2, let $\tau$ be a partial isomorphism of $N_r$
and let $\theta$ be any finite surjection onto a partial isomorphism of $N_r$ such that 
$dom(\theta)\cap nodes(N_r)= dom\tau$. \pe's response to \pa's move $(N_r, \theta)$ is necessarily 
$N\theta.$ Let $N_{r+1}$ be her response, using her wining strategy, to the subsequent \pa 
move $(N_r, N_r\theta).$ 
%The amalgamation move in $H(\alpha)$ means that this can be done.  We omit
%the details. 

Now let $N_a$ be the limit of this sequence.
This limit is well-defined since the hypernetworks are nested.  

Let $\theta$ be any finite partial isomorphism of $N_a$ and let $X$ be
any finite subset of $\nodes(N_a)$.  Since $\theta, X$ are finite, there is
$i<\omega$ such that $\nodes(N_i)\supseteq X\cup\dom(\theta)$. There
is a bijection $\theta^+\supseteq\theta$ onto $\nodes(N_i)$ and $j\geq
i$ such that $N_j\supseteq N_i, N_i\theta^+$.  Then $\theta^+$ is a
partial isomorphism of $N_j$ and $\rng(\theta^+)=\nodes(N_i)\supseteq
X$.  Hence, if $\theta$ is any finite partial isomorphism of $N_a$ and
$X$ is any finite subset of $\nodes(N_a)$ then
\begin{equation}\label{eq:theta}
\exists \mbox{ a partial isomorphism $\theta^+\supseteq \theta$ of $N_a$
 where $\rng(\theta^+)\supseteq X$}
\end{equation}
and by considering its inverse we can extend a partial isomorphism so
as to include an arbitrary finite subset of $\nodes(N_a)$ within its
domain.
Let $L$ be the signature with one $\mu$ -ary predicate symbol ($b$) for
each $b\in\alpha$, and one $k$-ary predicate symbol ($\lambda$) for
each $k$-ary hyperlabel $\lambda$.  

%[Notational point: if $\lambda$ is
%$k$-ary and $l$-ary for $k\neq l$ then make one $k$-ary predicate
%symbol $\lambda$ and one $l$-ary predicate symbol $\lambda'$, so that
%every predicate symbol has a unique arity.]  We have alos equlatiy and a sequence of variable of order type $\omega$.
 
%Rename the nodes of the
%hypernetworks so that $a\neq b\in \alpha \Rightarrow
%\nodes(N_a)\cap\nodes(N_b)=\emptyset$.  For each $a\in \alpha$ pick
For fixed $f_a\in\;^\omega\!\nodes(N_a)$, let
$U_a=\set{f\in\;^\omega\!\nodes(N_a):\set{i<\omega:g(i)\neq
f_a(i)}\mbox{ is finite}}$.
Notice that $U_a$ is weak unit ( a set of sequences agrreing cofinitely with a fixed one)


% For any function
%$g\in\:^\gamma\!\nodes(N_a)$ write $g\approx f_a$ and say ``$g$ is finitely
%close to $f_a$'' if% $\rng(g)\subseteq\nodes(N_a)$ and
%$\set{i<\gamma:g(i)\neq f_a(i)}$ is finite.  
%Let
%$U=\bigcup_{a\in\alpha}\set{f\in\;^\gamma\!\nodes(N_a):\exists a\in\alpha, \; f\approx f_a}$.

We can make $U_a$ into the base of an $L$ relativized structure $\c N_a$.
Satisfiability for $L$ formluas at assignments $f\in U_a$ is defined the usual Tarskian way.

For $b\in\alpha,\;
l_0, \ldots l_{\mu-1}, i_0 \ldots, i_{k-1}<\omega$, \/ $k$-ary hyperlabels $\lambda$,
and all $L$-formulas $\phi, \psi$, let
\begin{eqnarray*}
\c N_a, f\models b(x_{l_0}\ldots  x_{n-1})&\iff&N_a(f(l_0),\ldots  f(l_{n-1}))=b\\
\c N_a, f\models\lambda(x_{i_0}, \ldots,x_{i_{k-1}})&\iff&  N_a(f(i_0), \ldots,f(i_{k-1}))=\lambda\\
\c N_a, f\models\neg\phi&\iff&\c N_a, f\not\models\phi\\
\c N_a, f\models (\phi\vee\psi)&\iff&\c N_a,  f\models\phi\mbox{ or }\c N_a, f\models\psi\\
\c N_a, f\models\exists x_i\phi&\iff& \c N_a, f[i/m]\models\phi, \mbox{ some }m\in\nodes(N_a)
\end{eqnarray*}
For any $L$-formula $\phi$, write $\phi^{\c N_a}$ for the set of assighnments satisfying it; that is
$\set{f\in\;^\omega\!\nodes(N_a): \c N_a, f\models\phi}$.  Let
$D_a = \set{\phi^{\c N_a}:\phi\mbox{ is an $L$-formula}}.$ 
Then this is the universe of the following weak set algebra 
\[\c D_a=(D_a,  \cup, \sim, {\sf D}_{ij}, {\sf C}_i)_{ i, j<\omega}\] 
then 
$\c D_a\in\RCA_\omega$. (Weak set algebras are representable).

Let $\phi(x_{i_0}, x_{i_1}, \ldots, x_{i_k})$ be an arbitrary
$L$-formula using only variables belonging to $\set{x_{i_0}, \ldots,
x_{i_k}}$.  Let $f, g\in U_a$ (some $a\in \alpha$) and suppose
is a partial isomorphism of $N_a$.  We can prove by induction over the
quantifier depth of $\phi$ and using (\ref{eq:theta}), that
\begin{equation}
\c N_a, f\models\phi\iff \c N_a,
g\models\phi\label{eq:bf}\end{equation} 

Let $\c C=\prod_{a\in
\alpha}\c \c D_a$.  Then  $\c C\in\RCA_\omega$, and $\c C$ is the desired generalized weak set algebra.
Note that unit of $\c C$ is the disjoint union of the weak spaces.
We set out to prove our claim. 
We shall show that $\alpha\cong \At\Nr_n\c C.$

An element $x$ of $\c C$ has the form
$(x_a:a\in\alpha)$, where $x_a\in\c D_a$.  For $b\in\alpha$ let
$\pi_b:\c C\to\c \c D_b$ be the projection defined by
$\pi_b(x_a:a\in\alpha) = x_b$.  Conversely, let $\iota_a:\c D_a\to \c
C$ be the embedding defined by $\iota_a(y)=(x_b:b\in\alpha)$, where
$x_a=y$ and $x_b=0$ for $b\neq a$.  Evidently $\pi_b(\iota_b(y))=y$
for $y\in\c D_b$ and $\pi_b(\iota_a(y))=0$ if $a\neq b$.

Suppose $x\in\Nr_{\mu}\c C\setminus\set0$.  Since $x\neq 0$, 
it must have a non-zero component  $\pi_a(x)\in\c D_a$, for some $a\in \alpha$.  
Say $\emptyset\neq\phi(x_{i_0}, \ldots, x_{i_k})^{\c
 D_a}= \pi_a(x)$ for some $L$-formula $\phi(x_{i_0},\ldots, x_{i_k})$.  We
 have $\phi(x_{i_0},\ldots, x_{i_k})^{\c D_a}\in\Nr_{\mu}\c D_a)$.  Pick
 $f\in \phi(x_{i_0},\ldots, x_{i_k})^{\c D_a}$ and let $b=N_a(f(0),
 f(1), \ldots f_{n-1})\in\alpha$.  We will show that 
$b(x_0, x_1, \ldots x_{n-1})^{\c D_a}\subseteq
 \phi(x_{i_0},\ldots, x_{i_k})^{\c D_a}$.  Take any $g\in
b(x_0, x_1\ldots x_{n-1})^{\c D_a}$, 
so $N_a(g(0), g(1)\ldots g(n-1))=b$.  The map $\set{(f(0),
g(0)), (f(1), g(1))\ldots (f(n-1), g(n-1))}$ is 
a partial isomorphism of $N_a$.  By
 (\ref{eq:theta}) this extends to a finite partial isomorphism
 $\theta$ of $N_a$ whose domain includes $f(i_0), \ldots, f(i_k)$. Let
 $g'\in U_a$ be defined by
\[ g'(i) =\left\{\begin{array}{ll}\theta(i)&\mbox{if }i\in\dom(\theta)\\
g(i)&\mbox{otherwise}\end{array}\right.\] By (\ref{eq:bf}), $\c N_a,
g'\models\phi(x_{i_0}, \ldots, x_{i_k})$. Observe that
$g'(0)=\theta(0)=g(0)$ and similarly $g'(n-1)=g(n-1)$, so $g$ is identical
to $g'$ over $\mu$ and it differs from $g'$ on only a finite
set of coordinates.  Since $\phi(x_{i_0}, \ldots, x_{i_k})^{\c
\ D_a}\in\Nr_{\mu}(\c C)$ we deduce $\c N_a, g \models \phi(x_{i_0}, \ldots,
x_{i_k})$, so $g\in\phi(x_{i_0}, \ldots, x_{i_k})^{\c D_a}$.  This
proves that $b(x_0, x_1\ldots x_{\mu-1})^{\c D_a}\subseteq\phi(x_{i_0},\ldots,
x_{i_k})^{\c D_a}=\pi_a(x)$, and so 
$$\iota_a(b(x_0, x_1,\ldots x_{n-1})^{\c \ D_a})\leq
\iota_a(\phi(x_{i_0},\ldots, x_{i_k})^{\c D_a})\leq x\in\c
C\setminus\set0.$$  Hence every non-zero element $x$ of $\Nr_{n}\c C$ 
is above
a an atom $\iota_a(b(x_0, x_1\ldots n_1)^{\c D_a})$ (some $a, b\in
\alpha$) of $\Nr_{n}\c C$.  So
$\Nr_{n}\c C$ is atomic and $\alpha\cong\At\Nr_{n}\c C$ --- the isomorphism
is $b \mapsto (b(x_0, x_1,\dots x_{n-1})^{\c D_a}:a\in A)$.
\end{proof}

We can use such games to show that for $n\geq 3$, there is a representable $\A\in \CA_n$ 
with atom structure $\alpha$ such that $\forall$ can win the game $F^{n+2}(\alpha)$.
However \pe\ has a \ws\ in $H_n(\alpha)$, for any $n<\omega$.
It will follow that there a countable cylindric algebra $\c A'$ such that $\c A'\equiv\c
A$ and \pe\ has a \ws\ in $H(\c A')$.
So let $K$ be any class such that $\Nr_n\CA_{\omega}\subseteq K\subseteq S_c\Nr_n\CA_{n+2}$.
$\c A'$ must belong to $\Nr_n(\RCA_\omega)$, hence $\c A'\in K$.  But $\c A\not\in K$
and $\c A\preceq\c A'$. Thus $K$ is not elementary. From this it easily follows that the class of completely representable cylindric algebras
is not elementary, and that the class $\Nr_n\CA_{n+k}$ for any $k\geq 0$ is not elementary either. 
Furthermore the constructions works for many variants of cylindric algebras
like Halmos' polyadic equality algebras and Pinter's substitution algebras.
Formally we shall prove:
\begin{theorem}\label{r} Let $3\leq n<\omega$. Then the following hold:
\begin{enumroman}
\item Any $K$ such that $\Nr_n\CA_{\omega}\subseteq K\subseteq S_c\Nr_n\CA_{n+2}$ is not elementary.
\item The inclusions $\Nr_n\CA_{\omega}\subseteq S_c\Nr_n\CA_{\omega}\subseteq S\Nr_n\CA_{\omega}$ are all proper
\end{enumroman}
\end{theorem}

%\end{demo}

 
\begin{thebibliography}{100}

\bibitem{AGMNS} Andr\'eka, H., Givant, S., Mikulas, S. N\'emeti, I., Simon A., 
{\it Notions of density
that imply representability in algebraic logic.} 
Annals of Pure and Applied logic, {\bf 91}(1998) p. 93 --190. 


%\bibitem{AGMNS}

\bibitem{1} H. Andreka, M. Ferenczi, I. Nemeti (editors) {\it Cylindric-like algebras and algebraic logic} Bolyai Society, Mathematical Studies, Springer
(2013).

\bibitem{ANT} Andr\'eka, H., N\'emeti, I., Sayed Ahmed, T., 
{\it Omitting types for finite variable fragments and complete representations of algebras.}
Journal of Symbolic Logic {\bf 73}(1) (2008) p.65-89

%\bibitem{B} Becker H {\it Polish group actions, dichotomies and generalized elementary embeddings} Journal of the Americal Mathematical Soceity, 11(2) 397-449 1998.

\bibitem{CF} Casanovas, E., and Farre, R. {\it Omitting Types in Incomplete Theories.}  
Journal of Symbolic logic, Vol. 61, Number 1, p. 236-245.
March 1996. 

%\bibitem{M80} Miller, A. {\it Covering $^{\omega}2$ with $\omega_1$ disjoint closed sets.} 
%The Kleene Symposuim (proceedings, Madison, Wisconsin, 1978). Studies
%in Logic and the Foundation of Mathematics, vol 101, North-Holland,
%Amsterdam, 1980 pp. 415--421.

%\bibitem{Miller1} Miller A. {\it A characterization of the least cardinal for which 
%the Baire Category Theorem 
%fails}. Proceedings of the Americal Mathematical Society, {\bf 86}(3) (1982) p. 498--502.
%\bibitem{Bulletin} T.Sayed Ahmed {\it A sufficient and necessary condition for omitting types}
%Bulletin section of logic {\bf 34}(1)(2005)  23-28

\bibitem{DM} Daigneault, A., and Monk,J.D., 
{\it Representation Theory for Polyadic algebras}. 
Fund. Math. {\bf 52}(1963) p.151-176.




\bibitem{HMT1}L. Henkin, D. Monk, A. Tarski {\it Cylindric algebras, part 1} 1970

\bibitem{HMT2}L. Henkin, D. Monk, A. Tarski {\it Cylindric algebras, part 2} 1985
%\bibitem{Hodkinson}  Cylindric algebras are not closed under completemath lohg 3 2008 294-306

\bibitem{r} R. Hirsch {\it Relation algebra reducts of cylindric algebras and complete representations}. Journal of Symbolic Logic
(72) (2007) p. 673-703

\bibitem{Hodkinson} Hodkinson{\it Atom structures of relation and cylindric algebas} Appals(1997)p. 117-148.

\bibitem {Hirsch} Hirsch and Hodkinson  {\it  Complete representations in algebraic logic} JSL (62)(1997) p. 816-647


\bibitem{R} Hirsch R. {\it Relation algebra reducts of cylindric algebras and complete representations} Journal of Symbolic Logic (72) (2007) 673-703

\bibitem{HHbook} Hirsch, Hodkinson {\it Relation lgebras by Games}  147 Studies in Logic and Foundations of Mathematics Elsevier 
North Holland 2002

\bibitem{strong} Hirsch, Hodkinson {\it Strongly representable  atom structures of cylindric algebra} Journal of Symbolic Logic (74)(2009) 811-828

\bibitem{HHM}  Hirsch, Hodkinson and Maddux {\it On provability with finitely many varibales} Bulletin of Symbolic Logic 
(8) (2002) p. 329-347.

\bibitem {HHK} Hirsch Hodkinson and Kurucz {\it On modal logics between $K\times K\times K$ and $S5\times S5\times S5$.} 
Journal of Symbolic Logic(67) (2002) 221-234.


\bibitem{HH} Hirsch and Hodkinson {\it Completions and complete representations}
In \cite{1} p. 61-90.

\bibitem{Shelah} S. Shelah {\it Classification Theory } 1978

\bibitem{m} Maddux {\it Non finite axiomatizability results for cylindric and relation algebras} JSL, 54 (1989) 951-974.
\bibitem{N} Newelski, L. {\it Omitting types and the real line.}
Jornal of Symbolic Logic, {\bf 52}(1987),  p.1020-1026.

\bibitem{IGPL} Sayed Ahmed {\it The class of neat reducts is not elementary} Logic Journal of $IGPL$(9) (2001) 593-628

\bibitem{MLQ} Sayed Ahmed {\it A model theoretic solutin to a problem of Tarski} 
Mathematical Logic quarterly (48) (2002) 343-355.

\bibitem{FM} T. Sayed Ahmed {\it The class of 2 dimensional neat reducts of polyadic algebras is not elementary}. 
Fundementa Mathematica (172) (2002) p.61-81

\bibitem{MLQ} T. Sayed Ahmed {\it A model theoretic solution to a problem of Tarski} Mathematical Logic Quarterly (48) (2002)
p.343-355

\bibitem{Bulletin}  Sayed Ahmed,T., {\it Algebraic Logic, where does it stand today?}
Bulletin of Symbolic Logic. {\bf 11}(4)(2005), p.465-516.

\bibitem{w} Sayed Ahmed {\it  Weakly representable atom structures that are not strongly representable with an application to first order logic}
Math Logic Quarterly (2008) p. 294-306


\bibitem{OTT} Sayed Ahmed T., {\it Completions, complete representations and omiting types} In \cite{1} p. 205-222

\bibitem{OTT2} Sayed Ahmed, T. and Samir B., {\it Omitting types for first order logic with infinitary predicates}
Mathematical Logic Quaterly  {\bf 53}(6) (2007) p.564-576.


\bibitem{KS} Khaled, Sayed Ahmed {\it On complete representations of algebras of logic} 
Logic Journal of $IGPL$, (2009)   p.267-272

\bibitem{IGPl} Sayed ahmed {\it  The class of neat reducts is not elementary} Logic Journal of IGPL (9) (2001)p. 593-628

%\bibitem{MLQ} A model theoretic MLQ 48 (2002) 343-355.

\bibitem{OTT2} Sayed Ahmed, T. and Samir B., {\it Omitting types for first order logic with infinitary predicates}
Mathematical Logic Quaterly  {\bf 53}(6) (2007) p.564-576.

\bibitem{hv} Hodkinson Venema{ \it Canonivcal varities with no canonical axiomatizations} Tarns 357 (2005) 4579-4605
 
\bibitem{ghv} {\it Erdos graphs resolves Fine's canonicity problem} Bull 10(2) (2004) 186-208.

\bibitem{Kechris} A.S. Kechris, \emph{Classical Descriptive Set Theory}, Springer Verlag, New York, 1995.


\bibitem{sep} M. Assem,  \emph{Separating Models by Formulas and the Number of Countable Models}, submitted (2013).
\bibitem{BeckerKechris} H. Becker and A.S. Kechris, \emph{The Descriptive Set Theory of Polish Group Actions}, Cambridge University Press, 1996. CMP 97:

\end{thebibliography}

%\bibitem{hv} Barely canonical


\end{document}

\subsection*{Details of the above idea}

Fix finite $n>2$. We use a rainbow construction for cylindric algebras.
We shall construct a cylindric atom structure based on graphs.
A coloured graph is an undirected irreflexive graph $\Gamma$
such that every edge of $\Gamma$ is coloured by a unique 
edge colour and some $n-1$ tuples
have a unique colour too. $\Z$ denotes the set of integers.
Let $P$ be the set of partial order preserving functions $f:\Z\to \N$ with $|\dom(f)|\leq 2$.

The edge colours (or future atoms) are

\begin{itemize}

\item greens: $\g_i$ ($i<n-1)$, $\g_0^i$, $i\in \Z$.  

\item whites : $\w, \w_f: f\in P$
\item yellow : $\y$
\item black :  $\bb$
\item reds:  $\r_{ij}$ $(i,j\in \N)$,

\item shades of yellow : $\y_S: S\subseteq_{\omega} \N$ or $S=\N$

\end{itemize}

\begin{definition} 
Let $i\in \Z$, and let $\Gamma$ be a coloured graph  consisting of $n$ nodes
$x_0,\ldots  x_{n-2}, z$. We call $\Gamma$ an $i$ - cone if $\Gamma(x_0, z)=\g^0_i$   
and for every $1\leq j\leq n-2$ $\Gamma(x_j, z)=\g_j$, 
and no other edge of $\Gamma$
is coloured green.
$(x_0,\ldots x_{n-2})$ 
is called the centre of the cone, $z$ the apex of the cone
and $i$ the tint of the cone.
We define a class $\bold J$ consisting of coloured graphs 
with the following properties:

\begin{enumarab}

\item $\Gamma$ is a complete graph.

\item $\Gamma$ contains no triangles (called forbidden triples) 
of the following types:

\vspace{-.2in}
\begin{eqnarray}
&&\nonumber\\
%(1', x, y)&&\mbox{unless }x=y\label{forb:id}\\
(\g, \g^{'}, \g^{*}), (\g_i, \g_{i}, \w), 
&&\mbox{any }i\in n-1\;  \\
(\g^j_0, \y, \w_f)&&\mbox{unless }f\in P, i\in dom(f)\\
(\g^j_0, \g^k_0, \w_0)&&\mbox{ any } j, k\in \Z\\
\label{forb:pim}(\g^i_0, \g^j_0, \r_{kl})&&\mbox{unless } \set{(i, k), (j, l)}\mbox{ is an order-}\\
&&\mbox{ preserving partial function }\Z\to\N\nonumber\\
%\label{forb:pim2}(\g_i, \g_j, \r_{kl})&&\mbox{if } i=j\mbox{ but }k\neq l\\
\label{forb: black}(\y,\y,\y), (\y,\y,\bb)\\ 
\label{forb:match}(\r_{ij}, \r_{j'k'}, \r_{i^*k^*})&&\mbox{unless }i=i^*,\; j=j'\mbox{ and }k'=k^*
\end{eqnarray}
and no other triple of atoms is forbidden.

%Let $\c A$ be the complex algebra 
%over $\alpha$ (so the domain of $\c A$ consists of arbitrary sets of atoms).
%\end{definition}

\item If $a_0,\ldots   a_{n-2}\in \Gamma$ are distinct, and no edge $(a_i, a_j)$ $i<j<n$
is coloured green, then the sequence $(a_0, \ldots a_{n-2})$ 
is coloured a unique shade of yellow.
No other $(n-1)$ tuples are coloured shades of yellow.

\item If $D=\set{d_0,\ldots  d_{n-2}, \delta}\subseteq \Gamma$ and 
$\Gamma\upharpoonright D$ is an $i$ cone with apex $\delta$, inducing the order 
$d_0,\ldots  d_{n-2}$ on its base, and the tuple 
$(d_0,\ldots d_{n-2})$ is coloured by a unique shade 
$y_S$ then $i\in S.$

\end{enumarab}
\end{definition}
\begin{demo}{Proof} 
The proof is very simiar to Hirsch's proof for reation algebras, 
except that we lift the rainbow construction to cylindric algebras. This is highly non- trivial, 
it was first done in \cite{complete}. The idea is to use labelled graphs as atoms; that is algebras will be based on atom structures consisting of labelled 
graph. Cylindrfiers are stimulated by shades of yellow,
which has to do with the $n$ tuples, and their an additional complexity the presence of cones. 

We define a cylindric algebra of 
dimension $n$. We first specify its atom structure.
Let $$K=\{a: a \text { is a surjective map from $n$ onto some } \Gamma\in \bold J
\text { with nodes } \Gamma\subseteq \omega\}.$$
We write $\Gamma_a$ for the element of $K$ for which 
$a:\alpha\to \Gamma$ is a surjection.
Let $a, b\in K$ define the following equivalence relation: $a \sim b$ if and only if
\begin{itemize}
\item $a(i)=a(j)\text { and } b(i)=b(j)$

\item $\Gamma_a(a(i), a(j))=\Gamma_b(b(i), b(j))$ whenever defined

\item $\Gamma_a(a(k_0)\dots a(k_{n-2}))=\Gamma_b(b(k_0)\ldots b(k_{n-1}))$ whenever  
defined
\end{itemize}
Let $\mathfrak{C}$ be the set of equivalences classes. Then define
$$[a]\in E_{ij} \text { iff } a(i)=a(j)$$
$$[a]T_i[b] \text { iff }a\upharpoonright n\sim \{i\}=b\upharpoonright n\sim \{i\}.$$
This, as easily checked, defines a $\CA_n$ 
atom structure.
Let $3\leq n<\omega$. 
Let $\c C_{n}$ be the complex algebra over $\mathfrak{C}$. We will show that $\c C_{n}$ is 
not in $S_c\Nr_n\CA_{n+2}$ 
but an elementary extension of $\c A$ belongs to
$\Nr_{n}\CA_{\omega}.$
But first we translate our games to games on coloured graphs:
Let $N$ be an atomic $\c C_{n}$ network. 
Let $x,y$ be two distinct nodes occuring in the
$n$ tuple $\bar z$. $N(\bar z)$ is an atom of $\c C_{n}$ 
which defines an edge colour of 
$x,y$. Using the fact that the dimension is at least $3$, 
the edge colour depends only on $x$ and $y$
not on the other elements of 
$\bar z$ or the positions of $x$ and $y$ in $\bar z$.
Similarly $N$ defines shades of white for certain $(n-1)$ tuples.  In this way $N$ 
translates
into a coloured graph. 
This translation has an inverse. More precisely we have:
Let $\Gamma\in \bold J$ be arbitrary. Define $N_{\Gamma}$ 
whose nodes are those of $\Gamma$
as follows. For each $a_0,\ldots a_{n-1}\in \Gamma$, define 
$N_{\Gamma}(a_0,\ldots  a_{n-1})=[\alpha]$
where $\alpha: n\to \Gamma\upharpoonright \set{a_0,\ldots a_{n-1}}$ is given by 
$\alpha(i)=a_i$ for all $i<n$. 
Then, as easily checked,  $N_{\Gamma}$ is an atomic $\c C_{n}$ network.
Conversely , let $N$ be any non empty atomic $\c C_n$ network. 
Define a complete coloured graph $\Gamma_N$
whose nodes are the nodes of $N$ as follows:  
\begin{itemize}
\item For all distinct $x,y\in \Gamma_N$ and edge colours $\eta$, $\Gamma_N(x,y)=\eta$
if and only if  for some $\bar z\in ^nN$, $i,j<n$, and atom $[\alpha]$, we have
$N(\bar z)=[\alpha]$, $z_i=x$ $z_j=y$ and the edge $(\alpha(i), \alpha(j))$ 
is coloured $\eta$ in the graph $\alpha$.

\item For all $x_0,\ldots x_{n-2}\in {}^{n-1}\Gamma_N$ and all yellows $\y_S$, 
$\Gamma_N(x_0,\ldots x_{n-2})= \y_S$ if and only if 
for some $\bar z$ in $^nN$, $i_0, i_{n-2}<n$
and some atom $[\alpha]$, we have
$N(\bar z)=[\alpha]$, $z_{i_j}=x_j$ for each $j<n-1$ and the $n-1$ tuple 
$\langle \alpha(i_0),\ldots \alpha(i_{n-2})\rangle$ is coloured 
$\y_S.$ Then $\Gamma_N$ is well defined and is in $\bold J$.
\end{itemize}
The following is then, though tedious and long,  easy to  check:
For any $\Gamma\in \bold J$, we have  $\Gamma_{N_{\Gamma}}=\Gamma$, 
and for any $\c C_n$ network
$N$ $N_{{\Gamma}_N}=N.$
This translation makes the following equivalent formulation of the 
games $F^m(\At\c C_n),$ originally defined on networks. 
%Now we define them on coloured graph; such games  have also $\omega$ rounds,
The new  game builds a nested sequence $\Gamma_0\subseteq \Gamma_1\subseteq \ldots $.
of coloured graphs. 
Let us start with the game $F^m(\alpha)$. 
\pa\ picks a graph $\Gamma_0\in \bold J$ with $\Gamma_0\subseteq m$ and 
$|\Gamma_0|=m$. $\exists$ makes no response
to this move. In a subsequent round, let the last graph built be $\Gamma_i$.
$\forall$ picks 
\begin{itemize}
\item a graph $\Phi\in \bold J$ with $|\Phi|=m$
\item a single node $k\in \Phi$
\item a coloured garph embedding $\theta:\Phi\sim \{k\}\to \Gamma_i$
Let $F=\phi\smallsetminus \{k\}$. Then $F$ is called a face. 
\pe\ must respond by amalgamating
$\Gamma_i$ and $\Phi$ with the embedding $\theta$. In other words she has to define a 
graph $\Gamma_{i+1}\in C$ and embeddings $\lambda:\Gamma_i\to \Gamma_{i+1}$
$\mu:\phi \to \Gamma_{i+1}$, such that $\lambda\circ \theta=\mu\upharpoonright F.$
\end{itemize} 
Let us consider the possibilities. There may be already a point $z\in \Gamma_i$ such that
the map $(k\to z)$ is an isomorphism over $F$.
In this case \ \pe does not need to extend the 
graph $\Gamma_i$, she can simply let $\Gamma_{i+1}=\Gamma_i$
$\lambda=Id_{\Gamma_i}$, and $\mu\upharpoonright F=Id_F$, $\mu(\alpha)=z$.
Otherwise, without loss of generality, 
let $F\subseteq \Gamma_i$, $k\notin \Gamma_i$.
Let ${\Gamma_i}^*$ be the colored graph with nodes $\nodes(\Gamma_i)\cup\{k\}$,
whose edges are the combined edges of $\Gamma_i$ and $\Phi$, 
such that for any $n-1$ tuple $\bar x$ of nodes of 
${\Gamma_i}^*$, the color ${\Gamma_i}^*(\bar x)$ is
\begin{itemize}
\item $\Gamma_i(\bar x)$ if the nodes of $x$ 
all lie in $\Gamma$ and $\Gamma_i(\bar x)$ is defined 
\item $\phi(\bar x)$ if the nodes of $\bar x$ all lie in 
$\phi$ and $\phi(\bar x)$ is defined
\item undefined, otherwise.
\end{itemize}
\pe\ has to complete the labeling of $\Gamma_i^*$ by adding 
all missing edges, colouring each edge $(\beta, k)$
for $\beta\in \Gamma_i\sim\Phi$ and then choosing a shade of 
white for every $n-1$ tuple $\bar a$ 
of distinct elements of ${\Gamma_i}^*$
not wholly contained in $\Gamma_i$ nor $\Phi$, 
if non of the edges in $\bar a$ is coloured green.
She must do this on such a way that the resulting graph belongs to $\bold J$.
If she survives each round, \pe\ has won the play.
Notice that \pe\ has a \ws in the in $F^m(\At(\C_n))$ if and only if 
and only if she has a wininng strategy in the 
graph games defined above. Th is tedious and rather long to verify but basically routine. 
We now show that the rainbow algebra $\c A$ (definition above) 
is not in ${\bf S_c}\Nr_{\mu}\CA_{n+2}$.

For that we show $\forall$ can win the game $F^{n+2}(\At(\C_n))$. In his zeroth move, $\forall$ plays a graph $\Gamma \in \bold J$ with
nodes $0, 1,\ldots, n-1$ and such that $\Gamma(i, j) = \w (i < j <
n-1), \Gamma(i, n-1) = \g_i ( i = 1,\ldots, n), \Gamma(0, n-1) =
\g^0_0$, and $ \Gamma(0, 1,\ldots, n-2) = \y_\omega $. This is a $0$-cone
with base $\{0,\ldots , n-2\}$. In the following moves, $\forall$
repeatedly chooses the face $(0, 1,\ldots, n-2)$ and demands a node (possibly used before)
$\alpha$ with $\Phi(i,\alpha) = \g_i (i = 1,\ldots,  n-2)$ and $\Phi(0, \alpha) = \g^\alpha_0$,
in the graph notation -- i.e., an $\alpha$-cone on the same base.
$\exists$, among other things, has to colour all the edges
connecting nodes. The idea is that by the rules of the game 
only permissible colours would be red. Using this, $\forall$ can force a
win eventually for else we are led to a a decreasing sequence in $\N$.
In more detail,
In the initial round $\forall$ plays a graph $\Gamma$ with nodes $0,1,\ldots n-1$ such that $\Gamma(i,j)=\w$ for $i<j<n-1$ 
and $\Gamma(i,n-1)=\g_i$
$(i=1, \ldots n-2)$, $\Gamma(0,n-1)=\g_0^0$ and $\Gamma(0,1\ldots n-2)=\y_{N}$.
$\exists$ must play a graph with $\Gamma_1(0,\ldots n-1)=\g_0$.
In the following move $\forall$ chooses the face $(0,\ldots n-2)$ and demands a node $n$ 
with $\Gamma_2(i,n)=\g_i$ and $\Gamma_2(0,n)=\g_0^{-1}$
$\exists$ must choose a label for the edge $(n,n-1)$ of $\Gamma_2$. It must be a red atom $r_{mn}$. Since $-1<0$ we have $m<n$.
In the next move $\forall$ plays the face $(0, \ldots n-2)$ and demands a node $n+1$ such that  $\Gamma_3(i,n+1)=\g_i^{-2}$.
Then $\Gamma_3(n+1,n)$ $\Gamma_3(n+1,n-1)$ both being red, the indices must match.
$\Gamma_3(n+1,n)=r_{ln}$ and $\Gamma_3(n+1, n-1)=r_{lm}$ with $l<m$.
In the next round $\forall$ plays $(0,1\ldots n-2)$ and reuses the node $n-2$ such that $\Gamma_4(0,n-2)=\g_0^{-3}$.
This time we have $\Gamma_4(n,n-1)=\r_{jl}$ for some $j<l\in N$.
Continuing in this manner leads to a decreasing sequence in $\N$.

Recall from definition~\ref{def:games} that $H_k(\alpha)$ 
is the hypernetwork game with $k$ rounds.
The translation of the games $H$ and $H_k$ to graphs is as follows.

\begin{itemize}
\item
Fix some hyperlabel $\lambda_0$.  $H_{\mu}(\alpha)$ is  a 
game the play of which consists of a sequence of
$\lambda_0$-neat hypernetworks 
$N_0, N_1,\ldots$ where $\nodes(N_i)$
is a finite subset of $\omega$, for each $i<\omega$.  
%The other main
%difference is that \pa\ can play a more general kind of amalgamation
%move.  
A neat hypernetwork, now, is a pair $(\Gamma, N^h)$ with $\Gamma$ a coloured 
graph. $N^h$ are the hyperlabels, these we forget for a while and identify the pair $(\Gamma, N^h)$ with $\Gamma.$
Hyperedges will be dealt with later, and we shall see that they are easior to deal with.
$\forall$ picks a graph $\Gamma_0\in \bold J$ with $\Gamma_0\subseteq_{\omega} \omega$ and 
here we do not require that $|\Gamma_0|=n$. 
$\exists$ make no response
to this move. In a subsequent round, let the last graph built be $\Gamma_i$.
$\forall$ picks 
\begin{itemize}
\item a graph $\Phi\in \bold J$ with $|\phi|=|\Gamma|$
\item a single node $k\in \phi$
\item a colored garph embedding $\theta:\Phi\sim\{k\}\to \Gamma_i$
Let $F=\phi\sim \{k\}$. Then $F$ is called a face. 
\pe must respond by amalgamating
$\Gamma_i$ and $\phi$ with the embedding $\theta$ as before. 
In other words she has to define a 
graph $\Gamma_{i+1}\in C$ and embeddings $\lambda:\Gamma_i\to \Gamma_{i+1}$
$\mu:\phi \to \Gamma_{i+1}$, such that $\lambda\circ \theta=\mu\upharpoonright F.$


\end{itemize} 
Now we may write $N_{\Gamma}$ or simply $N$ instead of $\Gamma$, but in all cases we are dealing with {\it coloured graphs} 
that is {\it the translation} of networks. That is when we writ $N$ then, $N$ will be viewed as a coloured graph.
Alternatively, \pa\ can play a \emph{transformation move} by picking a
previously played graph $N$ and a partial, finite surjection
$\theta:\omega\to\nodes(N)$, this move is denoted $(N, \theta)$.  \pe\
must respond with $N\theta$.  Finally, \pa\ can play an
\emph{amalgamation move} by picking previously played graphs
$M, N$ such that $M\equiv^{\nodes(M)\cap\nodes(N)}N$ and
$\nodes(M)\cap\nodes(N)\neq \emptyset$  
This move is denoted $(M,
N)$.  To make a legal response, \pe\ must play a $\lambda_0$-neat
hypernetwork $L$ extending $M$ and $N$, where
$\nodes(L)=\nodes(M)\cup\nodes(N)$.
Again, \pe\ wins $H(\alpha)$ if she responds legally in each of the
$\omega$ rounds, otherwise \pa\ wins. 


It will simplify things a bit if we alter the rules of the game
$H(\alpha)$ slightly.
We impose certain restrictions on \pa.
First a piece of notation. A triangle move $\Gamma$ in the graph game will be denoted
by $(\Gamma, F, k, b, l),$  $\Phi=F\cup\{k\}$ and 
$b\leq {\sf c}_lN_{\Gamma}(f_0,\dots  f_{n-1})$.  
\begin{itemize}
\item
\pa\ is only allowed to play a triangle move $(\Gamma, F, k, b, l)$ if
there does not exist $l\in\nodes(\Gamma)$ such that 
$N_{\Gamma)}(f_0,\ldots f_i,l\ldots f_{n-2})=b.$ 

\item
\pa\ is only allowed to play transformation moves $(N, \theta)$ if $\theta$ is injective.

\item
\pa\ is only allowed to play an amalgamation move $(M, N)$ if for all
$m\in\nodes(M)\setminus\nodes(N)$ and all
$n\in\nodes(N)\setminus\nodes(M)$ the map $\set{(m, n)}\cup\set{(x,
x):x\in\nodes(M)\cap\nodes(N)}$ is not a partial isomorphism.  I.e. he
can only play $(M, N)$ if the amalgamated part is `as large as
possible'.
\end{itemize}  If, as a result of these restrictions, \pa\ cannot move at
some stage then he loses and the game halts.

It is easy to check that \pa\ has a \ws\ in $H(\alpha)$ iff he has a
\ws\ with these restrictions to his moves.  Also, if \pa\ plays with
these restrictions to his moves, if \pe\ has a \ws\ then she has a
\ws\ which only directs her to play strict hypernetworks.  The same
holds when we consider $H_n(\alpha)$.  
We will assume that \pa\ plays 
according to these restrictions.
\end{itemize}




Now we deal with hypernetworks.
Notice that in this game $\forall$ is not allowed to reuse nodes and so his winning strategy above does not work.
In a play of $N(\alpha)$ $\exists$ is required to play $\lambda_0$ neat hypernetworks, so she has no choice about the 
hyperedges for short edges, these are labelled by $\lambda_0$. In response to a cylindrifier move 
$(N, F, k,l)$ all long hyperedges not incident with $k$ necessarily keep the hyperlabel they had in $N$. 
All long hyperedges incident with $k$ in $N$
are given unique hyperlabels not occuring as the hyperlabel of any other hyperedge in $M$. 
We assume, without loss of generality, that we have infinite supply of hyperlabels of all finite arities so this is possible.
In reponse to an amalgamation move $(M,N)$ all long hyperedges whose range is contained in $\nodes(M)$ 
have hyperlabel determined by $M$, and those whose range is contained in nodes $N$ have hyperlabel determined
by $N$. If $\bar{x}$ is a long hyperedge of $\exists$'s response $L$ where 
$\rng(\bar{x})\nsubseteq \nodes(M)$, $\nodes(N)$ then $\bar{x}$ 
is given 
a new hyperabel, not used in any previously played hypernetwork and not used within $L$ as the label of any hyperedge other than $\bar{x}$.
This completeles her strategy for labelling hyperedges.

Now we give $\exists$'s strategy for edge labelling. We need some notation and terminology. 
Every irreflexive edge of any hypernetwork has an owner $\forall$ or $\exists$ namely the one who played this edge.
We call such edges $\forall$ edges or $\exists$ edges. Each long hyperedge $\bar{x}$ in a hypernetwork $N$ 
occuring in the play has an envelope $v_N(\bar{x})$ to be defined shortly. 
In the initial round of $\forall$ plays $a\in \alpha$ and $\exists$ plays $N_0$ 
then all irreflexive edges of $N_0$ belongs to $\forall$. There are no long hyperdeges in $N_0$. If in a later move, 
$\forall$ plays the transformation move $(N,\theta)$ 
and $\exists$ responds with $N\theta$ then owners and envelopes are inherited in the obvious way. 
If $\forall$ plays a cylindrifier move $(N, F, k, b,l)$ and $\exists$ responds with $M$ then the owner
in $M$ of an edge not incident with $k$ is the same as it was in $N$
and the envelope in $M$ of a long hyperedge not incident with $k$ is the same as that it was in $N$. 
The edges $(f,k), (k,f)$ belong to $\forall$ in $M$ all edges $(l,k)(k,l)$
for $l\in nodes(N)\sim F$ belong to $\exists$ in $M$.
if $\bar{x}$ is any long hyperedge of $M$ with $k\in \rng(\bar{x}$, then $v_M(\bar{x})=\nodes(M)$.
If $\forall$ plays the amalgamation move $(M,N)$ and $\exists$ responds with $L$ 
then for $m\neq n\in nodes(L)$ the owner in $L$ of a edge $(m,n)$ is $\forall$ if it belongs to
$\forall$ in either $M$ or $N$, in all other cases it belongs to $\exists$ in $L$. If $\bar{x}$ is a long hyperedge of $L$
then $v_L(\bar{x}=v_M(x)$ if $range(x)\subseteq nodes(M)$ $v_L(x)=v_N(x)$ and  $v_L(x)=\nodes(M)$ otherwise.
This completes the definition of owners and envelopes.
By induction on the number of rounds one can show


\begin{athm}{Claim} Let $M, N$ ocur in a play of $H_n(\alpha)$ in which $\exists$ uses default labelling
for hyperedges. Let $\bar{x}$ be a long hyperedge of $M$ and let $\bar{y}$ be a long hyperedge of $N$.
\begin{enumarab}
\item For any hyperedge $\bar{x}'$ with $\rng(\bar{x}')\subseteq _M(\bar{x})$, if $M(\bar{x}')=M(\bar{x})$
then $\bar{x}'=\bar{x}$.
\item if $\bar{x}$ is a long hyperedge of $M$ and $\bar{y}$ is a long hyperedge of $N$, and $M(\bar{x})=N(\bar{y})$ 
then there is a local isomorphism $\theta: v_M(\bar{x})\to v_N(\bar{y})$ such that
$\theta(x_i)=y_i$ for all $i<|x|$.
\item For any $x\in \nodes (M)\sim v_M(\bar{x})$ and $S\subseteq v_M(\bar{x})$, if $(x,s)$ belong to $\forall$ in $M$
for all $s\in S$, then $|S|\leq 2$.
\end{enumarab}
\end{athm}  
Now we define $\exists$'s straetegy for choosing the labels for edges and yelows colours for $n-1$ hyperedges.
Let $N_0, N_1,\ldots N_r$ be the start of a play of $H_k(\alpha)$ just before round $r+1$. 
$\exists$ computes partial functions $\rho_s:Z\to N$, for $s\leq r$. Inductively
for $s\leq r$ suppose

I. If $N_s(x,y)$ is green or yellow then $(x,y)$ belongs to $\forall$ in $N_s$.

II. $\rho_0\subseteq \ldots \rho_r$

III. $\dom(\rho_s)=\{i\in Z: \exists t\leq s, x,y\in \nodes(N_t), N_t(x,y)=\g_0^i\}$

IV. $\rho_s$ is order preserving: if $i<j$ then $\rho_s(i)<\rho_s(j)$. The range
of $\rho_s$ is widely spaced: if $i<j\in \dom\rho_s$ then $\rho_s(i)-\rho_s(j)\geq  3^{n-r}$, where $n-r$ 
is the number of rounds remaining in the game.

V. For $u,v,x,y\in \nodes(N_s)$, if $N_s(u,v)=\r_{\mu,\delta}$, $N_s(x,u)=\g_0^i$, $N_s(x,v)=\g_0^j$
$N_s(y,u)=N_s(y,v)=\y$ then

(a) if $N_s(x,y)\neq \w_f$ then $\rho_s(i)=\mu$ and $\rho_s(j)=\delta$

(b) If $N_s(x,y)=\w_f$ for some $f\in P$ , the $\mu=f(i)$ , $\delta=f(j)$.


VI. $N_s$ is a strict $\lambda_0$ neat hypernetwork.

To start with if $\forall$ plays $a$ in the intial round then $\nodes(N_0)=\{0,1,\ldots n-1\}$, the 
hyperedge labelling is defined by $N_0(0,1,\ldots n)=a$.

In reponse to a traingle move $(N_s, F, k, \g_0^p,l)$ by $\forall$, for some $s\leq r$ and some $p\in Z$, 
$\exists$ must extend $\rho_r$ to $\rho_{r+1}$ so that $p\in \dom(\rho_{r+1})$
and the gap between elements of its range is at least $3^{n-r-1}$. Inductively $\rho_r$ is order preserving and the gap betwen its elements is 
at least $3^{n-r}$, so this can be maintained in a further round.  
If $\forall$ chooses non green atoms, green atoms with the same suffix, or green atom whose suffices 
already belong to $\rho_r$, there would be fewer 
elements to add to the domain of $\rho_{r+1}$, which makes it easy for $\exists$ to define $\rho_{r+1}$.
Tis establishe properties $II-IV$ for round $r+1$.

Let us assume that  $\forall$ play the triangle move $(N_s, F, k, a, l)$ in round $r+1$. 
$\exists$ has to choose labels for $\{(x,k), (k,x)\}$ 
$x\in \nodes(N_s)\sim F$. She chooses labels for the edges
$(x,k)$ one at a time and then determines the reverse edges $(k,x)$ uniquely.
Property $I$ is clear since in all cases the only atoms $\exists$ chooses are white, black or red.

Now we distinguish between two case.

If $x$ and $k$ are both apexes of cones on $F$, then $\exists$ has no choice but to pick a red atom. The colour she chooses is uniquely defined.
Other wise, this is not the case, so for some $i<n-1$ there is no $f\in F$ such 
that $N_s(\beta, f), N_s(f,x)$ are both coloured $\g_i$ or if $i=0$, they are coloured
$\g_0^l$ and $\g_0^{l'}$ for some $l$ and $l'$.

Now we distinguish between several subcases:

\begin{enumarab}

\item Suppose that it is not the case that there exists distinct $i,j\in F$ $N_s(x,i)$ and $a$ are both green
and $N_s(x,j)$ and $a$ are both green.
Let $S=\{p\in Z: (N_s(x,i)=\g_0^p\land a =\y)\lor N_s(x,i)=\y \land a=\g_0^p)\}$
Then $|S|\leq 2$. $\exists$ lets $N_{s+1}(x,k)=\w_f$ for some $f$ with $\dom(f)=S$.

Suppose that there exists $i,j$ distinct in $F$ such that $N_s(i,j)=\r_{\beta, \mu}$, $N_s(x,i)=\g_p$, $N_s(x,j)=\g_q$ for some $p,q\in Z$. 
By property $(IV)$ $f=\{(p,\beta), (q,\mu)\}$ is order preserving.
$\exists$ lets $N_{s+1}(x,k)=\w_f$ in this case.

In all other cases: either there are $i,j\in F$ such that $N_s(i,j)$ is not red, or if it is 
then it is not the case that $N_s(x,i)$ $N_s(x,j)$ are both green, and it is not the case that $N_s(x,i)=N_s(x,j)=\y$, 
$\exists$ lets $f:S\to N$ an arbitrary order preserving function.
The only forbidden triangles involving $\w_f$ are avoided. Since $\exists$ does not change green or yellow atoms to label new edges
and $N_{r+1}(x,k)=\w_f$, all traingles involving the new edge $(x,k)$ are consistent in $N_{r+1}$.
Clearly propery $VI$ holds after round $r+1$.

\item Else if it is not the case that $N_s(x,i)=a$ and not the case that $N_s(x,j)=y$, $\exists$ lets $N_r(x,k)=\bb$. 
Property $V$ is not applicable in this case.
The only forbidden triple involving the atom $\bb$ is avoided, so all triangles 
$(x,y,k)$ are consistent in $N_{r+1}$ and property
$VI$ holds after round $r+1$lets $N_{r+1}(x,k)=a$.

\item If neither case above applies, then for all distinct $i,j\in F$ either $N_s(x,i)=\g_p$, $a=\g_q$ (for some $p$ and $q$)
and $N_s(x,j)=N_s(x,i)=a=\y$ and $N_s(x,j)=\g_p$, $b=\g_q$.
Assume the first alternative.  There are two subcases.
\begin{enumroman}
\item $N_s(i,j)\neq w_f$ for all $f\in P$. $\exists$ lets $\mu=\rho_{r+1}(p)$, $\delta=\rho_{r+1}(q)$,
maintaining property $Va$. The only forbidden triples of atoms involving $r_{\mu,\delta}$ are avoided. The triple
of atoms form a traingle $(x,y,k)$ will not be forbidden since the only green edge incident 
with $k$ is $(i,k)$ and since $\rho_{r+1}$ is order preserving. To check forbidden triple (16)
suppose $N_s(x,y), N_{r+1}(k)$ are both red for some $y\in \nodes(N_r).$
We have $y\notin \{i,j\}$ so $\exists$ chose the red label $N_{r+1}(y,k)$. By her strategy we have
$N_s(i,y)=\g_t$ and $N_s(j,y)=\y$ . By property $Va$ for $N_{r+1}$ we have 
$N_{r+1}(x,y)=r_{\rho_{r+1}(p)\rho_{r+1}(t)}$ and $N_{r=1}(y,k)=r$ 
The property $VI$ holds for $N_{r+1}$

\item $N_s(i,j)=\w_f$ for some $f\in F$. By consistency of $N_s$ and forbidden triple (14), we have $p\in dom(f)$and since $\forall$'s move 
we have $q\in \dom(f)$. $\exists$ lets $\mu=f(p)$ $\delta=f(q)$
maintaining property
$V$ for round $r+1$. As above, the only forbidden triples of atoms involving $r_{\mu,\delta}$ are (15) and (16).
Since $f$ is order preserving and since the only green edge incident
with $k$ is $(i,k)$ in $N_{r+1}$  tariangles involving the new edge $(x,k)$ cannot give a forbidden triple. 
For forbidden triple (16) let $y\in \nodes(N_s)$ and suppose
$N_{r+1}(x,y)$$ N_{r+1}(y,k)$ are both red. As above, by her strategy we must have $N_s(y,i)=\g_t$ for some $t$ and $N_s(y,j)=\y$.
By consistency of $N_s$ we have $t\in \dom(f)$ and the current part of her strategy she lets $N_{r+1}(y,k)=\r_{f(t),f(q)}.$ 
By property $Vb$ for $N_s$
we have $N_{r+1}(x,y)=\r_{f(p), f(t)}$. So the triple of atoms from the triangle $(x,y,k)$ is not forbidden by (16). This establishes propery $(VI)$ for
$N_{r+1}$
\end{enumroman}
\end{enumarab}
We have finished with triangle moves. Now we move to amalgamation moves.
Although our hypernetworks are all strict, it is not necessarily the case that hyperlabels label unique hyperedges - amalgmation moves can force that 
the same hyperlabel can label more than one hyperedge. However, within the envelope of a hyperdege $\bar{x}$, the hyperlabel $N(\bar{x})$ is unique.

We consider an amalgamation move $(N_s,N_t)$ chosen by $\forall$ in round $r+1$.
$\exists$ has to choose a label for each edge $(i,j)$ where $i\in \nodes(N_s)\sim \nodes(N_t)$ and $j\in \nodes(N_t)\sim (N_s)$.
This determines the label for the reverse edge.
Let $\bar{x}$ enumerate $\nodes(N_s)\cap \nodes(N_t)$
If $\bar{x}$ is short, then there are at most two nodes in the intersection 
and this case is similar to the trangle move. if not, that is if $\bar{x}$ is long in $N_s$, then by the claim 
there is a partial isomorphism $\theta: v_{N_s}(\bar{x})\to v_{N_t}(\bar{x})$ fixing
$\bar{x}$. We can assume that $v_{N_s}(\bar{x})=\nodes(N_s)\cap \nodes N_t)=rng(\bar{x})=v_{N_t}(bar{x})$.
It remains to label the edges $(i,j)\in N_{r+1}$ where $i\in \nodes(N_s)\sim \nodes N_t$ and $j\in \nodes(N_t)\sim \nodes(N_s)$. 
Her startegy is similar to the triangle move. if $i$ and $j$ are tints of the same cone she choose a red. If not she 
chooses  white atom if possible, else the black atom if possible, otherwise a red atom.
She never chooses a green atom, she lets $\rho_{r+1}=\rho_r$ and properies $II$, $III$, $IV$ remain true
in round $r+1$.

\begin{enumarab}  

\item There is no $x\in \nodes(N_s)\cap \nodes(N_t)$ such that $N_s(i,x)$ and $N_t(x,j)$ are both green. 
If there are nodes $u,v\in \nodes(N_s)\cap \nodes(N_t)$ such that
$N_s(u,v)=\r_{\beta,\mu}$, $N_s(i,u)=\g_p$, $N_s(i,v)=\g_q$, $N_t(u,j)=N_t(v,j)=y$ for some $\beta, \mu\in N$, $p,q\in Z$
or the roles of $i,j$ are swapped, she lets $f=\{(p,\beta), (q,\mu)\}$ and sets $N_{r+1}(i,j)=w_f$.
Since all edges labelled by green or yellow atoms belong to $\forall$, we can apply the above claim to 
show that the points $u,v$ are unique so $f$ is well defined.
This s also true if $\bar{x}$ is short, since in this csase there are only
two nodes
in $\nodes(N_s)\cap \nodes(N_t)$.

If there are no such $u,v$ as described then let $S=\{p\in Z: \exists y\in nodes(N_s)\cap Nodes(N_t), (N_s(i,y)=g_p\land N_t(y,j)=y)\lor 
(N_s(i,y)=y\land N_t(y,j=g_p)\}$. Then $|S|\leq 2$. 
Let $f$ be any order preserving function and $\exists$ let $N_{r+1}=w_f$.
Property $(VI)$ holds for $N_{r+1}$ as for traingle moves.

\item Otherwse if there is no such $x$, then she lets $N_r(i,j)=b$. As with traingle moves all properties are maintained.

\item Otherwise, there are $x,y\in \nodes(N_s)\cap \nodes(N_t)$ such that $N_s(i,x)=g_k$, $N_s(x,j)=g_l$ 
for some $k,l\in N$ and $N_s(i,y)=N_t(y,j)=y$. By claim 3 $x,y$ are unique. She labels $(i,j)$in $N_r$ with a red atom $r_{\beta,\mu}$ where
\begin{enumroman}
\item If $N_s(x,y)\neq \w_f$ for all $f\in P$, then $\beta=\rho_{r+1}(k)$, $\mu=\rho_{r+1}(l)$.
This maintains property $Va$.
\item Otherwise $N_s(x,y)=\w_f$ for some $f\in F$ and $\beta=f(k)$ $\mu=f(l)$.


\end{enumroman}
\end{enumarab}

Now we turn to coluring of $n-$ tuples. For each tuple $\bar{a}=a_0,\ldots a_{n-2}\in (\Gamma)^{n-1}$ with no edge 
$(a_i, a_j)$ coloured green, $\exists$ colours $\bar{a}$ by $y_S$, where
$S=\{i\in N: \text { there is an $i$ cone in $\Gamma$ with base $a_0,\ldots a_{n-2}$}\}$
We need to check that such labeling works.

Let us check that $(n-1)$ tuples are labeled correctly, by yellow colours. Let $D$ be  set of $n$ nodes, and supose that $\Gamma\upharpoonright D$ 
is an $i$ cone with apex
$\delta$ and base $\{d_0,\ldots d_{n-2}\}$ , and that the tuple $(d_0,\ldots d_{n-2})$ is labelled $y_S$ in $\Gamma$. 
We need to show that $i\in S$. If $D\subseteq \Gamma$, then inductively the graph
$\Gamma$ constructed so far is in $\bold J$, and therefore
$i\in S$. If $D\subseteq \Phi$ then as $\forall$ chose $\Phi$ in $\bold J$ we get also $i\in S$. If neither holds, then $D$ contains $\alpha$ 
and also some
$\beta\in \Gamma\sim \Phi$. $\exists$ chose the colour $\Gamma^+(\alpha,\beta)$ and her strategy ensures her that it is green. 
Hence neither $\alpha$ or $\beta$ can be the apex of the cone $\Gamma^+\upharpoonright D$, 
so they must both lie in the base $\bar{d}$. 
This implies that 
$\bar{d}$ is not yet labelled in $\Gamma^*$, so $\exists$ has applied her strategy to choose the colour $y_S$ to label $\bar{d}$ in $\Gamma^+$. 
But this strategy will have chosen $S$ containing $i$ since $\Gamma^*\upharpoonright D$ is already a cone
in $\Gamma^*$. Also $\exists$ never chooses a green edge, so all green edges of $\Gamma^+$ lie in $\Gamma^*$. 

That leaves one (hard) case, where there are two nodes $\beta, \beta',
\in \Gamma$, $\exists$ colours both $(\beta, \alpha)$ and $(\beta',
\alpha)$ red, and the old edge $(\beta, \beta')$ has already been
coloured red (earlier in the game).
If $(\beta, \beta')$ was coloured by $\exists$, then there is no problem. 
So suppose, for a contradiction, that $(\beta, \beta')$ was coloured by
$\forall$. Since $\exists$ chose red colours for $(\alpha, \beta)$
and $(\alpha, \beta')$, it must be the case that there are cones in
$\Gamma^*$ with apexes $\alpha, \beta, \beta'$ and the same base,
$F$, each inducing the same linear ordering $\bar{f} = (f_0,\ldots,
f_{n-2})$, say, on $F$. Of course, the tints of these cones may all
be different. Clearly, no edge in $F$ is labeled green, as no cone base can contain
green edges. It follows that $\bar{f}$  must be labeled by some
yellow colour, $\y_S$, say. Since $\Phi\in \bold J$, it obeys its definition, so
the tint $i$ (say) of the cone from $\alpha$ to $\bar{f}$ lies in
$S$. Suppose that $\lambda$ was the last node of $ F \cup \{ \beta,
\beta' \}$ to be created,as the game proceeded. As $ |F \cup \{
\beta, \beta' \}| = n + 1$, we see that $\exists$ must have chosen
the colour of at least one edge in this : say, $( \lambda, \mu )$.
Now all edges from $\beta$ into $F$ are green, and so chosen by
$\forall$, and the edge  $(\beta, \beta')$ was also chosen by him.
The same holds for edges from $\beta'$ to $F$. Hence $\lambda, \mu
\in F$. We can now see that it was $\exists$ who chose the colour $\y_S$ of
$\bar{f}$. For $\y_S$ was chosen in the round when $F$'s last node,
i.e., $\lambda$ was created. It could only have been chosen by
$\forall$ if he also picked the colour of every edge in $F$
involving $\lambda$. This is not so, as the edge $(\lambda, \mu)$
was coloured by $\exists$, and lies in $F$.
As $i \in S$, it follows from the definition of $\exists$'s strategy
that at the time when $\lambda$ was added, there was already an
$i$-cone with base $\bar{f}$, and apex $\gamma$ say.
We claim that $ F \cup \{ \alpha \}$ and $ F \cup \{ \gamma \}$ are
isomorphic over $F$. For this, note that the only $(n - 1)$-tuples of
either $ F \cup \{ \alpha \}$ or $ F \cup \{ \gamma \}$ with a
yellow colour are in $F$ ( since all others involve a green edge
). But this means that $\exists$ could have taken $\alpha = \gamma$ in
the current round, and not extended the graph. This is contrary to
our original assumption, and completes the proof.



Let $\c A$ be the rainbow algebra defined above.
There is a countable cylindric  algebra $\c A'$ such that $\c A'\equiv\c
A$ and \pe\ has a \ws\ in $H(\c A')$.
We have seen that for $n<\omega$ \pe\ has a \ws\ $\sigma_n$ in $H_n(\c A)$.  We can assume that $\sigma_n$ is deterministic.
Let $\c B$ be a non-principal ultrapower of $\c A$.  We can show that
\pe\ has a \ws\ $\sigma$ in $H(\c B)$ --- essentially she uses
$\sigma_n$ in the $n$'th component of the ultraproduct so that at each
round of $H(\c B)$ \pe\ is still winning in co-finitely many
components, this suffices to show she has still not lost.  
Now use an
elementary chain argument to construct countable elementary
subalgebras $\c A=\c A_0\preceq\c A_1\preceq\ldots\preceq\c B$.  For this,
let $\c A_{i+1}$ be a countable elementary subalgebra of $\c B$
containing $\c A_i$ and all elements of $\c B$ that $\sigma$ selects
in a play of $H_\omega(\c B)$ in which \pa\ only chooses elements from
$\c A_i$. Now let $\c A'=\bigcup_{i<\omega}\c A_i$.  This is a
countable elementary subalgebra of $\c B$ and \pe\ has a \ws\ in
$H(\c A')$.
\end{demo}

\section{The class of neat reducts}
 Here we study the class of neat reducts. 

\begin{theorem}
\begin{enumarab}
\item  Let $n>1$ and $\alpha\geq \omega$. Then the class $\Nr_n\CA_{\alpha}=\Nr_n\RCA_{\alpha}$ is pseudo-elementary, but is not elementary.
Furthermore, $El\Nr_n\CA_{\omega}\subset \RCA_n$, $EL\Nr_n\CA_{\omega}$ is recursively enumerable, and for $n>2$ is 
not finitely axiomatizable.
\item The class $\CCA_n$ of dimension $n\geq 3$, is also psuedo elementary and furthermore, its elementary closure
is  not finitely axiomatixable
\end{enumarab}
\end{theorem}

\begin{demo}{Proof}  We first show that for any infinite $\alpha$, $\Nr_n\CA_{\omega}=\Nr_n\CA_{\alpha}$. Let $\A\in \Nr_n\CA_{\omega}$, 
so that $\A=\Nr_n\B'$, $\B'\in \CA_{\omega}$. Let $\B=\Sg^{\B'}A$. Then $\B\in \Lf_{\omega}$, and $\A=\Nr_n\B$. 
But $\Lf_{\omega}=\Nr_{\omega}\Lf_{\alpha}$ and we are done.
To show that $\Nr_n\CA_{\omega}\subseteq \Nr_n\RCA_{\omega}$, let $\A\in \Nr_n\CA_{\omega}$, then by the above argument
there exists  
then $\B\in \Lf_{\omega}$ such that $\A=\Nr_n\B$. by $\Lf_{\omega}\subseteq \RCA_{\omega},$ we are done. 


Hence It is known that class $\Nr_n\CA_{\omega}$ is not elementary. In fact, there is an algebra $\A\in \Nr_n\CA_{\omega}$ 
having a complete subalgebra $\B$, and $B\notin \Nr_n\CA_{n+1}$ 

To show that it is pseudo-elementary, we use a three sorted defining theory, with one sort for a cylindric algebra of dimension $n$ 
$(c)$, the second sort for the Boolean reduct of a cylindric algebra $(b)$
and the third sort for a set of dimensions $(\delta)$. We use superscripts $n,b,\delta$ for variables 
and functions to indicate that the variable, or the returned value of the function, 
is of the sort of the cylindric algebra of dimension $n$, the Boolean part of the cylindric algebra or the dimension set, respectively.
The signature includes dimension sort constants $i^{\delta}$ for each $i<\omega$ to represent the dimensions.
The defining theory for $\Nr_n\CA_{\omega}$ incudes sentences demanding that the consatnts $i^{\delta}$ for $i<\omega$ 
are distinct and that the last two sorts define
a cylindric algenra of dimension $\omega$. For example the sentence
$$\forall x^{\delta}, y^{\delta}, z^{\delta}(d^b(x^{\delta}, y^{\delta})=c^b(z^{\delta}, d^b(x^{\delta}, z^{\delta}). d^{b}(z^{\delta}, y^{\delta})))$$
represents the cylindric algebra axiom ${\sf d}_{ij}={\sf c}_k({\sf d}_{ik}.{\sf d}_{kj})$ for all $i,j,k<\omega$.
We have have a function $I^b$ from sort $c$ to sort $b$ and sentences requiring that $I^b$ be injective and to respect the $n$ dimensional 
cylindric operations as follows: for all $x^r$
$$I^b({\sf d}_{ij})=d^b(i^{\delta}, j^{\delta})$$
$$I^b({\sf c}_i x^r)= {\sf c}_i^b(I^b(x)).$$
Finally we require that $I^b$ maps onto the set of $n$ dimensional elements
$$\forall y^b((\forall z^{\delta}(z^{\delta}\neq 0^{\delta},\ldots (n-1)^{\delta}\rightarrow c^b(z^{\delta}, y^b)=y^b))\leftrightarrow \exists x^r(y^b=I^b(x^r))).$$
For ${\A}\in \CA_n,$ $\Rd_3{\A}$ denotes the $\CA_3$
obtained from $\A$ by discarding all operations indexed by indices in $n\sim 3.$
$\Df_n$ denotes the class of diagonal free cylindric algebras.
$\Rd_{df}{\A}$ denotes the $\Df_n$ obtained from $\A$
by deleting all diagonal elements. To prove the non-finite axiomatizability result we use Monk's algebras.
For $3\leq n,i<\omega$, with $n-1\leq i, {\C}_{n,i}$ denotes
the $\CA_n$ associated with the cylindric atom structure as defined on p. 95 of 
\cite{HMT1}.
Then by \cite[3.2.79]{HMT1}
for $3\leq n$, and $j<\omega$, 
$\Rd_3{\C}_{n,n+j}$ can be neatly embedded in a 
$\CA_{3+j+1}$.  (1)
By \cite[3.2.84]{HMT1}) we have for every $j\in \omega$, 
there is an $3\leq n$ such that $\Rd_{df}\Rd_{3}{\cal C}_{n,n+j}$
is a non-representable $\Df_3.$ (2)
Now suppose $m\in \omega$. By (2), 
choose $j\in \omega\sim 3$ so that $\Rd_{df}\Rd_3{\C}_{j,j+m+n-4}$
is a non-representable $\Df_3$. By (1) we have
$\Rd_{df}\Rd_3{\C}_{j,j+m+n-4}\subseteq \Nr_3{\B_m}$, for some 
${\B}\in \CA_{n+m}.$
Put ${\A}_m=\Nr_n\B_m$.  
$\Rd_{df}{\A}_m$ is not representable, 
a friotri, ${\A}_m\notin \RCA_{n},$ for else its $\Df$ reduct would be 
representable. Therefore $\A_m\notin EL\Nr_n\CA_{\omega}$.
Now let $\C_m$ be an algebra similar to $\CA_{\omega}$'s such that $\B_m=\Rd_{n+m}\C_m$.
Then $\A_m=\Nr_n\C_m$. Let $F$ be a non-principal ultrafilter on $\omega$. Then
$$\prod_{m\in \omega}\A_m/F=\prod_{m\in \omega}(\Nr_n\C_m)/F=\Nr_n(\prod_{m\in \omega}\C_m/F)$$
But $\prod_{m\in \omega}\C_m/F\in \CA_{\omega}$. Hence $\CA_n\sim El\Nr_n\CA_{\omega}$ is not closed under ultraproducts.
It follows that the latter class is not finitely axiomatizable.
In \cite{IGPL} it is proved that for $1<\alpha<\beta$, $El\Nr_{\alpha}\CA_{\beta}\subset S\Nr_{\alpha}\CA_{\beta}$.
\end{demo}
From the above proof it follows that
\begin{corollary} Let $K$ be any class such that $\Nr_n\CA_{\omega}\subseteq K\subseteq \RCA_n$. Then $ELK$ is not finitely axiomatizable
\end{corollary}


\subsection{Weakly but not strongly neat atom structures}

\begin{definition} An atom structure of dimension $\alpha$ is strongly neat, 
if any algebra having this atom structure is in $\Nr_{\alpha}\CA_{\alpha+\omega}$.
It is weakly neat if one algebra based on it is neat reduct, very weakly neat if no algebra based on it is a neat reduct.
\end{definition}

Since there are representable  algebras that are not neat reducts all algebras we are about to construct will be variation on themes
of representable algebras constructed in previous publications.
For completely representable algebras an atom structure is completely representable if it has at least one algebra having this atom structure
that is completely representable. This implies that all algebras based on its atom structure are comletely representable.

Here we show that this not the case with neat atom structures. For this puropse, we construct a weakly neat atom structure, that is not strongly neat.
Similar constructions were used to show that the class of 
neat reduct is not closed under forming subalgebras, and that there are isomorphic algebras that generate 
non -isomorphic ones in extra dimensiona, and that neat subreduct may not be full neat reducts, 
Such  construction, and several related ones, were also used to confirm unresolved  a conjecture of Tarski. 
The construction is joint with Istvan N\'emeti.

We put this construction to another new use,
and therefore we find it appropriate to give the details with several variants that  appeared in three  
publications of the author. 
The latter also shows in addition that there are representable cylindric algebras, satisfying the merry go round identies 
but cannat be a reduct of a $QEA$s.

\begin{theorem} For every ordinal $\alpha>1$, there exists a weakly neat atom structure of dimension $\alpha$, 
that is not strongly neat.
\end{theorem}
\begin{demo}{Proof}
Let $\alpha>1$ and $\F$ is field of characteristic $0$. 
Let 
$$V=\{s\in {}^{\alpha}\F: |\{i\in \alpha: s_i\neq 0\}|<\omega\},$$
%$$W=\{s\in {}^{\beta}\F: |\{i\in \alpha: {\sf }_i\neq 0\}<\omega\}$$ 
Note that $V$is a vector space over the field $\F$. 
We will show that $V$ is a weakly neat atom structure that is not strongly neat.
Indeed $V$ is a concrete atom structure $\{s\}\equiv _i \{t\}$ if 
$s(j)=t(j)$ for all $j\neq i$, and
$\{s\}\equiv_{ij} \{t\}$ if $s\circ [i,j]=t$.

Let $\C$ be the full complex algebra of this atom structure, that is
$${\C}=(\wp(V),
\cup,\cap,\sim, \emptyset , V, {\sf c}_i,{\sf d}_{ij}, {\sf s}_{ij})_{i,j\in \alpha}.$$  
Then clearly $\wp(V)\in \Nr_{\alpha}\CA_{\alpha+\omega}$.
Indeed Let $W={}^{\alpha+\omega}\F^{(0)}$. Then
$\psi: \wp(V)\to \Nr_{\alpha}\wp(W)$ defined via
$$X\mapsto \{s\in W: s\upharpoonright \alpha\in X\}$$
is an isomomorphism from $\wp(V)$ to $\Nr_{\alpha}\wp(W)$.
We shall construct an algebra $\A$ such that $\At\A\cong V$ but $\A\notin \Nr_{\alpha}\CA_{\alpha+1}$.

Let $y$ denote the following $\alpha$-ary relation:
$$y=\{s\in V: s_0+1=\sum_{i>0} s_i\}.$$
Note that the sum on the right hand side is a finite one, since only 
finitely many of the $s_i$'s involved 
are non-zero. 
For each $s\in y$, we let 
$y_s$ be the singleton containing $s$, i.e. $y_s=\{s\}.$ 
Define 
${\A}\in WQEAs_{\alpha}$ 
as follows:
$${\A}=\Sg^{\C}\{y,y_s:s\in y\}.$$
We shall prove that  
$$\Rd_{SC}\A\notin \Nr_{\alpha}SC_{\alpha+1}.$$ 
That is for no $\mathfrak{P}\in SC_{\alpha+1}$, it is the case that $\Sg^{\C}X$ exhausts the set of all $\alpha$ dimensional elements 
of $\mathfrak{P}$. 
So assume, seeking a contradiction,  that $\Rd_{SC}{\A}\in \Nr_{\alpha}SC_{\alpha+1}$. 
Let $X=\{y_s:s\in y\}$. 
Of course every element of $X,$ being a singleton, is an atom.
Next we show that $\A$ is atomic, i.e evey non-zero element contains a minimal non-zero element.  
Towards this end, let $s\in {}^{\alpha}\F^{(\bold 0)}$ be an arbitrary sequence. 
Then 
$$\langle s_0,s_0+1-\sum_{i>1}s_i,s_i\rangle_{i>1}$$ 
and 
$$\langle \sum_{0< i<\alpha}s_i-1,s_i\rangle_{i\geq 1}$$ are elements in y.
Since 
$$\{s\}={\sf c}_1\{\langle s_0,s_0+1-\sum_{i>1}s_i,s_i\rangle_{i>1}\}\cap
{\sf c}_0\{\langle \sum_{0\neq i<\alpha}s_i-1,s_i\rangle_{i\geq 1}\},$$ 
it follows
that $\{s\}\in A$. 
We have shown that $A$ contains all singletons, hence is atomic. Call an element rectangular if ${\sf c}_0x\cap {\sf c}_1x=x$.
As easily checked, every singleton is rectangular.
Also $y=\bigcup X$, and so $y=\sup X$ exists in ${\A}$.
Finally, let $x=\{s\}$ be an atom of ${\A}$.
Then 
$${\sf c}_0x\cap y=\{\langle \sum_{i>0}s_i-1,s_i\rangle_{i>0}\}$$ 
and
$${\sf c}_1x\cap y=\{\langle s_0,s_0+1-\sum_{i>1}s_i,s_i\rangle_{i>1}\};$$ 
which are singletons, hence atoms.
%For the sake of brevity let 
%$$Y=\{\tau(a,b):a,b\in X\}.$$
%Having shown that ${\A}$.
%We now show that $k=\sup Y$ exists in $\A$. 
Let $$\tau(x,y) ={\sf c}_1({\sf c}_0x\cdot {\sf s}_1^0{\sf c}_1y)\cdot {\sf c}_1x\cdot {\sf c}_0y.$$
Let $$Y=\{\tau(a,b):a,b\in X\}.$$
We now show that, given that $\A$ is a neat reduct, $k=\sup Y$ exists in $\A$. 
Let $$\tau_{\alpha}(x,y) = {\sf c}_{\alpha}({\sf s}_{\alpha}^1{\sf c}_{\alpha}x\cdot {\sf s}_{\alpha}^0{\sf c}_{\alpha}y).$$
Then $\tau_{\alpha}(x,y)\leq \tau({\sf c}_{\alpha}x,{\sf c}_{\alpha}y)$
and $x,y$ are $\alpha$-closed and rectangular, then
$\tau_{\alpha}(x,y)=\tau(x,y)$.
Indeed, 
computing we get
$$\tau_{\alpha}(x,y)={\sf c}_{\alpha}
({\sf s}_{\alpha}^1{\sf c}_{\alpha}x\cap {\sf s}_{\alpha}^0{\sf c}_{\alpha}y)$$ 
$$\leq {\sf c}_{\alpha}({\sf s}_{\alpha}^1
({\sf c}_0{\sf c}_{\alpha}x\cap {\sf c}_1{\sf c}_{\alpha}x)\cap {\sf s}_{\alpha}^0
({\sf c}_0{\sf c}_{\alpha}y\cap {\sf c}_1{\sf c}_{\alpha}y))$$
(Here, equality holds if $x,y$ are $\alpha$-closed and rectangular)
$$={\sf c}_{\alpha}({\sf s}_{\alpha}^1{\sf c}_0{\sf c}_{\alpha}x\cap 
{\sf s}_{\alpha}^1{\sf c}_1{\sf c}_{\alpha}x\cap {\sf s}_{\alpha}^0
{\sf c}_0{\sf c}_{\alpha}y\cap {\sf s}_{\alpha}^0{\sf c}_1{\sf c}_{\alpha}y).$$
$$={\sf c}_{\alpha}({\sf s}_{\alpha}^1{\sf c}_0{\sf c}_{\alpha}x\cap {\sf c}_1{\sf c}_{\alpha}x\cap {\sf c}_0{\sf c}_{\alpha}y\cap {\sf s}_{\alpha}^0{\sf c}_1{\sf c}_{\alpha}y)$$
$$={\sf c}_{\alpha}({\sf s}_{\alpha}^1{\sf c}_0{\sf c}_{\alpha}x\cap {\sf s}_{\alpha}^0{\sf c}_1{\sf c}_{\alpha}y)\cap {\sf c}_1{\sf c}_{\alpha}x\cap {\sf c}_0{\sf c}_{\alpha}y$$
$$={\sf c}_{\alpha}({\sf s}_{\alpha}^1{\sf c}_0{\sf c}_{\alpha}x\cap {\sf s}_{\alpha}^1{\sf s}_1^0{\sf c}_1{\sf c}_{\alpha}y)\cap {\sf c}_1{\sf c}_{\alpha}x\cap {\sf c}_0{\sf c}_{\alpha}y$$
$$={\sf c}_{\alpha}{\sf s}_{\alpha}^1({\sf c}_0{\sf c}_{\alpha}x\cap {\sf s}_1^0{\sf c}_1{\sf c}_{\alpha}y)\cap {\sf c}_1{\sf c}_{\alpha}x\cap {\sf c}_0{\sf c}_{\alpha}y$$
$$={\sf c}_1{\sf s}_1^{\alpha}({\sf c}_0{\sf c}_{\alpha}x\cap {\sf s}_1^0{\sf c}_1{\sf c}_{\alpha}y)\cap {\sf c}_1
{\sf c}_{\alpha}x\cap {\sf c}_0{\sf c}_{\alpha}y$$
$$={\sf c}_1{\sf s}_1^{\alpha}{\sf c}_{\alpha}({\sf c}_0{\sf c}_{\alpha}x\cap {\sf s}_1^0{\sf c}_1{\sf c}_{\alpha}y)\cap {\sf c}_1
{\sf c}_{\alpha}x\cap {\sf c}_0{\sf c}_{\alpha}y.$$
$$={\sf c}_1{\sf c}_{\alpha}({\sf c}_0{\sf c}_{\alpha}x\cap {\sf s}_1^0{\sf c}_1{\sf c}_{\alpha}y)
\cap {\sf c}_1{\sf c}_{\alpha}x\cap {\sf c}_0{\sf c}_{\alpha}y$$
$$={\sf c}_1({\sf c}_0{\sf c}_{\alpha}x\cap {\sf s}_1^0{\sf c}_1{\sf c}_{\alpha}y)
\cap {\sf c}_1{\sf c}_{\alpha}x\cap {\sf c}_0{\sf c}_{\alpha}y$$
$$=\tau({\sf c}_{\alpha}x,{\sf c}_{\alpha}y).$$  
Let $k=\tau_{\alpha}^{\B}(y,y)$, 
where ${\B}$ is any algebra in $SC_{\alpha+1}$
such that $\Rd_{SC}{\A}=\Nr_{\alpha}\B$. Now $k\in A$.
Let $a,b\in X$, and $i\in \{0,1\}.$ 
SiSince ${\sf s}_{\alpha}^i$ is a boolean endomorphism for $i<\alpha$
%\footnote {This term is evaluated in the big algebra ${\B}$, so strictly speaking we should 
%write ${\sf s}_{\alpha}^i{}^{\B}$; but from now on we omit superscripts.
%It will be clear from context which terms are evaluated in the fixed (but arbitrary)
%$\B.$},   
we have $${\sf s}_{\alpha}^0a\leq {\sf s}_{\alpha}^0y\text { and }
{\sf s}_{\alpha}^1b\leq {\sf s}_{\alpha}^1y.$$
But $a,b$ are rectangular and together with $y$ are $\alpha$-closed,
we obtain that
$$\tau(a,b)={\sf c}_{\alpha}({\sf s}_{\alpha}^1a\cap {\sf s}_{\alpha}^0b)
\leq {\sf c}_{\alpha}({\sf s}_{\alpha}^1y\cap {\sf s}_{\alpha}^0y)=k.$$
This shows that $k\in B$, is an upper bound of $Y$.
We claim  that $k$ is in fact the least such upper bound, i.e. that $k=\sup Y.$
Towards this end, suppose that $t$ is an upper bound of $Y$ in $\A$.
We have to show that $k\leq t$.
Assume, to the contrary, that this is not the case, i.e. that $k-t\neq 0$.
Since ${\A}$ is atomic, there is an atom $z$ say, below $k-t$ 
i.e.  $$z\leq k\text { and }z\leq -t.$$                            
From $0\neq z\leq w={\sf c}_{\alpha}({\sf s}_{\alpha}^1y\cap {\sf s}_{\alpha}^0y)$, and $z$
is $\alpha$-closed, it readily follows that 
$$z\cap ({\sf s}_{\alpha}^1y\cap {\sf s}_{\alpha}^0y)\neq 0$$ thus 
$${\sf c}_0z\cap {\sf c}_1z\cap ({\sf s}_{\alpha}^1y\cap {\sf s}_{\alpha}^0y)\neq 0.$$
From ${\sf s}_{\alpha}^i{\sf c}_iz={\sf c}_iz$, and the fact that the ${\sf s}_{\alpha}^i$'s are
boolean endomorphisms for each $i<\alpha$, we get  
$${\sf s}_{\alpha}^1({\sf c}_1z\cap y)\cap {\sf s}_{\alpha}^0({\sf c}_0z\cap y)\neq 0.$$
Now let $a= {\sf c}_1z\cap y$. Then  
$a$ is a singleton, hence $a\in At{\C}$. Also $a\leq y$.
But $y=\sup X$ and $a\cap y=a\neq 0$, therefore $a\cap x\neq 0$, for some
atom $x\in X$, and so $a=x\in X$.
Analogously, if $b={\sf c}_0z\cap y$, then $b\in X$.
Being atoms, $a$ and $b$ are rectangular, and therefore we have from
$$\tau(a,b)={\sf c}_{\alpha}({\sf s}_{\alpha}^1({\sf c}_1z\cap y)\cap {\sf s}_{\alpha}^0({\sf c}_0z\cap y)).$$
and that  
$\tau(a,b) > 0.$

On the other hand, by invoking definitions and the fact that $z$ is 
rectangular a straightforward computation yields:
$$\tau(a,b)=\tau({\sf c}_1z\cap y, {\sf c}_0z\cap y)$$
$$= {\sf c}_1({\sf c}_0({\sf c}_1z\cap y)\cap {\sf s}_1^0{\sf c}_1({\sf c}_0z\cap y))\cap {\sf c}_1({\sf c}_1z\cap y)\cap 
{\sf c}_0({\sf c}_0z\cap y)).$$
$$={\sf c}_1({\sf c}_0({\sf c}_1z\cap y)\cap {\sf s}_1^0{\sf c}_1({\sf c}_0z\cap y))\cap 
({\sf c}_1z\cap {\sf c}_1y)\cap ({\sf c}_0z\cap {\sf c}_0y).$$
$$={\sf c}_1({\sf c}_0({\sf c}_1z\cap y)\cap {\sf s}_1^0{\sf c}_1({\sf c}_0z\cap y))\cap 
({\sf c}_0z\cap {\sf c}_1z)\cap ({\sf c}_0y\cap {\sf c}_1y).$$
$$={\sf c}_1({\sf c}_0({\sf c}_1z\cap y)\cap {\sf s}_1^0{\sf c}_1({\sf c}_0z\cap y))\cap z\cap ({\sf c}_0y\cap {\sf c}_1y).$$
That is, $0<\tau(a,b)\leq z$. But $z$ is an atom, and so $z=\tau(a,b)$.
Since $t$ is an upper bound for $Y$, we get $z\leq t$.
But we have $z\leq -t$. Hence $z=0$ which is impossible,
since $z$ is an atom. 
This means that $k\leq t$, and so $k=\sup Y$ as required. 
We will determine $k.$ 
We start by evaluating $\tau(a,b)$ for $a,b\in X$.
Towards this end, let $a,b\in X$. Assume that $a=y_r$ and $b=y_t$ 
with 
$$r=\langle r_i:i<\alpha\rangle\text { and }t=\langle t_i:i<\alpha\rangle$$ 
are elements in 
$y$.
$$\text { If } r_1\neq t_0\text { or } r_i\neq t_i \text { for some }i>1, \text { then }
\tau(y_r,y_s)=0.$$
Else, $r_1=t_0$ and $r_i=t_i$ for all $i>1$.
In this case, 
$$\tau(y_r,y_t)=\{s\}\text { where }s_0=r_0, \text { and } s_i=t_i \text { for all } i>0.$$
Moreover, an easy computation shows that 
$$s_0+2=s_1+2\sum_{i>1}s_i.$$
Now let $w$ be the set of all such elements, i.e.
$$w=\{s\in {}^{\alpha}\F^{(\bold 0)}: s_0+2=s_1+2\sum_{i>1}s_i\}.$$
Let $s\in w$. Then putting 
$$r=\langle s_0,s_0+1-\sum_{i>0}s_i,s_i\rangle_{i>1}$$ 
and 
$$t=\langle s_0+1-\sum_{i>0}s_i,s_1,s_i\rangle_{i>1},$$ 
we get $r,t\in y$ and $\tau(y_r,y_s)=\{s\}$.
We have shown that $$Y\sim \{0\}=\{\{s\}:s\in w\},$$ 
and so $$w=\bigcup Y.$$
Now clearly $w\leq k$. We will show that $k\leq w$.
But this easily follows from the fact that $A$ contains all singletons.
In more detail, let $z\in {}^{\alpha}\F^{(\bold 0)}\sim w$. 
Then $\{z\}$, and hence $^{\alpha}\F^{(\bold 0)}\sim \{z\}\in A$. 
Since $w\subseteq ^{\alpha}\F^{(\bold 0)}\sim \{z\}$, 
we have $\tau(a,b)\subseteq {}^{\alpha}\F^{(\bold 0)}\sim \{z\}$, 
for every $a,b\in X$, i.e. 
$^{\alpha}\F^{(\bold 0)}\sim \{z\}$ is an upper bound of of $Y$. 
But this means that $k\subseteq  V\sim \{z\}$, i.e. $z\notin k$.
We have proved that $k\leq w$.  
Next we proceed to show that, in fact, $w\notin A$. 
(That is $w$ is a new $\alpha$ dimensional element).
This contradiction will show that $\Rd_{SC}\A\notin \Nr_{\alpha}SC_{\beta}.$
Let
$$Pl=\{\{s\in {}^{\alpha}\F^{(\bold 0)}:t+\sum (r_is_i)=0\}:\{t,r_i:i<\alpha\}\subseteq \F\}.$$
$Pl$ consists of all hyperplanes of dimension $\alpha$. 
For $i<\alpha$, let 
$$q_i=\{s\in {}^{\alpha}\F^{(\bold 0) }:s_i+1=\sum_{j\neq i}s_j\}.$$
Let $i\in \alpha$ and $k,l\in \alpha$ be distinct.
Then if $i\in \{k,l\}$, $i=k$ say, then 
$${\sf s}_{kl}q_i=q_l.$$
Else, we have $i\notin \{k,l\}$, so that $i,k,l$ are pairwise distinct,
in which case we have 
$${\sf s}_{kl}q_i=q_i.$$ 
Now let 
$$Pl^S=\{q_i:i<\alpha\}.$$ 
Then by the above $Pl^S$ is closed under the operations
${\sf s}_{kl}$ for all $k,l\in \alpha$. 
Notice that $y=q_0\in Pl^S$. Thus $Pl^S$
consists of all hyperplanes obtained by interchanging the $k,l$ co-ordinates
of $y$, for all $k,l\in \alpha$, i.e. implementing the operation ${\sf s}_{kl}$. 
Here the superscript
$S$ is short for substitutions (corresponding to replacements) 
reflecting the fact that  $Pl^S$ consists of {\it substituted}  versions of $y$. Let
$$Pl^<=\{p\in Pl: {\sf c}_ip=p, \text { for some }i<\alpha\}.$$
If $p\in Pl^<$ and ${\sf c}_ip=p$, then 
$p$ is a hyperplane parallel to the $i$-th axis.

Note  that for $p\in Pl$, $p=\{s\in {}^{\alpha}\F^{(\bold 0)}:t+\sum_ir_is_i=0\}$ say, then
${\sf c}_ip=p$ (i.e. $p$ is parallel to the $i$-th axis) iff $r_i=0$. 
Note too, that if $p\in Pl^<$ and $k,l\in \alpha$ then ${\sf s}_{kl}p\in Pl^<$.

In the following we summarize the above, 
and state some other facts easy to check as well.
$$(1)\ \ y=q_0\in Pl^S.$$
$$(2)\  \ Pl^S\cup \{w,{\sf d}_{ij}:i,j\in \alpha\}\subseteq Pl.$$
$$(3)\ \text { If }q\in Pl^S (Pl^<) \text { and }i,j\in \alpha, 
\text { then } {\sf s}_{ij}q\in Pl^S(Pl^<).$$
$$(4)\  \ Pl^S\cap Pl^<=\emptyset, w\notin Pl^<\text { and }{}^{\alpha}\F^{(\bold 0)}\in Pl^<.$$
$$(5)\ \{{\sf d}_{ij}:i\neq j, i,j\in \alpha\}\subseteq Pl^<\text { iff }\alpha\geq 3.$$ 


For further use, we shall need:

\begin{athm}{Notation} 

\begin{enumroman}

\item Let $X$ be a set. Then $Z\subseteq_{\omega}X$ abbreviates 
``$Z$ is a finite subset $X$". $\wp_{\omega}X$ denotes all finite subsets of $X$, i.e.
$$\wp_{\omega}X=\{Z\in \wp(X): Z\subseteq_{\omega} X\}.$$

\item Let ${\A}\in K_{\alpha}$, $a\in A$ and $\Delta\subseteq_{\omega}\alpha$.
Then ${\sf c}_{(\Delta)}a={\sf c}_{{i_0}}\cdots {\sf c}_{i_{n-1}}a$, 
where $\Delta=\{i_0,\cdots i_{n-1}\}$. 
Because cylindrifications commute, 
the definition of ${\sf c}_{(\Delta)}$ does not depend on any
linear order defined on $\Delta$.  
When $n=0$, i.e. $\Delta$ is empty, then ${\sf c}_{(\Delta)}a={\sf c}_{\emptyset}a:=a$. 
\item To simplify notation, from now on we may write $-a$ 
instead of $^{\alpha}\F^{(\bold 0)}\sim a.$
It will be clear from context whether $-$ refers to the operation of substraction in $\F$
or the operation of complementation in $\A.$
We also write $1^{\A}$, or just $1,$ 
instead of the more cumbersome 
$^{\alpha}\F^{(\bold 0)}.$

\end{enumroman}

\end{athm}


{\bf Some more definitions.}
$$G=\{q,-q,\ p,-p,\ {\sf c}_{(\Delta)}\{\bold 0\},
-{\sf c}_{(\Delta)}\{\bold 0\}:\ q\in Pl^S,\ p\in Pl^<\cup \{{\sf d}_{01}\},\ \Delta\subseteq_{\omega} \alpha,\  0\in \Delta\}.$$
Forming finite intersections of elements in $G$, we let
$$G^*=\{\bigcap_{i\in n}g_i:n\in \omega,g_i\in G\},$$
and forming finite unions of elements in $G^*$, we let
$$G^{**}=\{\bigcup_{i\in n}g_i:n\in \omega, g_i\in G^*\}.$$
It is easy to see that $\{y,y_s:s\in y\}\subseteq G^{**}$, and $G^{**}$ 
is a boolean field of sets.
We start by proving that $w\notin G^{**}$. To this end, we set:
$$L =\{p\in Pl^<: {\sf c}_0p\neq p\}\text { and } P(0)=L\cup \{{\sf d}_{01}\}.$$
Notice that $L = P(0)$ iff $\alpha>2$.
$L$ stands for the set of ``lines" 
not parallel to the $0$th axis. 
Being in $Pl^<$, any such line is parallel to some other axis. 
When $\alpha=2$, i.e. in the  plane $\F\times \F$, these 
are precisely the lines that are parallel to the ``$y$ axis".
Next we define
$$G_1=\{g\in G^*:g\subseteq q,\text{ for some }q\in Pl^S\}$$ and
$$G_2=\{g\in G^*:g\not\subseteq q,\text{ for all }q\in Pl^S \text 
{ and }g\subseteq p, \text{ for some } p\in P(0)\}.$$ 
Note that $Pl^S\subseteq G_1$ and that $P(0)\subseteq G_2$.
Note too, that $G_1\cap G_2=\emptyset$.
Now let 
$$G_3=\{p_1\cap p_2\ldots \cap p_k:\ k\in \omega,\ 
\{p_1,p_2,\ldots, p_k\}\subseteq G\sim (Pl^S\cup P(0))\}.$$
It is easy to see that $G^*=G_1\cup G_2\cup G_3$.
To prove that $w\notin G^{**}$ we need:

\begin{athm}{Claim} If $g\in G_3$ and $0\neq g$, then $g\not\subseteq w$.
\end{athm}


{\bf Proof of Claim}

Assume that $g=p_1\cap p_2\ldots  \cap p_k$ say, with 
$p_i\in G$ and $p_i\notin (Pl^S\cup P(0))$
for $1\leq i\leq k$, and let $z\in g.$
Then for $1\leq i\leq k$, 
we have in addition that 
$$p_i\in \{p,-p, {\sf c}_{(\Delta)}\{\bold 0\}, -{\sf c}_{(\Delta)}\{\bold 0\}, -q: p\in Pl^<\cup \{{\sf d}_{01}\}, q\in Pl^S, \Delta\subseteq_{\omega}\alpha,
0\in \Delta\}.$$ 
Let $[{}]$ be the function from $G$ into $\wp(\F)$ defined as follows: 
$$[p] = \{1/r_0(-t-\sum_{0\neq i<\alpha} r_iz_i)\} \text { if }p=-\{s\in ^{\alpha}\F^{(\bold 0)}:t+\sum_{i<\alpha} r_is_i=0\}, 
r_0\neq 0.$$ 
Else  
$$[p]=0.$$
Let $$r\in \F\sim ((\bigcup_{1\leq i\leq k}[p_i])\cup [-w])$$ 
be arbitrary,
and let $$z_r^0=z\sim \{(0,z_0)\}\cup \{(0,r)\}.$$
By the choice of $r\notin [-w]$ we have
$$r\neq -2+z_1+2\sum_{i>1} z_i.$$ 
Hence $z_r^0\notin w$. If $p_i=-q$ with  $q\in Pl^S$, then similarly, $r\notin [-q]$ hence $z_r^0\notin p_i$.
Now assume that $p_i\in Pl^<$. Then we have $p_i\notin P(0)$ hence ${\sf c}_0p_i=p_i.$
Since $z\in p_i$ it follows that $z_r^0\in {\sf c}_0p_i.$ But $p_i\notin P(0)$ hence $z_r^0\in {\sf c}_0p_i=p_i.$
If $p_i=-p$ where $p\in Pl^<$, then $z\notin p$ and so $z_r^0\notin p$ since again ${\sf c}_0p=p$.
That is if $z_r^0\in p$ then $z$ differing from $z_r^0$ in atmost the $0$th place would also be in $p$.
Now assume that $p_i={\sf c}_{(\Delta)}\{\bold 0\}$. Then if $z\in p_i$ then $z_r^0\in p_i$ since ${\sf c}_0p_i=p_i$ by $0\in \Delta$.
Finally if $p_i=- {\sf c}_{(\Delta)}\{\bold 0\}$ then $z\notin {\sf c}_{(\Delta)}\{\bold 0\}$ iff $z_r^0\notin {\sf c}_{(\Delta)}\{\bold 0\}$ by $0\in \Delta$. 
We have shown that 
$$z_r^0\in g\sim w\text { i.e. }g\not\subseteq w.$$
%(Here we are using that 
%when ${\sf c}_{(\Delta)}\{\bold 0\}\in G$, then $0\in \Delta$.)


We now proceed to show that $w\notin G^{**}$.
Assume to the contrary, that 
$$w=\bigcup\{g_i^1:i<n_1\}\cup\bigcup\{g_i^2:i<n_2\}\cup\bigcup \{g_i^3:i<n_3\}$$ 
where 
$$\{g_i^j:i<n_j\}\subseteq G_j\text { for }j\in \{1,2,3\}.$$
By the above, we have $g_i^3=0$, for all $i<n_3$.
From the definition of $G_1$ and $G_2$ , it follows that
$$w\subseteq \bigcup\{q_j:j<n_1\}\cup\bigcup \{p_j:j<n_2\}$$ 
where 
$$\{q_j:j<n_1\}\subseteq
Pl^S$$ 
and 
$$\{p_j:j<n_2\}\subseteq P(0).$$
Since $w\subseteq -{\sf d}_{01}$ for $\alpha=2$, and $P(0)=L$ for $\alpha>2$ 
we can, and will, assume that
$$\{p_j:j<n_2\}\subseteq L.$$
Before proceeding, let us take a special case as an illustration. Assume that $\alpha=3$.
Then each $p_j$ is a hyperplane that is determined by an equation.
Assume that the following equations determine the planes $p_j$ $j<n_2$.
$$(1)\ \ \ \ t_0+r_{00}x_0+r_{01}x_1+r_{02}x_2=0$$
$$\ \ \ \ t_1+r_{10}x_0+r_{11}x_1+r_{12}x_2=0$$
$$\cdots$$
$$\cdots$$
$$\ \ \ \ t_{n_2-1} +r_{n_2-1,0}x_0+ r_{n_2-1, 2}x_2=0.$$
In each of these equation one of the coefficients other than the zeroth is equal to $0$. The zeroth coefficient is {\it not} zero.
Therefore, these equations determine hyperplanes parallel to one of the axis
other than the zeroth axis.
Now consider the equations
$$(2)\ \ \ \ x_0+2=x_1+2x_2$$ 
and
$$(3)\ \ \ \ x_{\pi(0)}+1=x_{\pi(1)}+x_{\pi(2)}$$
where $\pi$ is a permutation of $\{0,1,2\}$.
$(3)$ consists of $6$ equations, only three of which are distinct (because of commutativity). Such equations 
represent the $q_i$'s.
Then it can be easily seen that there is an $s$ that satisfies 
(2) but does not satisfy (the equations in) $(1)$ and $(3)$.


And indeed in the general case, it can be seen by implementing easy linear algebraic 
arguments that, for every $n\in \omega$, 
for every $m\in \omega\cap (\alpha+1)$,
and for every  system
$$t_0+\sum(r_{0i}x_i)=0$$
$$\cdots$$
$$\cdots$$
$$t_n+\sum(r_{ni}x_i)=0,$$
of equations, such that for all $j\leq n$, there exists $i<\alpha$, such that
$$r_{ji}=0\text { and }r_{j0}\neq 0,$$
the equation 
$$x_0+2=x_1+2\sum x_j$$ 
has a solution $s$ in the weak space
$^{\alpha}\F^{(\bold 0)}$, such that
$s\notin q_i$ for every $i<m$, and such that $s$ is not a solution of
$$t_j+\sum (r_{ji}x_i)=0,$$ for every $j\leq n$.
Geometrically (and intuitively) 
a finite union of hyperplanes 
cannot cover a hyperplane, 
unless it is one of them.
By taking $n=n_1$ and $m=n_2$ we get 
the desired conclusion.
Let us prove this  formally.  It clearly suffices to show that for all positive integers $m$, $n$ the equation 
$$x_0+2=x_1+2\sum_{1< i \leq m} x_i$$
 has a solution which is not a solution of any of the following equations:
$$(4)\ \ \ \ t_k+\sum_{j\leq m}r_{kj}x_j=0,$$
$$(5)\ \ \ \ x_l+1=\sum_{1\neq i\leq m}x_i,$$
where $k=1,\ldots , n$, each $r_{k0}\neq 0$, for each $k=1,\ldots , n$ there is a $j$ with $0<j\leq m$ such that $r_{kj}=0$, and $0\leq l\leq m.$
To do this, first substitute $x_0=-2+x_1+2\sum_{1<i\leq m}x_i$ into each of the equations $(4)$ and $(5)$, obtaining
$$(6) \ \ \ \ t_k-2r_{k0}+(r_{k0}+r_{k1})x_1+\sum_{1<j\leq m}(2r_{k0}+r_{kj})x_j=0,$$
$$(7) \ \ \ \ -1+\sum_{1<j\leq m} x_j=0,$$
$$(8) \ \ \ \ 3+\sum_{1<j\leq m}(- 3 x_j)=0,$$
$$(9)\ \ \ \ 3-2x_1-x_k+\sum_{j\leq m, j\notin \{0,1,k\}}(-3x_j)=0\text { for } k>1.$$
Now we define $s_1,\ldots s_m$ by recursion. Choose $s_1$ so that $$t_k-2r_{k0}+(r_{k0}+r_{k1})s_1\neq 0$$
for each $k$ such that $r_{k0}+r_{k1}\neq 0$. Having defined $s_t$ with $t<m$ choose $s_{t+1}$ so that
$$t_k-2r_{k0}+(r_{k0}+r_{k1})s_1+\sum_{1<j\leq t+1}(2r_{k0}+r_{kj})s_j\neq 0$$
for each $k$ for which $2r_{k0}+r_{k,t+1}\neq 0$; also, if $t+1=m$, assure that the equations $(7)- (9)$ all fail.
Finally let $s_0=-2+s_1+2\sum_{1<i}s_i.$
Clearly the desired conclusion holds. 
We have proved that $w\notin G^{**}$. .

To show that 
$w\notin A$, we will  show that $G^{**}$ is closed under 
the polyadic set operations.
It only remains to show that $G^{**}$ 
is closed under cylindrifications and substitutions, 
since by definition, it is a boolean field of sets and 
contains the diagonal elements. ( Recall that for $\alpha>2$, ${\sf d}_{ij}\in Pl^<$.)



\begin{enumarab}

\item $G^{**}$ is closed under cylindrifications.

It is enough to show that (since the ${\sf c}_i$'s are additive), that
for $j\in \alpha$ and $g\in G^*$ arbitrary, we have ${\sf c}_jg\in G^{**}$.
For this purpose, put for every $p \in Pl$
$$p(j|0)={\sf c}_j\{s\in p: s_j=0\}\text { and } (-p)(j|0)=-p(j|0).$$  
Then it is not hard to see that 
$$p(j|0)=\{s\in {}^{\alpha}\F^{(\bold 0)}:t+\sum_{i\neq j}(r_is_i)=0\},$$
if $$p=\{s\in {}^{\alpha}\F^{(\bold 0)}:t+\sum_{i<\alpha}(r_is_i)=0\},$$ 
 
%$$p(j|0)\in Pl^<\text { for every }p\in Pl.$$
Indeed assume that $s_1\in {\sf c}_j\{s\in p: s_j=0\}$.
Then there exists $s_2$ in $p$ such that $s_1$ and  $s_2$ agree at all components except the $j$th where 
$s_2(j)=0$.
Then $$0=t+\sum_{i\neq j} r_is_1(i)=t+\sum_{i\neq j}r_is_2(i).$$
The other inclusion is analogous. Assume that $s_1\in p$ and $r_j\neq 0$ such that $t+\sum_{i\neq j}r_is_1(i)=0.$
Then $s_1(j)=0$. Hence $s_1\in {\sf c}_j\{s\in P: s_j=0\}.$ 
It follows thus that  
$$p(j|0)\in Pl^<\text { for every }p\in Pl.$$

Now let $j$ and $g$ be as indicated above. We can assume that
$$g=q_1\cdots \cap q_l\cap p_1\cap\cdots \cap p_n\cap  -Q_1\cdots \cap 
-Q_L$$
$$\cap -P_1\cdots \cap -P_m\cap y\cap -{\sf c}_{(\Delta_1)}
\{\bold 0\}\cdots\cap -{\sf c}_{(\Delta_N)}\{\bold 0\},$$ 
where
$$l, L, n, m, N\in \omega, q_i, Q_i\in Pl^S, 
p_i, P_i\in Pl^<\cup \{{\sf d}_{01}\},$$ 
$${\sf c}_jp_i\neq p_i,\  {\sf c}_jP_i\neq P_i,\ {\sf c}_jq_i\neq q_i,\  {\sf c}_jQ_i\neq Q_i,$$ 
$$y\in \{{\sf c}_{(\Delta)}\{\bold 0\},1:\Delta\in \wp_{\omega}\alpha,\ 0\in \Delta,\ j\notin \Delta\},$$ 
and 
$$\{\Delta_1,\cdots,\Delta_n\}\subseteq \{x\in \wp_{\omega}\alpha:\ j\notin x, 0\in x\}.$$
This is so because $${\sf c}_j(x\cap {\sf c}_jy)={\sf c}_jx\cap {\sf c}_jy,$$
$$\text { if } j\notin \Delta x\text { then }{\sf c}_jx=x$$ 
and $${\sf c}_{(\Delta)}\{\bold 0\}\cap {\sf c}_{(\Gamma)}\{\bold 0\}={\sf c}_{(\Delta \cap \Gamma)}\{\bold 0\}.$$

We distinguish between $2$ cases:

{\bf Case 1.}
$$y={\sf c}_{(\Delta)}\{\bold 0\}\text { and }j\notin \Delta.$$ 
Then a lenghly but routine computation gives
$${\sf c}_j(q_1\cdots \cap q_l\cap -Q_1\cdots \cap 
-Q_L\cap p_1\cdots\cap p_n\cap -P_1\cdots \cap -P_m$$
$$\cap {\sf c}_{(\Delta)}\{\bold 0\}\cap -{\sf c}_{(\Delta_1)}\{\bold 0\}\cdots \cap 
-{\sf c}_{(\Delta_N)}\{\bold 0\})$$
$$=q_1(j|0)\cap q_l(j|0)\cap -Q_1(j|0)\cdots \cap 
-Q_L(j|0)\cap p_1(j|0)\cdots $$
$$\cap p_n(j|0)\cap -P_1(j|0)\cdots \cap -P_m(j|0)$$
$$\cap {\sf c}_j
{\sf c}_{(\Delta)}\{\bold 0\}\cap -{\sf c}_j{\sf c}_{(\Delta_1)}\{\bold 0\}\cdots \cap -
{\sf c}_j{\sf c}_{(\Delta_N)}\{\bold 0\}.$$
Indeed let $$s\in 
{\sf c}_j(q_1\cdots \cap q_l\cap -Q_1\cdots \cap 
-Q_L\cap p_1\cdots\cap p_n\cap -P_1\cdots \cap -P_m$$
$$\cap {\sf c}_{(\Delta)}\{\bold 0\}\cap -{\sf c}_{(\Delta_1)}\{\bold 0\}\cdots \cap 
-{\sf c}_{(\Delta_N)}\{\bold 0\}).$$
For a sequence $t$, we write $t_u^j$ for the sequence that agrees with $t$ except at $j$ where $t_u^j(j)=u$.
Now, by definition of cylindrifications, there exists $u\in \F$ such that 
$$s_u^j\in 
(q_1\cdots \cap q_l\cap -Q_1\cdots \cap 
-Q_L\cap p_1\cdots\cap p_n\cap -P_1\cdots \cap -P_m$$
$$\cap {\sf c}_{(\Delta)}\{\bold 0\}\cap -{\sf c}_{(\Delta_1)}\{\bold 0\}\cdots \cap 
-{\sf c}_{(\Delta_N)}\{\bold 0\}).$$
Now since $s_u^j\in q_1$ and $s_u^j\in {\sf c}_{(\Delta)}\{\bold 0\}$ and $j\notin \Delta$ it readily follows that $u=0$.
But then $s\in q_1(j|0).$ Similarly $s\in q_i(j|0)$ for every $i\leq l$ and $s\in p_i(j|0)$ for each $i\leq n$. It is clear that $s\in {\sf c}_j{\sf c}_{(\Delta)}\{\bold 0\}$.
Also $s\notin Q_i(j|0)$ for any $i\leq L$ for else $s_0^j\in Q_i$
which is not the case.
Same reasoning gives $s\notin P_i(j|0)$ for all $i\leq m$. 
Finally $s\notin {\sf c}_j{\sf c}_{(\Delta_i)}\{\bold 0\}$ for $i\leq N$, because $s_u^j\notin {\sf c}_{(\Delta_j)}\{\bold 0\}$.
Now conversely if we start with 
$$s\in q_1(j|0)\cap q_l(j|0)\cap -Q_1(j|0)\cdots \cap 
-Q_L(j|0)\cap p_1(j|0)\cdots $$
$$\cap p_n(j|0)\cap -P_1(j|0)\cdots \cap -P_m(j|0)$$
$$\cap {\sf c}_j
{\sf c}_{(\Delta)}\{\bold 0\}\cap -{\sf c}_j{\sf c}_{(\Delta_1)}\{\bold 0\}\cdots \cap -
{\sf c}_j{\sf c}_{(\Delta_N)}\{\bold 0\}.$$
Then $$s_0^j\in 
(q_1\cdots \cap q_l\cap -Q_1\cdots \cap 
-Q_L\cap p_1\cdots\cap p_n\cap -P_1\cdots \cap -P_m$$
$$\cap {\sf c}_{(\Delta)}\{\bold 0\}\cap -{\sf c}_{(\Delta_1)}\{\bold 0\}\cdots \cap 
-{\sf c}_{(\Delta_N)}\{\bold 0\}.)$$



{\bf Case 2.}
$$y=1$$
This case is harder, so we start with some special cases before embarking on the most general case.
Consider the case when
$$g=q \cap p
\cap - P\cap  -{\sf c}_{(\Delta)}\{\bold 0\},$$ 
where $q\in Pl^S$, and $p,P\in Pl^<$. Then 
$${\sf c}_jg={\sf c}_j(q\cap p)\cap {\sf c}_j(q-P)\cap {\sf c}_j(q-{\sf c}_{(\Delta)}\{\bold 0\}).$$
$$\cap {\sf c}_j(p -P)\cap {\sf c}_j(p-{\sf c}_{(\Delta)}\{\bold 0\}).$$
Let us prove this special case to visualise matters. The general case will be a little bit more involved, but all the same an immediate generalization.
Let $$s\in {\sf c}_j(q\cap p)\cap {\sf c}_j(q-P)\cap {\sf c}_j(q-{\sf c}_{(\Gamma)}\{\bold 0\})$$
$$\cap {\sf c}_j(p -P)\cap {\sf c}_j(p-{\sf c}_{(\Delta)}\{\bold 0\}).$$
Then there exist $u_1, u_2, u_3, u_4, u_5 \in \F$ such that 
$$s^j_{u_1}\in q\cap p, s^j_{u_2}\in q-P, s^j_{u_3}\in q-{\sf c}_{(\Gamma)}\{\bold 0\},$$ 
$$s^j_{u_4}\in p-P\text {  and  }s^j_{u_5}\in p-{\sf c}_{(\Delta)}\{\bold 0\}.$$
Now ${\sf c}_jp\neq p$ it follows that $u_4=u_5$. 
Similarly $u_1$ and $u_2$ and $u_3$ are also equal because ${\sf c}_jq\neq q$.
But note that $s^j_{u_1}\in p\cap q$ and ${\sf c}_jq\neq q$, 
it follows that all of the $u_i$'s are in fact equal to $u$ say.
It readily follows that $$s^j_u\in q \cap p
\cap - P\cap  -{\sf c}_{(\Delta)}\{\bold 0\},$$
hence
$$s\in {\sf c}_j(q \cap p
\cap - P\cap  -{\sf c}_{(\Delta)}\{\bold 0\}).$$
The other inclusion is much easior, in fact it is absolutely straightforward.
Start with  $$s\in {\sf c}_j(q \cap p
\cap - P\cap  -{\sf c}_{(\Delta)}\{\bold 0\}).$$
Then $$s^j_u\in (q \cap p
\cap - P\cap  -{\sf c}_{(\Delta)}\{\bold 0\}).$$
It follows that 
$$s^j_{u}\in q\cap p, s^j_{u}\in q-P, s^j_{u}\in q-{\sf c}_{(\Gamma)}\{\bold 0\},$$ 
$$s^j_{u}\in p-P\text {  and  }s^j_{u}\in p-{\sf c}_{(\Delta)}\{\bold 0\}.$$
The required follows.
Now assume (still considering a special case) that 
$$g=q\cap p_1\cap\cdots \cap p_n\cap \cdots \cap -P_1\cdots \cap -P_m$$
$$\cap -{\sf c}_{(\Delta_1)}
\{\bold 0\}\cdots\cap -{\sf c}_{(\Delta_N)}\{\bold 0\},$$ 
then by the same token
 $${\sf c}_jg=\cap_{k\leq n}{\sf c}_j(q\cap p_k)\cap \cap_{i\leq m}{\sf c}_j(q-P_i)\cap_{i\leq N}{\sf c}_j(q-{\sf c}_{\Delta_N)}\{\bold 0\})$$
$$\cap_{k\leq n}((\cap_{i\leq n}{\sf c}_j(p_k\cap p_i)\cap_{i\leq m}{\sf c}_j(p_k-P_i)\cap_{i\leq N}{\sf c}_j(p_k- {\sf c}_{(\Gamma)}\{\bold 0\})).$$
To illustrate matters further, paving the way for the general case, consider another special case
that is essentially different than the one just considered.
$$g=-Q\cap p_1\cdots p_n
\cap -P_1\cdots \cap -P_m$$
$$\cap -{\sf c}_{(\Delta_1)}
\{\bold 0\}\cdots\cap -{\sf c}_{(\Delta_N)}\{\bold 0\}.$$ 
Then 
$${\sf c}_jg=\cap_{k\leq n}{\sf c}_j(p_k-Q)$$
$$\cap_{k\leq n}((\cap_{i\leq n}{\sf c}_j(p_k\cap p_i)\cap_{i\leq m}{\sf c}_j(p_k-P_i)\cap_{i\leq N}{\sf c}_j(p_k-{\sf c}_{(\Delta_i)}\{\bold 0\})).$$
Now consider the general case.
We assume for better readability that 
 $$g=p_1\cap\cdots \cap p_n\cap \cdots \cap -P_1\cdots \cap -P_m$$
$$\cap -{\sf c}_{(\Delta_1)}
\{\bold 0\}\cdots\cap -{\sf c}_{(\Delta_N)}\{\bold 0\},$$ 
where $p_i, P_i\in Pl^<\cup Pl^S$. Here we have collected the planes in $Pl^<$ and $Pl^S$ together to simplify matters.
Now we have by the same reasoning as above
$${\sf c}_jg=\cap_{k\leq n}((\cap_{i\leq n}{\sf c}_j(p_k\cap p_i)\cap \cap_{i\leq m}{\sf c}_j(p_k-P_i)$$
$$\cap_{i\leq N}{\sf c}_j(p_k-{\sf c}_{(\Delta_i)}\{\bold 0\})).$$
%$$\cap_{k\leq n}((\cap_{i\leq n}{\sf c}_j(p_k.p_i).\cap_{i\leq m}p_k-P_i)\cap_{i\leq N}{\sf c}_j(p_k-{\sf c}_{(\Delta_i)}\{\bold 0\})).$$
Now for every 
$p,q\in Pl$, there are $p',q',p''$ and $q''\in Pl^<$
such that (*)
$${\sf c}_j(p\cap q)=p'\cap q',$$ 
$${\sf c}_j(p\sim q)=p''\sim q''$$ 
and if  $j\in \Delta p\sim \Gamma,$ 
then (**)
$${\sf c}_j(p \sim {\sf c}_{(\Gamma)}\{\bold 0\})=
{}^{\alpha}\F^{(\bold 0)}\sim p(j|0)
\cup (p(j|0)\sim {\sf c}_j{\sf c}_{(\Gamma)}\{\bold 0\}).$$
Let us consider (*). We give an illustration when $\alpha=2$
and $\F=\R$. In this case elements in $Pl^<$ are either the whole plane or horizontal lines or vertical lines.
If $p$ and $q$ are any two straight lines that are not parallel, then they intersect
at a point. If we cylindrify (this point) with respect to one of the co-ordinates we get a line that is indeed parallel to one of the axis.
Next, if we have two straight lines $p$ and $q$, say, that intersect and neither is parallel to one of the axis, then $p-q$ will be $p$ ``minus" one point (the point of intersection). 
If we cylindrify with respect to one of the co-ordinates then we get the plane
minus the straight line that is parallel to one of the axis; call this straight line $p'$. 
But this is nothing more than $1-p'$.  On the other hand, if $p$ is parallel to one of the axis and we cylindrify $p$
in the direction of this axis, we get $p-q$ where $q$ is the line orthogonal to $p$ and passes by the missing point, hence $q$ is parallel to the other axis.
If $p$ and $q$ are parallel
then $p-q=p$ and so for $j\in \{0,1\}$, we have ${\sf c}_j(p-q)={\sf c}_jp\in Pl^<$.

Now we turn to (**). Assume that $p=\{s\in {}^0\F^{(\bold 0)}:  t+\sum_{i}r_is_i=0\}$. Let $j\in \Delta p\sim \Gamma$. 
Let $s\in {\sf c}_j(p-{\sf c}_{(\Gamma)}\{\bold 0\})$. Then there is some
$u\in \F$ such that $s_u^j\in p$ and $s_u^j\notin  {\sf c}_{(\Gamma)}\{\bold 0\}$.
Since $j\in \Delta p$ we have ${\sf c}_jp\neq p$ hence $r_j\neq 0$. Let $u=1/r_j(t+\sum_{j\neq i}r_is_i)$.
Then we have two cases. If $u\neq 0$, then $s\notin p(j|0)$ for else there exists $q\in p$ such that $q$ differs from $s$ in at most the $j$th component and $q(j)=0$.
But $q(j)=-1/r_0(t+\sum_{i\neq j}r_is_i)=u$. This would imply that $u=0$ which is a contradiction.
Else $u=0$. Then $s\in p(j|0)$ because $s_u^j\in p$ and $s_u^j(j)=0$.
Also $s\notin {\sf c}_j{\sf c}_{(\Gamma)}\{\bold 0\}$ for else $s_u^j\in {\sf c}_{(\Gamma)}\{\bold 0\}.$
We leave the other inclusion to the reader.
We have proved that ${\sf c}_jg\in G^{**}$.
\item $G^{**}$ is closed under substitutions.

It is enough to consider the ${\sf s}_{ij}$'s, since for all $i,j\in \alpha$
${\sf s}_i^j(x)={\sf c}_i(x\cap {\sf d}_{ij})$ i.e. 
the ${\sf s}_i^j$'s are term definable, $G^{**}$ is 
closed under the ${\sf c}_i$'s and ${\sf d}_{ij}\in G^{**}$.
For this purpose, let 
$$H= Pl^S\cup Pl^<\cup \{{\sf c}_{(\Delta)}{\bold \{\bold 0\}}:\Delta\subseteq_{\omega}\alpha\}.$$ 
Here we are not requiring that $0\in \Delta$,
so that $H$ is bigger than $G$.
However, since $G^{**}$ is closed under cylindrifications we still have
$G\subseteq H\subseteq G^{**}$. 
Now let $i,j\in \alpha$, and $x\in H$. 
Then it not hard to check that ${\sf s}_{ij}x\in G^{**}$. 
Indeed $Pl^S$ is closed under ${\sf s}_{ij}$ and so is $Pl^<$.
Assume that $y={\sf c}_{(\Delta)}\{\bold 0\}$, $\Delta\subseteq_{\omega} \alpha$,  and $i,j<\alpha$ are distinct. If $\{i,j\}\subseteq \Delta$ or $\{i,j\}\cap \Delta=\emptyset$, then
${\sf s}_{ij}{\sf c}_{(\Delta)}\{\bold 0\}={\sf c}_{(\Delta)}\{\bold 0\}$. 
Now suppose that $i\in \Delta$, $j\notin \Delta$ and that ${\sf c}_{(\Delta}\{\bold 0\}=V_1\times V_2\times V_l\ldots $ where $V_l\in \{\F, \{0\}\}$
and $V_l=\F$ only for finitely many $l$'s.
Assume further that $i<j$,  $V_i=\F$ and $V_j=\{0\}$ so that 
${\sf c}_{(\Delta}\{\bold 0\}=V_1\times\ldots  \F\times \ldots \{0\}\times \ldots.$
Then ${\sf s}_{ij}{\sf c}_{(\Delta)}\{\bold 0\}$ is obtained from ${\sf c}_{(\Delta)}\{\bold 0\}$ by interchanging
the $ith$ and $jth$ co-ordinates, that is it is equal to  $V_1\times\ldots  \{0\}\times \ldots \F\times \ldots.$, which is of the form ${\sf c}_{(\Gamma)}\{\bold 0\}$
for some $\Gamma\subseteq_{\omega}\alpha$. In fact $\Gamma=(\Delta\sim \{i\})\cup \{j\}.$
By noting that the ${\sf s}_{ij}$'s are boolean endomorphisms
we get that $G^{**}$, generated as a boolean algebra by $H$ 
(since $G\subseteq H$ is a set of generators),  
is closed under ${\sf s}_{ij}$. 
\end{enumarab}
It thus follows that $w\notin A$ and we are done.
\end{demo}

\section{The class of weakly neat atom structures is not elementary}

From the title, it is clear that there exists atom structures such that are very weakly neat. 
%The atom structure above is strongly neat, and this implies.

\begin{theorem} The class of weakly neat atom structures is not elementary, for finite and infinite dimensions.
\end{theorem}
\begin{proof} For the first part there is an atomic $\A\in \Nr_n\CA_{\omega}$ 
and an atomic $\B$ that is not a neat reduct, elementary equivalent to $\A$. $\At\A$ is srongly neat, but $\At\B$ is not.
But $\At\A\equiv \At\B$, 
because atom structure are first order interpretable in
the algebras, and we are done.
\end{proof}

The second part is harder. The construction is similar, but not identical to the one used in \cite{IGPL}. We prove our result for the harder infinite dimensional case.
The finite dimensional case can be easily destilled from the finite version of our algebras constructed in 
\cite{IGPL}.
Throughout this section $\alpha$ is a fixed arbitrary ordinal with 
$\alpha\geq \omega$ and $R$ is an uncountable set with $|R|>|\alpha|$.

\begin{athm}{Notation}
Let $s$ be an $\alpha$-ary sequence, and $i<\alpha$. Then $s(i|w)$ denotes the  
$\alpha$-ary sequence for which:
$s(i|w)(i)=w$ and $s(i|w)(x)=s(x)$ for $x\neq i$.
For a set $U$, $w\in W$ and $\beta$ an ordinal, 
we let $\bold w$ denote the $\beta$-ary sequence $\beta\times \{w\}$, i.e the constant 
sequence  for which $\bold w_i=w$, for all $i\in \beta$.
The arity of $\bold w$, we hope, will be clear from context.
\end{athm}
%We start by formulating infinite versions of conditions (iii) and (v) of Lemma 1.
\begin{lemma}
There are a set $U$, $w\in U$, and $\alpha$-ary relations 
$C_r\subseteq {}^{\alpha}U^{(\bold w)}$ 
for all $r\in R$, such that conditions (i) and (ii) below hold:
\begin{enumroman}
\item For all $r\in R$, for all $i<\alpha$, for all $s\in {}^{\alpha}U^{(\bold w)}$, 
there exists $y\in U-Rgs$ such that $s(i|y)\in C_r$. 
\item The $C_r$'s ($r\in R$) are pairwise disjoint.
\end{enumroman}
\end{lemma}
%The proof can be recovered from that of  Lemma $1^{**}$ below.
%\par 
\begin{athm}{Notation} Let $\bold U=\langle U, C_r\rangle_{r\in R}$, with 
$C_r\subseteq {}^{\alpha}U^{(\bold w)}$, 
be the structure described in Lemma 1*. 
Let $W=\cup_{i\in \alpha} U\times \{i\}$.
Then $W$ is simply $|\alpha|$ disjoint copies of $U$.
Let $s=\langle (w,0),\cdots,(w,j)\cdots\rangle_{j<\alpha}$. Then $s\in {}^{\alpha}W$
with  $s(i)\in W_i=U\times \{i\}$. 
(Recall that $\bold w=\langle w:i<\alpha \rangle$).
Let $V$ be the weak space ${}^{\alpha}W^{(s)}$ and let 
$C(\alpha) =\langle Sb(V),\cup,\cap,\-c_i,d_{ij}\rangle$ be the full $Ws_{\alpha}$ with unit $V$.
$S_{\alpha}$ denotes the set of one to one functions (equivalently permutations) 
which are almost everywhere identity, i.e, which are in $^{\alpha}\alpha^{(Id)}$.
\end{athm}
%\subsubsection*{Defining our algebras}

Let $u=\langle u_0,u_1,\cdots u_j\cdots \rangle_{j\in \alpha}\in S_{\alpha}$, and $r\in R$. Then 
$p(u,r)$ denotes the following $\alpha$-ary relation on $V$:
$$\{\langle (a_0,u_0),\cdots ,(a_j,u_j)\cdots \rangle_{j<\alpha} \in 
V:\quad\langle a_0,\cdots a_j,\cdots \rangle_{j<\alpha} \in C_r\}.$$
$$P(u)=\{p(u,r): r\in R\}.$$
Note that $P(u)$ consists of $|R|$ many $\alpha$ relations
on $V$. Let $N$ be a fixed countably infinite subset of  $R$.
Then $P_{\omega}(u)=\{p(u,r):r\in N\}$.
Again $P_{\omega}(u)$ countably infinite.
Now we define $\A(\alpha)\in Ws_{\alpha}$ and $\B(\alpha)\subseteq \A(\alpha)$ as follows:
$$\A(\alpha)=\Sg^{C(\alpha)}(\cup \{P(u):u\in S_{\alpha}\})$$
$$\B(\alpha)=\Sg^{A(\alpha)}((\cup \{P(u): u\in S_{\alpha}-\{Id\})\cup P_{\omega}(Id)).$$

\begin{theorem} $\A(\alpha)$ and $\B(\alpha)$ are atomic algebras with elementay equivalent atom structures.
$\At\B$ is not weakly neat, but $\At\A$ is.
\end{theorem}
\begin{proof}The thing to notice here, is that $\A(\alpha)$ is a disjoint union of algebras whose boolean part are isomorphic to the finite-cofinite
Boolean algebra on an uncountable set $R$. We have $^{\alpha}{\alpha}^{(Id)}$ many copies. Hence it is atomic.
Also $\B$ is atomic. We will show that that $\B\equiv \A$ , hence $\At\A\equiv \At\B$, and that any algebra having an atom 
structure isomorphic to $\At\B$ is not a 
neat reduct.
\end{proof}
%\end{theorem}

We shall start by showing that $\A(\alpha)$ and $\B(\alpha)$ 
are elementary equivalent, so that their atom structures will also be elementary equivalent,
and that $\B(\alpha)$ is not a neat reduct; not only that, but any algebra based on its atom structure is not a neat reduct.

Then we shall impose further restrictions on 
the structure $\langle U, C_r\rangle_{r\in R}$, that will $force$ 
$A(\alpha)$ to be a neat reduct ( and not just a subneat reduct). 

\begin{athm}{Notation}
Let $u=\langle u_0,u_1,\cdots u_j\cdots \rangle_{j\in \alpha}\in {}^{\alpha}\alpha^{(Id)}$. 
Then $1_u$ denote the  $\alpha$-ary relation on $U$:
$$\{\langle (a_0,u_0),(a_1,u_1),\cdots (a_j,u_j)\cdots \rangle_{j<\alpha}\in V: \langle a_0,\cdots a_j\cdots 
\rangle_{j\in \alpha}\in {}^{\alpha}U^{(\bold w)}\}.$$
$L$ stands for $1_{Id}$.
\end{athm}  
\begin{theorem}
\begin{enumroman}
\item For all $u\in S_{\alpha}$, $1_u\in A(\alpha)$.
\item Let $Rl_LA(\alpha)=\{x\in A(\alpha):x\leq L\}$. Then $Rl_LA(\alpha)$ 
is atomic and its set of atoms is equal to $P(Id)$.
Furthermore, $Rl_LA(\alpha)$ is generated as a $BA$ by $P(Id)$.
\item For all non zero $a\in Rl_LA(\alpha)$, for all $i<\alpha$, $c_ia=c_iL$.
\item For all $a\in A(\alpha)$, for all $i<\alpha$, $c_ia\cap L\in \{0,L\}$.
\item Let $P=\cup \{P(u): u\in S_{\alpha}\}$. Let $p$ be a permutation of $P(Id)$. Then 
there exists an automorphism $f$ of $A(\alpha)$,
such that $p\subseteq f$ and $f(x)=x$, for each $x\in P-P(Id)$.
\end{enumroman}
\end{theorem}
\begin{demo}{Proof}
\begin{enumroman}
\item Let $u\in S_{\alpha}$. Let $r_1,r_2\in R$. Then as easily checked  
$c_0(p(u,r_1))\cap c_1(p(u,r_2))=1_{u}$.
\item Since the $C_r$'s are pairwise disjoint, we have $p(Id,r_1)\cap p(Id,r_2)=0$ for distinct $r_1$ and
$r_2$. Thus the set of atoms of $Rl_LA(\alpha)$ is precisely $P(Id)$. 
By $c_i(p(Id,r))\cap 1_{Id}=1_{Id}$ and $p(Id,r)\cap d_{ij}=0$, for distinct $i,j<\alpha$, it follows that $Rl_LA(\alpha)$
is indeed generated as a $BA$ by the set $P(Id)$. 
\item Let $a\in Rl_LA(\alpha)$ be non zero. By (ii) there exists an atom $p(Id,r)$ below $a$. 
Let $i<\alpha$. 
Then $c_i(p(Id,r))=$(by Lemma 1 (i)) $ \cup\{1_u:u\in {}^{\alpha}\alpha^{(Id)}$ and $u(j)=j$ for all 
$j\neq i\}= c_iL$. From $p(Id,r)\leq a$, (iii) readily follows. 
\item follows from (iii).
\item We first, like we did in the finite case, extend $p$ to a $BA$ automorphism $p'$, say,  
of $Rl_LA(\alpha)$. 
This is possible by (i). Then we define $f(x)=p'(x\cap L)\cup (x\cap -L)$.
The rest is completely analogous to the proof of fact 4.2.
\end{enumroman}
\end{demo}
\begin{theorem} $\B(\alpha)$ is an elementary subalgebra of $\A(\alpha)$.
\end{theorem}
\begin{proof} This can be proved using the Tarski-Vaught test.
\end{proof}

Now we show that $B(\alpha)$ is not a neat reduct. 
We proceed basically as we did in the finite case, the trick being 
``a(n) (infinite) cardinality twist" that first order logic cannot witness.

\begin{theorem} Let $\Rl_L\B(\alpha)=\{x\in B(\alpha): x\leq L\}$. 
Then $|\Rl_L\B(\alpha)|\leq \alpha$.
\end{theorem}
\begin{demo}{Proof}
First of all note that by 6.1 (i) we have $L\in B$. 
Now let $$X=\cup \{P(u):u\in S_{\alpha}-\{Id\}\}\cup P_{\omega}(Id),$$ 
$$Y=\{c_ia:a\in A\}\cup \{d_{ij}: i,j\in \alpha\}\text { , and }$$ 
$$D(\alpha)=Sg^{Bl{A}(\alpha)}(X\cup Y).$$
Then $|Y|=\alpha$. This can be seen by noting that for every $x\in Y$, if $x$ is non zero, 
and $x$ is not a diagonal element, then by Lemma 1*(ii) 
there is an $S$ with $S\subseteq {}^{\alpha}\alpha^{(Id)}$ such that $x=\cup \{1_u: u\in S\}$.
Also $D(\alpha)$ is a subuniverse of $A(\alpha)$ because of the following:
$D(\alpha)$ is closed under the boolean set operations , contains the 
diagonals, and for $a\in D(\alpha)$, 
and $i<\alpha$, we have $a\in A(\alpha)$ and so $c_ia\in D(\alpha)$, 
that is $D(\alpha)$ is closed under cylindrifications.  
Since $X\subseteq D(\alpha)$, $X$ generates $B(\alpha)$ and 
$D(\alpha)\in CA_{\alpha}$, we have $B(\alpha)\subseteq D(\alpha)$.
Let $rl_L(a)=a\cap L$. Then we have $$Rl_LD(\alpha)=\{x\in D(\alpha): x\leq L\}=
 Sg^{BlA(\alpha)}rl_L(X\cup Y).$$
But $|rl_L(X\cup Y)|=\alpha$, because $P_{\omega}(Id)$ is countable
and $|Y|=\alpha$.
Therefore $|rl_L(X\cup Y)|=\alpha$, and so $|Rl_LD(\alpha)|\leq \alpha$.
By $Rl_LB(\alpha)\subseteq Rl_LD(\alpha)$ we are done. 
\end{demo}
\begin{theorem}
Let $p[0,1]$ be the transposition on $\alpha$ 
interchanging $0$ and $1$. 
Let $v=p[0,1]\circ Id=\langle 1023\cdots \rangle$. Then $v \in S_{\alpha}$ and the 
following hold: 
\begin{enumroman}
\item $t^{B(\alpha)}(1_{v})=1_{Id}$
\item For all $\beta>\alpha$, $CA_{\beta}\models {}_{\alpha}s(0,1)c_{\alpha}x\leq t(c_{\alpha}x)$.
\item For all $\beta>\alpha$ and 
$A\in CA_{\beta}$, $_{\alpha}s(0,1)\in Ism(BlNr_{\alpha}A, BlNr_{\alpha}A)$.
\end{enumroman}
\end{theorem}
%The proof is identical to the proof of fact 4.2. Therefore we omit it.
\begin{theorem}
$\B(\alpha)\notin \Nr_{\alpha}CA_{\alpha+1}$.
\end{theorem}
\begin{demo}{Proof}
If $B(\alpha)$ were a neat reduct then 
we would have $|Rl_L(B(\alpha)|=|R|$. 
But this is a contradiction.
\end{demo}

$\A(\alpha)$ constructed so far may or may not be a neat reduct. 
We now impose further restrictions on the $C_r$'s (or rather on the structure 
$\bold U=\langle U, C_r\rangle$), described in Lemma 1*, that {\it force} 
$A(\alpha)$ to be in $\cap_{k\in \omega}Nr_{\alpha}Ws_{\alpha+k}$.
These restrictions are described in Lemma 1** below.
Forming a ``limit" out of the $Ws_{\alpha+k}$'s ($k\in \omega$) , the $neat- \alpha$ 
reduct of which is $A(\alpha)$, we will show that $A(\alpha)\in Nr_{\alpha}Ws_{\alpha+\beta}$ for 
infinite $\beta$ as well.
This should be  done , of course, in such a way that does 
not interfere with what we have already 
established, namely:\par\noindent
$B(\alpha)$ is $not$ a neat reduct.\par\noindent
$A(\alpha)$ and $B(\alpha)$ are elementary equivalent.\par\noindent


To formulate Lemma 1**  we fix some needed notation:
\begin{athm}{Notation} Let $1\leq k<\omega$. Then $S(\alpha,\alpha+k)$ or 
simply 
$$S(\alpha,k)=\{i\in {}^{\alpha}(\alpha+k): \alpha+k-1\in Rgi\quad\text{and }
\{m\in \alpha:|\{i(m)\neq m\}|<\omega\}.$$ 
That is $S(\alpha,k)=\{i\in {}^{\alpha}(\alpha+k)^{(Id)}: \alpha+k-1\in Rgi\}$.
\par\noindent
$Cof^+R$ denotes the set of all non empty finite or cofinite subsets of $R$.
Let $C_r$ be an $\alpha$-ary relation symbol for all $r\in R$.
\par\noindent
For any $X\subseteq R$, $X$ finite we define the infinitary formulas:
$$\eta(X)=\lor\{C_r(x_0,\cdots x_j\cdots )_{j<\alpha}\, r\in X\},\quad\text{and }$$
$$\eta(R-X)=\land \{\neg C_r(x_0,\cdots x_j\cdots )_{j<\alpha}:r\in X\}.$$
\end{athm}
\par\noindent
$\eta(X)$ and $\eta(R-X)$ are {\it restricted} formulas, in the sense of [HMT2] sec 4.3. 
Satisfiability for such formulas (by $\alpha$-ary sequences) are defined the usual way. 
Below we shall have occasion to deal with (infinitary) formulas that are not restricted. 
These however can be obtained from restricted ones using  the (logical interpretation of the) 
$CA$ operations, i.e quantification on finitely many variables and equality. 
Now we are ready to formulate the infinite version of Lemma 1.



\begin{theorem}
There are a set $W$, an equivalence relation 
$E$ with $\alpha$-many blocks on $W$, $s\in {}^{\alpha}W$ with 
$s_i\in W_i$ the $i$th block, $i\in \alpha$, 
and $\alpha$-ary relations $C_r$ on $W$ for all $r\in R$,
such that conditions (i)-(v) below hold:
\begin{enumroman}
\item (Blocks) $C_r(w_0,w_1\cdots w_i\cdots )_{i<\alpha}$
implies  $D_E(w_0,w_1\cdots w_i\cdots )_{i<\alpha},$ for all $r\in R$ and $w_i\in W$.
Here $D_E(w_0, w_1\cdots w_i\cdots)_{i<j<\alpha}$ means that for any $i<j<\alpha$, 
$w_i,w_j$ are not $E$-equivalent i.e. they are in distinct blocks. 

\item (Symmetry) For all $f\in {}^{\alpha}W^{(s)}$ for all $r\in R$, for all permutations
$\pi\in ^{\alpha}\alpha^{(Id)}$, if $f\in C_r$ then $f\circ \pi\in C_r.$ 

\item (Bigness) For all $r\in R$, for all $i<\alpha$ 
and $v\in {}^{\alpha}W^{(s)}$ such that $D[v_i]_{i<\alpha}$
there exists $x\in W$ such that $v[i|x]\in C_r$.

\item (Saturation) For all $1\leq k<\omega$, for all 
$v\in {}^{\alpha+k-1}W^{(s)}$ one to one,  for all $x\in W$, for any
function $g:S(\alpha,k)\to Cof^+R$ 
for which $\{i\in S(\alpha,k):|\{g(i)\neq R\}|<\omega\}$, 
there is a $v_{\alpha+k-1}\in W\smallsetminus Rgv$ such that 
$xEv_{\alpha+k-1}$, i.e. $v_{\alpha+k-1}$ is in the same block as $x$,  
and  
$$\bigwedge \{D(v_{i_j})_{j<\alpha}\implies \eta(g(i))[\langle v_{i_j}\rangle]: 
i\in S(\alpha,k)\}.$$
\item (Disjointness) The $C_r$'s are pairwise disjoint.
\end{enumroman}
\end{theorem}
\begin{demo}{Proof}
Let $k\geq 1$. For brevity set 
$$Y(\alpha,R,k)=\{g\in {}^{S(\alpha,k)}Cof^+R: \{i\in S(\alpha,k):|\{g(i)\neq R\}|<\omega\}.$$
Let
$$Q=\cup\{I(^{\alpha+k-1}|R|)\times Y(\alpha,R,k):1\leq k<\omega\},$$
then $|Q|=|R|=\kappa$, say. Here $I(^{\alpha+k-1}|R|)$ stands for 
the set of all one to one functions
from $\alpha+k-1$ into $R$. 
Let $\rho$ be an enumeration of $Q$ such that:
for all $l<\kappa$, for all $q\in Q$, there exists $j$ with $l<j<\kappa$ such that $\rho(j)=q$.
Fix a well ordering $\prec$ of $R$.
let $l<\kappa$ and suppose that for all $\mu<l$ we have already defined the element
$x_{\mu}$, and the $\alpha$-ary relation 
$C_r^{\mu}\subseteq {}^{\alpha}W_{\mu}^{(s)}$, where $W_{\mu}=\{x_k:k<\mu\}$.
Assume that $\rho(l)=\langle \langle \beta_j \rangle_{j<\alpha+k-1},f\rangle$.
Then $\beta_j\in |R|$ for all $j<\alpha+k-1$, and $f\in Y(\alpha,R,k)$.
Let $x_l$ be an element not in $\cup \{W_{\mu}:\mu<l\}$.
If there exists $\mu<\alpha+k-1$ such that $l\leq \beta_{\mu}$, 
then for all $r\in R$ we define
$C_r^l=\cup \{C_r^{\mu}: \mu<l\}$. Else,  $l>\beta_{\mu}$ for all $\mu<\alpha+k-1$.
Let $x_{\mu}=x_{\beta_{\mu}}$ for $\mu<\alpha+k-1$ and let $x_{\alpha+k-1}=x_l$.
For all $r\in R$ and $ i\in S(\alpha,k)$ we let 
$$\langle u_{i_j}\rangle_ {j<\alpha}\in X_r^l\text { iff }r \text { is the }\prec-
\text { least element of }f( i)$$
and
$$C_r^l=\cup \{C_r^i:i<l\}\cup \{\langle t_{\pi(0)},t_{\pi(1)},\cdots \rangle:t\in X_r^l\text 
{ and }\pi\in S_{\alpha}\}.$$
Finally set 
$$W=\cup \{W_l:l<\mu\}\quad\text{and } C_r=\cup\{C_r^l:l<\mu\}.$$ 
It is not hard to check just like we did in the finite dimensional case,  that 
$\bold W=\langle W,C_r\rangle_{r\in R}$ is as desired.
\end{demo}


\par\noindent

\bigskip

Let $\bold W=\langle W, C_r\rangle$ and $s\in {}^{\alpha}W$ 
be as specified above
Then $W$ is a disjoint union of $\alpha$ copies. We let $W_i$ denote the $i$th copy.
We shall construct our algebras having unit $^{\alpha}W^{(s)}.$
Recall that $S_{\alpha}$ stands for the set of permutations in $^{\alpha}\alpha^{(Id)}$.
Let $u\in S_{\alpha}$ and $r\in R$. Then we set
$$p(u,r)=C_r\cap W_{u_0}\times W_{u_1}\times \cdots W_{u_i}\times\cdots \cap {}^{\alpha}W^{(s)}.$$
$$P(u)=\{p(u,r): r\in R\}.$$
Let $N$ be a fixed countable subset of $R.$
Let $$P_{\omega}(u)=\{p(u,r):r\in R\}.$$
With a slight abuse of notation we let $A(\alpha)$ and $B(\alpha)$ be the (new)  algebras 
consructed out of the (new) structure  $\bold W=\langle W, C_r\rangle_{r\in R}$, 
described in the previous lemma
That is $A(\alpha)\in Ws_{\alpha}$ and $B(\alpha)\subseteq A(\alpha)$ 
are defined as follows:
$$ A(\alpha)=Sg^C(\cup \{P(u): u\in S_{\alpha}\});$$
and
$$B(\alpha)=Sg^C((\cup \{P(u): u\in S_{\alpha}\smallsetminus \{Id\})\cup P_{\omega}(Id)).$$
Here $C$ is the full weak set algebra with greatest element $^{\alpha}W^{(s)}$, with
$s$ as specified in Lemma $1^{**}$. Recall that $s(i)\in W_i$, the $i$ th copy.
It is easy to check that  $A(\alpha)$ and $B(\alpha)$, so defined,
are (still) elementary equivalent (copy the proof of facts 6.1-2) 
and $B(\alpha)$ is (still) not a neat reduct (copy the proof of facts 6.3-6.5).  
Like in the finite case, we will use the new condition of 
``saturation" or  ``elimination of quantifiers" 
expressed in Lemma $1^{**}$ (iv) to  show that 
$A(\alpha)\in \cap_{k\in \omega}Nr_{\alpha}Ws_{\alpha+k}$. 
To ``lift" $A(\alpha)$ to arbitrary finite extra dimensions, we shall further need:

\subsubsection*{Some more Definitions.} Let $0\leq k<\omega$. Then we let
$W_k=W=\cup_{i<\alpha} W_i$. Fix $w\in W$. 
Let $s_k=s\cup \{\langle i,w\rangle: \alpha\leq i<(\alpha+k) \}$. 
Then $s_k\in {}^{\alpha+k}W$. (When $k=0$, then $s_k$ is just $s$).
Let $V_k$ be the weak space $^{\alpha+k}W^{(s_k)}$ and let 
$C(\alpha+k)$ be the full $Ws_{\alpha+k}$ with unit $V_k$. \par\noindent
Let $u\in {}^{\alpha+k}\alpha^{(Id)}$. 
Then 
$$F_k(u)=\{C_r(x_{i_0},\cdots x_{i_{j}}\cdots )_{j<\alpha}, \neg C_r(x_{i_0},\cdots x_{i_{j}}\cdots )_{j<\alpha}: 
r\in R,\  i\in {}^{\alpha}\alpha+k^{(Id)}$$
$$\text{and }
\langle u_{i_0},\cdots u_{i_j}\cdots \rangle_{j<\alpha}\quad\text{is one to one }\}\quad\cup Eq(u).$$
Recall that $Eq(u)$ stands for $\{x_i=x_j: x_i\neq x_j: (i,j)\in ker(u)\}$.  
Like the finite case we assume  that $\{T,F\}\in F_k(u)$. We put
$$ F_k(u)^*=\{\land J: J\subseteq_{\omega} F_k(u)\}\text { and } $$ 
$$F_k(u)^{**}=\{\lor J: J\subseteq_{\omega} F_k(u)^*\}.$$
Let $u\in {}^{\alpha+k}\alpha^{(Id)}$ and $\phi\in F_k(u)^{**}$. 
\par\noindent
Then we let $E(u,\phi)$ denote the following $(\alpha+k)$-ary relation on $V_k$
$$\{s\in V_k: s_j\in W_{u_j} \text { for all $j<\alpha+k$ and } \bold W\models 
\phi[\langle s_j\rangle_{j<\alpha+k}]\}.$$ 
As noted before, satisfiability is defined the usual way. 
In particular, when $k=0$, $u\in S_{\alpha}$ 
and $r\in R$, then $E(u,C_r(x_0,x_1,\cdots )$ is the same as $p(u,r)$.
\par\noindent
$ind(\phi)$ denotes the set $\{i\in \alpha+k:x_i\in var(\phi)\}$. 
\par\noindent
We let $G(k)=\{c_{(\Delta)}E(u,\phi), -c_{(\Delta)}E(u,\phi):\Delta\subseteq_{\omega} \alpha+k$, 
$u\in {}^{\alpha+k}\alpha^{(Id)},\phi\in F_k(u)^{**},\ \Delta\cap ind(\phi)=0\}$ 
$\cup\{d_{ij}^{V_k},-d_{ij}^{V_k}:i,j\in \alpha+k\}$.
\par\noindent
Here $c_{(\Delta)}$, the unary operation referred to as generalized 
cylindrification  in \cite{HMT1}.  
That is $c_{\emptyset}x=x$ and if $\Delta=\{k_0,\cdots k_{m-1}\}$  
is a non-empty finite subset of $\alpha+k$,  
then $c_{(\Delta)}x=c_{k_0}\cdots c_{k_{m-1}}x$.  
\par\noindent
For $i,j\in \alpha+k$, $d_{ij}^{V_k}$, 
or simply $d_{ij}$ is the diagonal element  $\{s\in V_k:\ s_i=s_j\}$.
\par\noindent
Forming finite intersections of elements in $G(k)$ we let
$$G(k)^*=\{\cap J: J\subseteq_{\omega} G(k)\}$$  
and forming finite unions of elements in $G(k)^*$, we finally let
$$G(k)^{**}=\{\cup J: J\subseteq_{\omega} G(k)^*\}.$$
We will show that $G(k)^{**}$ is a $Ws_{\alpha+k}$ (with unit $V_k$), that 
$G(0)^{**}=Nr_{\alpha}G(k)^{**}$ for all $k\in \omega$, and finally that
$G(0)^{**}=A(\alpha)$. This will show that  
$A(\alpha)\in \cap_{k\in \omega}INr_{\alpha}Ws_{\alpha+k}$.
To prove that $G(k)^{**}$ is a $Ws_{\alpha+k}$ we shall need:

\begin{athm}{Fact 6.6}
\begin{enumroman}
\item Let $u\in {}^{\alpha+k}\alpha^{(Id)}$, 
$\Delta \subseteq_{\omega} \alpha+k$ and $\phi\in F_k(u)^{**}$.
Then $c_{(\Delta)}E(u,\phi)\in G(k)^{**}$.
\item Let $g\in G(k)^{**}$ and $i,j\in \alpha+k$ be distinct.  Then $s_i^j(g)=c_j(g\cap d_{ij})\in G(k)^{**}$.
\item Let $H(k)=\{g\in G(k): g=c_{(\Delta)}E(u,\phi):\Delta\subseteq_{\omega}\alpha+k,u\in {}^{\alpha+k}\alpha^{(Id)}$ 
and $\phi\in F_k(u)^{**}\}$. Then $H(k)$ is closed under finite intersections.
\end{enumroman}
\end{athm}
\begin{demo}{Proof}
\begin{enumroman}
\item We first prove (i) when $\Delta$ consists of one element $j$, say.
That is we will show that for $u$ and $\phi$ as indicated above 
and $j<\alpha+k$ we have $c_jE(u,\phi)\in G(k)^{**}$.
We can and will assume that $\phi\in F_k(u)^*$ since 
$$c_jE(u,\phi_1 \lor\cdots \phi_n)=c_jE(u,\phi_1)\cup\cdots c_jE(u,\phi_n)$$
and $G(k)^{**}$ is closed under finite unions.
Then $\phi=\land J$, where $J\subseteq_{\omega} F_k(u)$.
Next, we  break up $\phi$ into two parts just like we did in the finite dimensional case; that is let
$J^+=\{\psi\in J: x_j\in var(\psi)\}$ and $J^-=\{\psi\in J:x_j\notin var(\psi)\}$.
For brevity set $\phi^+=\land J^+$ and  $\phi^-=\land J^-$,  where by the empty conjunction we understand the formula 
$T$. Now  $\phi=\phi^+\land \phi^-$, $\phi^-\in  F_k(u)^*$
and $x_j$ occurs only in $\phi^-$.
\begin{athm}{Notation} For $i,l\in \alpha+k$ and $\phi\in F_k(u)^{**}$ $\phi[i|l]$ 
stands for the formula obtained from $\phi$ by replacing all
(free and bound) occurences of $x_i$ in $\phi$ by $x_l$. 
\end{athm}
\noindent{\bf Case (a): }
$x_j=x_m\notin J^+$ and $x_j=x_m\notin J^+$ for all $m\in \alpha+k-\{j\}$.\par\noindent
Then either $\bold W\models \phi^+\equiv F$, in which case 
$\bold W\models \exists x_j \phi^+\equiv F$ , and so $c_jE(u,\phi)=c_jE(u,F)=0\in G(k)^{**}$.
Else  $\phi^+$ is satisfiable, in which case , by applying the saturation condition described
in  Lemma $1^{**}$, we get
$$\bold W\models (\exists x_{\alpha+k-1}\phi^+[j|\alpha+k-1])\equiv T,$$ 
which is the same as  $\bold W\models (\exists x_j\phi^+) \equiv T$, 
and so $$c_jE(u,\phi)=c_jE(u,(\exists x_j\phi^+) \land \phi^-)=c_jE(u,\phi^-).$$
The latter is in $G(k)^{**}$, since now  $x_j$ does not occur in $\phi^-$, 
and $\phi^-\in F_k(u)^*$. 
Note that $\phi^-\in F_k(v)^*$ for any $v$ that differs from $u$ in at most the $j$-th place;
and for any such $v$ we have $c_jE(u,\phi)=c_jE(v,\phi^-)$.
\par\noindent{\bf Case (b): }$x_j=x_m\in J^+$ or $x_m=x_j\in J^+$ for some $m\in \alpha+k$ and $m\neq j$.
\par\noindent
Assume that $x_j=x_m\in J^+$. Then , by definition of $F_k(u)$, we have $u_j=u_m$ . 
Also $\phi=(\land I)\land x_j=x_m$, for some $I\subseteq_{\omega} F_k(u)$.
For brevity let $\psi=\land I$. Then of course $\psi\in F_k(u)^*$. 
Computing we get  $$c_jE(u,\phi)=c_jE(u,\psi\land x_j=x_m)=c_jE(u,\psi[j|m]).$$
The latter is in  $G(k)^{**}$ since $x_j$ does not occur in $\psi[j|m]$, 
and $\psi[j|m]\in F_k(u)^*$ because $\psi \in F_k(u)^*$ and $u_j=u_m$.
In fact, it is not hard to see that $\psi[j|m]\in F_k(v)^{*}$, for any $v$ that differs from $u$ in at most the $j$ th place.
The case when $\Delta$ is an arbitrary finite subset of $\alpha+k$, 
follows  by using the above reasoning eliminating the variables occuring in $\phi$, 
with indices in $\Delta$, one by one. Alternatively one can use 
a straightforward induction on $|\Delta|$.
Note that we have actually proved the following infinite analogue of (**) in 
fact 3.1 way above: 
\par\noindent
\begin{athm}{(***)}Let $u\in {}^{\alpha}\alpha+k^{(Id)}$ and $\phi\in F_k(u)^{**}$. Let $\Delta=\{i_0,\cdots i_{n-1}\}$ be a finite subset 
of $\alpha+k$. Let $v\in {}^{\alpha}\alpha+k^{(Id)}$ be such that $v(j)=u(j)$ for all $j\notin \Delta$.
Then there exists $\psi\in F_k(v)^{**}$ such that $var(\psi)\cap \{x_{i_0},\cdots ,x_{i_{n-1}}\}=\emptyset$. 
i.e., $ind(\psi)\cap \Delta=\emptyset$, and $\bold W\models \psi\equiv \exists x_{i_0}\cdots \exists x_{i_{n-1}}\phi$.
In particular, $c_{(\Delta)}E(u,\phi)=c_{(\Delta)}E(v,\psi)$.
\item  Let $g\in G(k)^**$ and $i,j\in \alpha+k$ be distinct.
\end{athm}

{\bf Case (a): }$g=c_{(\Delta)}E(u,\phi)$ with $u\in {}^{\alpha+k}\alpha^{(Id)}$, $\phi\in F_k(u)^{**}$. 
\par\noindent
{\bf Subcase (a) : }
$j\in \Delta$.\par\noindent 
Then by \cite{HMT1} 1.5.8 (i) and 1.7.3 we have 
$$s_i^j(g)=s_i^jc_j(c_{(\Delta-\{j\})})E(u,\phi)=
c_j(c_{(\Delta-\{j\})})E(u,\phi)=g$$
\par\noindent
{\bf Subcase (b) : }
$j\notin \Delta$ and  $i\notin \Delta$. \par\noindent
Then by \cite{HMT1}, 1.5.8 (ii)  we have 
$$s_i^jg=s_i^jc_{(\Delta)}E(u,\phi)=c_{(\Delta)}s_i^jE(u,\phi).$$
Let $g^*=E(u,\phi)\cap d_{ij}$. Then $g^*$ and hence $s_j^ig=c_{(\Delta)}c_jg^*$, 
is equal to zero
if $u_i\neq u_j$.
Else, as easily checked,  $g^*=E(u,\phi \land x_i= x_j)$.  
By $u_i=u_j$ and $\phi\in F_k(u)^{**}$, 
we have $\phi \land x_i= x_j\in F_k(u)^{**}$.
By fact 6.6 (i)  we get 
that $$s_j^ig=c_{(\Delta)}c_jg^*=c_{(\Delta)}c_jE(u,\phi\land x_i=x_j)\in G(k)^{**}.$$
\par\noindent
{\bf Subcase (c) : }
$j\notin \Delta$ and $i\in \Delta$.\par\noindent
By \cite{HMT1} 1.7.3 and 1.5.8 (ii) we have 
 $$s_i^jc_{(\Delta)}E(u,\phi)=s_i^jc_{(\Delta-\{i\})}c_iE(u,\phi)
=c_{(\Delta-\{i\})}s_i^jc_iE(u,\phi).$$
Let $g^*=s_i^jc_iE(u,\phi)$. Then by \cite {HMT1} 
1.5.1 we have $g^*=c_j(c_i(E(u,\phi)\cap d_{ij})$.
Let $v=u_{u(j)}^{i}=u\circ [i|j]$. 
Here and elsewhere $[i|j]$ denotes the replacement that maps $i$ to $j$
and otherwise coincides with the identity. Then $v(j)=u(j)=v(i)$. Also, it is easy to see that 
$$c_i(E(u,\phi)\cap d_{ij})=E(v,\phi\land x_i=x_j).$$
Since $i\in \Delta$, we can assume by fact 6.6 (i)  that $x_i$ does not occur in $\phi$. 
Since $v$ differs from $u$ in at most the $ith$ 
place, we have $\phi\in F_k(v)^{**}$. But  $v(i)=v(j)$, and so  $x_i=x_j\in F_k(v)^{**}$, thus 
$\phi\land x_i=x_j\in F_k(v)^{**}$. By fact 6.6 (i)  we get that $g^*$ hence  $g$ is in $G(k)^{**}$.
\par\noindent
{\bf Case (b) : }$g=d_{kl}$, where $k,l\in \alpha+k.$ 
\par\noindent
By \cite{HMT1}, 1.5.4 we have $s_i^k(d_{kl})=d_{il}$ if $k\neq l$ and  
$s_i^jd_{kl}=d_{kl}$ if $j\notin \{k,l\}$. In either case we have $s_i^jg\in G(k)^{**}$.  
\par\noindent
{\bf Case (c) : }$g$ is an element in $G(k)^{**}$.
\par\noindent
Follows from the two previous cases, since the $s_i^j$'s ($i,j\in \alpha+k$) 
are $BA$ endomorphisms of the boolean algebra $G(k)^{**}$ (cf. \cite{HMT1} 
1.5.3 (i)). 
\item Let $g=c_{(\Delta_1)}E(u_1,\phi_1)\cap c_{(\Delta_2)}E(u_2,\phi_2)$
where $\Delta_i\cap ind(\phi_i)=0$ for $i\in \{1,2\}$.
Then $g=0$ if there exists $j\in (\alpha+k)-(\Delta_1\cup \Delta_2)$ such that $u_1(j)\neq u_2(j)$.
Else, it is not difficult to check that $g= c_{(\Delta_1\cap \Delta_2)}E(w,\phi_1\land \phi_2)$, 
where $w$ is defined as follows:
$$w(j)=u_1(j)=u_2(j)\text{ for }j\in \alpha+k-(\Delta_1\cup \Delta_2)$$
$$w(j)=u_1(j)\text{ if }j\in \Delta_2-\Delta_1$$
$$w(j)=u_2(j)\text{ if }j\in \Delta_1-\Delta_2,\text{ and }w\text{ is defined arbitrarily on }\Delta_1\cap \Delta_2.$$
Also by $ind(\phi_i)\cap \Delta_i=0$, for $i\in \{1,2\}$ and $w(j)=u_1(j)=u_2(j)$ for all 
$j\in \alpha+k-(\Delta_1\cup \Delta_2)$ we can assume by 6.6 (i) 
that $\phi_i\in F_k(w)^{**}$, 
hence $\phi_1\land \phi_2\in 
F_k(w)^{**}$. 
\end{enumroman}  
\end{demo}
To economise on notation we recall a few notions from \cite{HMT1}. 
A {\it generalized diagonal element} is a finite intersection of diagonal elements.
A {\it co-diagonal element} is the complement of a diagonal element. 
A {\it generalized co-diagonal element} is a finite intersection
of co-diagonal elements. In particular the unit of an algebra can be 
viewed as both a generalized diagonal and a 
generalized co-diagonal. 
\begin{athm}{Fact 6.7}
For all $0\leq k<\omega$, $G(k)^{**}\in Ws_{\alpha+k}$.
\end{athm}
\begin{demo}{Proof}
Clearly  $G(k)^{**}$ is a boolean field of sets with greatest element $V_k$. 
Also $G(k)^{**}$, by definition, contains all diagonal elements.
We are  thus left to check cylindrifications, by the additivity of which, 
it suffices  to show that for every  $g\in G(k)^*$ and $j<\alpha+k$ we have $c_jg\in G(k)^{**}$.
By fact 6.6 (iii) a typical element $g$ of $G(k)^*$ is of the form
\par\noindent
$a_0\cap b\cap c\cap -a_1\cap -a_2\cdots \cap -a_n$, where $n\in \omega$, 
$a_0,a_1,\cdots a_{n}\in H(k)$,
$b$ is a generalized diagonal element, and $c$ is a generalized co-diagonal element.
Now fix  $j<\alpha+k$ and let $g$ be as indicated above. 
\par\noindent
{\bf Case (a) : }
$b=d_{jl}\cap b'$ where $b'$ is a generalized diagonal element and $j\neq l$.
\par\noindent
Let $g^*=a_0\cap b'\cap c\cap -a_1\cdots \cap -a_n$. 
Then, of course $g^*\in G(k)^*$ and $c_jg=s_l^jg^*$.
\par\noindent
By fact 6.6 (i) the latter is in $G(k)^{**}$.
\par\noindent
{\bf Case (b) : }
$a_0=b=V_k$.
\par\noindent
In this case we have 
 $g= -c_{(\Delta_1)}E(u_1,\phi_1)\cdots \cap -c_{(\Delta_n)}E(u_n,\phi_n)
\cap -d_{jl_1}\cdots \cap -d_{jl_m}$, where $n,m\in \omega$, 
$\Delta_k\subseteq_{\omega}\alpha +k$ for $1\leq k\leq n$ 
and since $c_j(x\cap c_jy)=c_jx\cap c_jy$, we can assume 
that $j\notin \Delta_1\cup \cdots \Delta_n$. Assume that $g\neq 0$, 
for else there is nothing more to prove. In particular, $j,l_1,\cdots l_m$ are pairwise distinct. 
We claim that in this case we have   $c_jg=V_k$.
\par\noindent

Indeed, let $s=\langle s_l\rangle_{l\in \alpha+k}$ be an arbitrary element in $V_k$.
We will show that $s\in c_jg$, by which we will be done.
Choose $k_j\notin \{u_1(j),\cdots ,u_n(j)\}\cup \{i_{l_1},\cdots i_{l_m}\}$, and let  $u$
be an arbitrary element in $W_{k_j}.$
Let  $z=s(j|u)$. Assume that  $1\leq i\leq n$. 
Then $z\notin c_{(\Delta_i)}E(u_i,\phi_i)$ because
$z(j)=s(j|u)(j)=u\notin \cup_{i<n}W_{u_i(j)}$.
Assume now that  $1\leq i\leq m$. 
Then $z(j)\neq z(l_i)$, since they are in distinct copies by the choice of $k_j$,
i.e. $z\notin d_{jl_i}$.
We have shown that $z\in g$. Since $s$ differs fom $z$ in at most the $jth$ place we get
that $s\in c_jg$.  We have proved that $c_jg=V_k$.
\par\noindent
{\bf Case (c) : }
$$g=c_{(\Delta_0)}E(u_0,\phi_0)\cap -c_{(\Delta_1)}
E(u_1,\phi)\cap\cdots c_{(\Delta_n)}E(u_n,\phi_n)
\cap -d_{jl_1}\cdots\cap -d_{jl_m},$$ 
where $n,m\in \omega$, $\Delta_k\subseteq_{\omega}\alpha+k$ for 
$0\leq k\leq n$
and as in the previous case, we can assume that $j\notin \Delta_0\cup\cdots\Delta_n$ 
and that $j,l_1,\cdots l_m$ 
are pairwise distinct.\par\noindent
We can further assume that $\{l_1,\cdots l_m\}\subseteq \Delta_0$, for if $l_k\notin \Delta_0$ 
for some $1\leq k\leq n$,
then we can simply ``ignore" the co-diagonal element with indices $j$ and $l_k$, 
i.e. the element $-d_{jl_k}$,
because of the following:
$$c_{(\Delta_0)}E(u_0,\phi_0)\cap -d_{jl_{k}}=c_{(\Delta_0)}E(u_0,\phi_0\land x_j\neq x_{l_k})$$ 
if $u_0(j)=u_0(l_k)$,
and is equal to $c_{(\Delta_0)}E(u_0,\phi_0)$ otherwise.
Note too, that $\phi_0\land x_j\neq x_{l_k}\in F_k(u_0)^{**}$, 
whenever $\phi_0\in F_k(u_0)^{**}$ 
and $u_0(j)=u_0(l_k)$. In either case, $g$ can be written in the form:
$$c_{(\Delta_0)}E(u_0,\psi)\cap \cdots -c_{(\Delta_n)}E(u_n,\phi_n)\cap \cap \{-d_{jl}:l\in \{l_1,\cdots l_m\}-{l_k}\},$$ 
where $c_{(\Delta_0)}E(u_0,\psi)\in H(k)$.
\par\noindent
Now let 
$$\Gamma =\Delta_0-\{l_1,\cdots l_m\}\quad\text{and }\Gamma_1=\{l_1,\cdots l_m\}.$$ 
Then $\Delta_0=\Gamma\cup \Gamma_1$.
Let $S=\{j\}\times \Gamma_1=\{(j,l): l\in \Gamma_1\}$ and let  $d_{S}$  be the co-diagonal element:
$\cap \{-d_{jl}: (j,l)\in S\}$.\par\noindent 
Let $g*=-c_{(\Delta_1)}E(u_i,\phi_1)\cap\cdots -c_{(\Delta_n)}E(u_n,\phi_n)$,
$g_1= c_{(\Gamma)}E(u_0,\phi_0)\cap g*$ and 
$g_2= c_{(\Gamma_1)}E(u_0,\phi_0)\cap d_{S}\cap g*$.
Then by \cite {HMT1} 1.7.3, and 
the additivity of $c_j$ we have $c_jg=c_jg_1\cup c_jg_2$. 
\par\noindent
We shall show that $c_jg_1\in G(k)^{**}$ 
and $c_jg_2\in G(k)^{**}$, by which we will be done.
\par\noindent
{\it Proof of $c_jg_1\in G(k)^{**}$}.
\par
Write $g_1=c_{(\Gamma)}E(u_0,\phi_0)\cap\cdots -c_{(\Delta_n)}E(u_n,\phi_n)$.
\par\noindent
We can assume that for $1\leq k\leq n$ and $i\notin \Gamma \cup \Delta_k$, 
we have $u_0(i)=u_k(i)$, for if 
$u_0(i)\neq u_k(i)$ for some $1\leq k\leq n$, and some $i\notin \Gamma\cup \Delta_k$, then
$c_{(\Gamma)}E(u_0,\phi_0)\cap -c_{(\Delta_k)}E(u_k,\phi_k)=c_{(\Gamma)}E(u_0,\phi_0)$.
\par\noindent
For the time being assume that $\phi_1=\phi_2=\cdots \phi_n=T$.
\par\noindent
{\it Claim} 
\par\noindent
$c_jg_1=c_jc_{(\Gamma)}E(u_0,\phi_0)\cap -c_jc_{(\Delta_1)}E(u_1,T)\cap\cdots -c_jc_{(\Delta_n)}E(u,T)$.
\par\noindent
{\it Proof of claim}
\par
It is straightforward to show that r.h.s is contained in $c_jg$. Now for the opposite inclusion:

Let $s\in c_jg_1$. Then there exists $b\in W$ such that
$s(j|b)\in c_{(\Gamma)}E(u_0,\phi_0)$ and $s(j|b)\notin c_{(\Delta_k)}E(u_k,T)$ 
for all  $1\leq k\leq n$.
Thus $s\in c_jc_{(\Gamma)}E(u_0,\phi_0)$. 
We will show that $s\notin c_jc_{(\Delta_k)}E(u_k,T)$ for
all $1\leq k\leq n$, by which case we will be done. 
Towards this end, fix $1\leq k\leq n$ and assume to the contrary that
$s\in c_jc_{(\Delta_k)}E(u_k,T)$. Then there exists $c\in W$ 
such that $s(j|c)\in c_{(\Delta_k)}E(u_k,T)$.
By $s_c^j\in c_{(\Delta_k)}E(u_k,T)$ and $j\notin \Delta_k$
we get $c=u_k(j)$. Similarly, by $s_b^j\in c_{(\Gamma)}E(u_0,\phi_0)$ 
and $j\notin \Gamma$ we get $b=u_0(j)$.
Recall that we assumed that $u_0(j)=u_k(j)$ 
since $j\notin \Gamma\cup \Delta_k$, thus $b=c$.
Since $s(j|b)$ and $s(j|c)$ differ in at most the $jth$ place , 
$b=c$ and $s(j|c)\in c_{(\Delta_k)}E(u_k,T)$ we get
$s(j|b)\in c_{(\Delta_k)}E(u_k,T)$. Contradiction proving  that 
$s$ is as required.
\par\noindent
Now the general case follows from this by noting that 
$$-c_{(\Delta)}E(u,\phi)=c_{(\Delta)}E(u,\neg \phi)\cup -c_{(\Delta)}E(u,T),$$ 
and that $H(k)$ is closed 
under finite intersections.
\par\noindent
We now turn to showing that $c_jg_2\in G(k)^{**}$. 
This will be done by showing that 
$a=c_{(\Gamma_1)}E(u_0,\phi_0)\cap d_S$, hence $g_2=a\cap g*$, can be written 
as a finite  union of elements of the form 
$a_0\cap -a_1\cdots \cap -a_n$ where $a_i\in H(k)$.
Then by the additivity of $c_j$, and the previous case we get the desired.
We start off when $\Gamma_1$ consists of a single element $m$ say.
We compute $a=c_mE(u_0,\phi_0)\cap -d_{jm}$. 
For brevity set $u=u_0$. Let $v=u(m|u(j))=u\circ [m|j]$. 
Then $v(m)=v(j)=u(j)$.
Moreover it is not hard to check that 
$$a=(c_mE(u,\phi_0)\cap -E(v,T))\cup E(v,\phi_0\land x_j\neq x_m).$$
Since $x_m$ does not occur in $\phi_0$, and $v$ 
differs from $u$ in at most the $m$-th place, we infer from fact 6.6 (i) 
that $\phi_0\in F_k(v)^{**}$. By $v(m)=v(j)$ we get that $x_j\neq x_m$ is 
also in $F_k(v)$ hence in $F_k(v)^{**}$.
It follows that  $\phi_0\land x_j\neq x_m\in F_k(v)^{**}$. We have shown 
that $E(v,\phi_0\land x_j\neq x_m)\in H(k)$.
Thus $a$ is as desired, since  $c_mE(u,\phi_0)$ and $E(v,T)$ are in $H(k)$, too. 
Now for the general case. Assume that $\Gamma_1=\{l_1,\cdots l_m\}$. 
Having treated the case when $m=1$
we now assume that $m>1$. 
Then $$a=c_{(\Gamma_1)}E(u,\phi)\cap -d_{S}=\cap \{a_k: 1\leq k\leq m\}$$ where 
$a_k=c_{(\Gamma_1-\{\l_k\})}c_{l_k}E(u,\phi)\cap -d_{jl_k}$. 
Therefore  $a$, hence $g_2= a\cap g^*$, is indeed the finite union of 
elements of the $a_0\cap -a_1\cdots \cap -a_n$ where $a_i\in H(k)$.
Thus $c_jg_2$  is in $G(k)^{**}$.  
By this the proof of fact 6.7 is complete.
\end{demo}
\begin{athm}{Fact 6.8}
$G(0)^{**}\cong Nr_{\alpha}G(k)^{**}$, for all $0\leq k<\omega$.\footnote {We should point out that the complete 
analogue of fact 3.3 is true: 
That is for $0\leq l<k<\omega$ we have $G(l)^{**}\cong Nr_{\alpha+l}G(k)^{**}$. Fact 6.8
is the  special case when $l=0$, which is all what we need to show that 
${\bf A}(\alpha)\in Nr_{\alpha}Ws_{\alpha+k}$, for all $k<\omega$.}  
\end{athm}
\begin{demo}{Proof}
Let $k\geq 0$. Define $i(k)$ like in fact 3.3.
That is for $a\in G(0)^{**}$, let $i(k)(a)=\{t\in V_k :t\upharpoonright \alpha\in a\}$.
Then $i(k)\in Ism(A(\alpha), Nr_{\alpha}C(\alpha+k))$, where $C(\alpha+k)$ is the full 
$Ws_{\alpha+k}$ with unit $V_k$.
We will show that $i(k)G(0)^{**} \subseteq G(k)^{**}$.
For that it clearly suffices to show that for all $u\in {}^{\alpha}\alpha^{(Id)}$, and 
$\phi\in F_0(u)$, we have $i(k)E(u,\phi)\in G(k)^{**}$. Let $u$ and $\phi$ be as indicated:
Then $i(k)E(u,\phi)=\cup\{E(v,\phi):v\in {}^{\alpha+k}\alpha^{(Id)}:v\upharpoonright \alpha=u\}
=c_{\alpha}\cdots c_{\alpha+k-1}E(u*,\phi)$ where $u*=u\cup 
\{\langle i,0\rangle:\alpha\leq i< \alpha+k\}$. 
Since $F_0(u)\subseteq F_k(v)$ whenever $v\in {}^{\alpha+k}\alpha^{(Id)}$ is such that 
$v\upharpoonright \alpha=u$, 
we get $\phi\in F_k(u*)^{**}$, hence $i(k)E(u,\phi)\in G(k)^{**}$.
We have shown that $i(k)\in Ism(G(0)^{**}, Nr_{\alpha}G(k)^{**})$.
We will now show that $i(k)$ is actually onto $Nr_{\alpha}G(k)^{**}$.
Since the $c_i$'s are additive, it suffices to show that for all $g\in G(k)^*$,
there exists $a\in G(0)^{**}$, such that $i(k)a=c_{\alpha}\cdots c_{\alpha+k-1}g$.
\par\noindent
{\bf Case (a): }
$g=E(v,\phi)$, $v\in {}^{\alpha+k}\alpha^{(Id)}$, and $\phi\in F_k(v)^{**}$. 
\par\noindent
Then by fact 6.6 (i)   there exists $\psi\in F_k(v)^*$ such that 
$var(\psi)\cap \{x_{\alpha},\cdots x_{\alpha+k-1}\}=\emptyset$ and
$\bold W\models \exists x_{\alpha}\cdots \exists x_{\alpha+k-1}\phi\equiv \psi$.
Let $u=v\upharpoonright \alpha\in {}^{\alpha}\alpha^{(Id)}$.
By noting that $F_0(u)^*=\{\phi\in F_k(v)^*: var(\phi)\subseteq \alpha\}$, 
we get that $\psi\in F_0(u)^*$. Moreover we have  
$i(k)E(u,\psi)=c_{\alpha}\cdots c_{\alpha+k-1}E(u,\phi)$.
\par\noindent
{\bf Case (b): }
$g=a\cap c$, where $a=-a_0\cap \cdots -a_n$, $a_i\in H(k)$ and $c$ is a generalized codiagonal.\par\noindent
If $c_{\alpha}\cdots c_{\alpha+k-1}g=g$, then $g\in G(0)^{**}$. 
Else
$c_{\alpha}\cdots c_{\alpha+k-1}g$ is either $0$ or $V_k$, pending on whether $g=0$ or not.
The choice of $a\in G(0)^{**}$ in this case is also obvious.    
\par\noindent
{\bf Case (c): }
$g=d_0\cap -d_1\cap\cdots -d_n\cap b\cap c$, where $d_i\in H(k)$, $b$ is a generalized diagonal, and $c$ a
generalized codiagonal.\par\noindent
Assume that  $1\leq k<\omega$; else there is nothing to prove. \par\noindent
Let $\Gamma=\{\alpha,\cdots \alpha+k-1\}$.
By the proof of fact 6.7 ,  we have $c_{(\Gamma)}g$ 
is the finite union of elements of the form $a_0\cap- a_1\cdots \cap- a_n\cap b\cap c$ where
$a_0, \cdots a_n\in H(k)$, $b$ is a generalized diagonal and $c$ a generalized codiagonal.
Further we can assume that if $b=\cap \{d_{ij}^{V_k}:i,j\in \Delta_1\}$ and $c=\cap \{-d_{ij}^{V_k}:i,j\in \Delta_2, i\neq j\}$
then $\Delta_i\subseteq_{\omega} \alpha$, for $i\in \{1,2\}$. 
By the additivity of generalized cylindrifications (namely of $c_{(\Gamma)}$), we can and will assume that
$$c_{(\Gamma)}g=c_{(\Gamma_0)}E(v_0,\phi_0)\cap c_{(\Gamma_1)}E(v_1,\phi_1)\cdots 
\cap -c_{(\Gamma_n)}E(v_n,\phi_n)\cap b\cap c,$$
where for each $i\leq n$ we have  
$\Gamma\subseteq \Gamma_i\subseteq_{\omega}\alpha+k$,  $\phi_i\in F_k(v_i)^{**}$, $b$ is a 
generalized diagonal and $c$ is a generalized codiagonal as indicated above. 
By  fact 6.6 (i)   we can assume that $var(\phi_i)\cap \{x_{\alpha},\cdots x_{\alpha+k-1}\}=\emptyset$.
For $i\leq n$ let $u_i=v_i\upharpoonright \alpha$. 
Then $u_i\in {}^{\alpha}\alpha^{(Id)}$, and by fact 6.6 (i) $\phi_i\in F_0(u_i)^{**}$.
Let $b'=\cap \{d_{ij}^{V_0}:i,j\in \Delta_1\}$ and $c'=\cap \{d_{ij}^{V_0}:i,j\in \Delta_2, i\neq j\}$.
\par\noindent
Let $a= c_{(\Gamma_0-\Gamma)}E(u_0,\phi_0)\cap\cdots c_{(\Gamma_n-\Gamma)}E(u_n,\phi_n)\cap b'\cap c'$.
Then $a\in G(0)^{**}$ and an easy checked $i(k)a=c_{(\Gamma)}g$.
\end{demo}
\begin{athm}{Fact 6.9}
$G(0)^{*}=A(\alpha)$.
\end{athm}
\begin{demo}{Proof}
Let $u\in {}^{\alpha}\alpha^{(Id)}$ be one to one and $r\in R$. Then by definition we have
$$p(u,r)=E(u,C_r(x_0,x_1,\cdots )).$$
Hence $p(u,r)\in F_0(u)$.  Since $G(0)^{**}$ is a $Ws_{\alpha}$ 
containing the generators of $A(\alpha)$ 
it follows that
that $A(\alpha)\subseteq G(0)^{**}$.
For the other inclusion it is enough to show that for $u\in {}^{\alpha}\alpha^{(Id)}$ and $\phi\in F_0(u)$, we have
$E(u,\phi)\in A(\alpha)$. We start by showing that $E(u,T)\in A(\alpha)$ for all such $u$.
\par\noindent
{\bf Case (a): }
$u$ is one to one.\par\noindent 
Let $r\in R$ . Then $c_0(p(u,r))\cap c_1(p(u,r))=E(u,T)\in A(\alpha)$.
\par\noindent
{\bf Case (b): }
$u$ is not one to one.\par\noindent
Let $J$ be a finite subset of $\alpha$, such that $u(J)\subseteq J$ and $u(i)=i$ whenever $i\notin J$.
Such a $J$ exists, but of course is not unique. Fix one such $J$.  Let $u'=u|J$. Then $u':J\to J$. Let $n=|Rgu'|$. Then $n<|J|$, because $u$ hence $u'$ is not one to one.
We next procced as in the finite dimensional case; correlating a permuation $v$ to $u'$ as follows:
Let $y\in {}^nJ$ such that $y(0)<\cdots <y(n-1)$ and $\{u'_{y(0)},\cdots u'_{y(n-1)}\}=Rgu'$. 
\par\noindent
For $i<l$ put $m_{y(i)}=\min (u' {}^{-1}(u'_{y(i)}))$. Choose $v\in S_J$ ( a permutation on $J$) 
such that $v(m_{y(i)})=u'_{y(i)}$.
Such a $v$ exists. Let $I=J-\{m_{y(0)},\cdots ,m_{y(l-1)}\}$; suppose that $I=\{j_0,\cdots j_k\}$, $k\in \omega$ 
and $j_0<j_1\cdots <j_k$.
Let $t_i^l(x)=c_lx\cap d_{il}$. Let $v*=v\cup Id_{\alpha-J}$. Then $v*$ is a permutation on $\alpha$, and
like the finite case, we have
$$t_{j(k)}^{u_{j(k)}}\circ \cdots \circ t_{j(0)}^{u_{j(0)}}E(v*,T)=E(u,T).$$
From case (a) we get that  $E(u,T)\in A(\alpha)$. Now let $u\in {}^{\alpha}\alpha^{(Id)}$ 
and $i,j\in ker(u)$. Then $E(u,x_i=x_j)=E(u,T)\cap d_{ij}\in A(\alpha)$.
Also for all  $u,v\in S_{\alpha}$, we have by Lemma 1**(i), 
$E(u,C_r(x_0,x_1,\cdots ))=E(u,C_r(x_{v(0)},\cdots )=p(u,r)$.
Finally by noting that $E(u,\neg\phi)=E(u,T)-E(u,\phi)$ we get the desired. 
\end{demo}
\begin{athm}{Fact 6.10}
$A(\alpha)\in INr_{\alpha}Ws_{\beta}$, for all $\beta>\alpha$.
\end{athm}
\begin{demo}{Proof} 
By fact 6.9, it suffices to show that $A(\alpha)\in Nr_{\alpha}Ws_{\alpha+\beta}$ for infinite $\beta$.
So let $\beta\geq \omega$. Let $V_0={}^{\alpha}W^{(s)}$ be the unit of $A(\alpha)$. 
\par\noindent
Let $w\in W$. Let $s_{\beta}=s\cup \{(i,w):\alpha\leq i<\beta\}$. 
Then $s_{\beta}\in {}^{\beta}W$.
Let $V_{\beta}$ be the weak space ${}^{\alpha+\beta}W^{(s_{\beta})}$ and let 
$C(\alpha+\beta)$, or simply $C(\beta)$,  be the full $Ws_{\alpha+\beta}$ 
with unit $V_{\beta}$.
For $k<\omega$ let $i(k,\beta)$ be the function with domain $G(k)^{**}$, such that
$i(k,\beta)a=\{s\in V_{\beta}: s\upharpoonright \alpha+k\in a\}$, $a\in G(k)^{**}$.
Then $i(k,\beta):G(k)^{**}\to C(\beta)$.
Let $G(\beta)=Sg^{C(\beta)}\{i(k,\beta)G(k)^{**}: 0\leq k<\omega\}$. 
Then $G(\beta)\in Ws_{\alpha+\beta}$.
Moreover, it is easy to show  that $G(\beta)=\cup \{i(k,\beta)G(k)^{**}:0\leq k<\omega\}$.
From this together with facts 6.8 and 6.9, 
we get (using the same ideas of the proof of fact 3.4) 
that 
$A(\alpha)\cong Nr_{\alpha}G(\beta)$, thus 
$A(\alpha)\in INr_{\alpha}Ws_{\alpha+\beta}$.  
\end{demo}
Finally we show
\begin{theorem} The atom structure $\At\B$ is not weakly neat.That is for any atomic $\D\in \CA_{\alpha}$ if $\At\D\cong \At\B$, 
then $\D\notin \Nr_{\alpha}\CA_{\alpha+1}$.
\end{theorem}

\begin{proof}
Let $\D$ be an atomic algebra such that $\At\D\cong \At\B$. Then 
$\D$ will contain the term algebra which will be the disjoint copies indexed by $^{\alpha}\alpha^{(Id)}$ except that $P_{Id}$ is countable. 
The same cardinality trick implemented above works. 
Had it been a neat reduct that ${}_{\alpha}s(0,1)P_r=P_{Id}$ 
would be uncountable and this cannot happen.
\end{proof}


%\end{enumarab}
 
\begin{thebibliography}{100}

\bibitem{AGMNS} Andr\'eka, H., Givant, S., Mikulas, S. N\'emeti, I., Simon A., 
{\it Notions of density
that imply representability in algebraic logic.} 
Annals of Pure and Applied logic, {\bf 91}(1998) p. 93 --190. 


%\bibitem{AGMNS}

\bibitem{1} H. Andreka, M. Ferenczi, I. Nemeti (editors) {\it Cylindric-like algebras and algebraic logic} Bolyai Society, Mathematical Studies, Springer
(2013).

\bibitem{ANT} Andr\'eka, H., N\'emeti, I., Sayed Ahmed, T., 
{\it Omitting types for finite variable fragments and complete representations of algebras.}
Journal of Symbolic Logic {\bf 73}(1) (2008) p.65-89

%\bibitem{B} Becker H {\it Polish group actions, dichotomies and generalized elementary embeddings} Journal of the Americal Mathematical Soceity, 11(2) 397-449 1998.

\bibitem{CF} Casanovas, E., and Farre, R. {\it Omitting Types in Incomplete Theories.}  
Journal of Symbolic logic, Vol. 61, Number 1, p. 236-245.
March 1996. 

%\bibitem{M80} Miller, A. {\it Covering $^{\omega}2$ with $\omega_1$ disjoint closed sets.} 
%The Kleene Symposuim (proceedings, Madison, Wisconsin, 1978). Studies
%in Logic and the Foundation of Mathematics, vol 101, North-Holland,
%Amsterdam, 1980 pp. 415--421.

%\bibitem{Miller1} Miller A. {\it A characterization of the least cardinal for which 
%the Baire Category Theorem 
%fails}. Proceedings of the Americal Mathematical Society, {\bf 86}(3) (1982) p. 498--502.
%\bibitem{Bulletin} T.Sayed Ahmed {\it A sufficient and necessary condition for omitting types}
%Bulletin section of logic {\bf 34}(1)(2005)  23-28

\bibitem{DM} Daigneault, A., and Monk,J.D., 
{\it Representation Theory for Polyadic algebras}. 
Fund. Math. {\bf 52}(1963) p.151-176.




\bibitem{HMT1}L. Henkin, D. Monk, A. Tarski {\it Cylindric algebras, part 1} 1970

\bibitem{HMT2}L. Henkin, D. Monk, A. Tarski {\it Cylindric algebras, part 2} 1985
%\bibitem{Hodkinson}  Cylindric algebras are not closed under completemath lohg 3 2008 294-306

\bibitem{r} R. Hirsch {\it Relation algebra reducts of cylindric algebras and complete representations}. Journal of Symbolic Logic
(72) (2007) p. 673-703

\bibitem{Hodkinson} Hodkinson{\it Atom structures of relation and cylindric algebas} Appals(1997)p. 117-148.

\bibitem {Hirsch} Hirsch and Hodkinson  {\it  Complete representations in algebraic logic} JSL (62)(1997) p. 816-647


\bibitem{R} Hirsch R. {\it Relation algebra reducts of cylindric algebras and complete representations} Journal of Symbolic Logic (72) (2007) 673-703

\bibitem{HHbook} Hirsch, Hodkinson {\it Relation lgebras by Games}  147 Studies in Logic and Foundations of Mathematics Elsevier 
North Holland 2002

\bibitem{strong} Hirsch, Hodkinson {\it Strongly representable  atom structures of cylindric algebra} Journal of Symbolic Logic (74)(2009) 811-828

\bibitem{HHM}  Hirsch, Hodkinson and Maddux {\it On provability with finitely many varibales} Bulletin of Symbolic Logic 
(8) (2002) p. 329-347.

\bibitem {HHK} Hirsch Hodkinson and Kurucz {\it On modal logics between $K\times K\times K$ and $S5\times S5\times S5$.} 
Journal of Symbolic Logic(67) (2002) 221-234.


\bibitem{HH} Hirsch and Hodkinson {\it Completions and complete representations}
In \cite{1} p. 61-90.

\bibitem{Shelah} S. Shelah {\it Classification Theory } 1978

\bibitem{m} Maddux {\it Non finite axiomatizability results for cylindric and relation algebras} JSL, 54 (1989) 951-974.
\bibitem{N} Newelski, L. {\it Omitting types and the real line.}
Jornal of Symbolic Logic, {\bf 52}(1987),  p.1020-1026.

\bibitem{IGPL} Sayed Ahmed {\it The class of neat reducts is not elementary} Logic Journal of $IGPL$(9) (2001) 593-628

\bibitem{MLQ} Sayed Ahmed {\it A model theoretic solutin to a problem of Tarski} 
Mathematical Logic quarterly (48) (2002) 343-355.

\bibitem{FM} T. Sayed Ahmed {\it The class of 2 dimensional neat reducts of polyadic algebras is not elementary}. 
Fundementa Mathematica (172) (2002) p.61-81

\bibitem{MLQ} T. Sayed Ahmed {\it A model theoretic solution to a problem of Tarski} Mathematical Logic Quarterly (48) (2002)
p.343-355

\bibitem{Bulletin}  Sayed Ahmed,T., {\it Algebraic Logic, where does it stand today?}
Bulletin of Symbolic Logic. {\bf 11}(4)(2005), p.465-516.

\bibitem{w} Sayed Ahmed {\it  Weakly representable atom structures that are not strongly representable with an application to first order logic}
Math Logic Quarterly (2008) p. 294-306


\bibitem{OTT} Sayed Ahmed T., {\it Completions, complete representations and omiting types} In \cite{1} p. 205-222

\bibitem{OTT2} Sayed Ahmed, T. and Samir B., {\it Omitting types for first order logic with infinitary predicates}
Mathematical Logic Quaterly  {\bf 53}(6) (2007) p.564-576.


\bibitem{KS} Khaled, Sayed Ahmed {\it On complete representations of algebras of logic} 
Logic Journal of $IGPL$, (2009)   p.267-272

\bibitem{IGPl} Sayed ahmed {\it  The class of neat reducts is not elementary} Logic Journal of IGPL (9) (2001)p. 593-628

%\bibitem{MLQ} A model theoretic MLQ 48 (2002) 343-355.

\bibitem{OTT2} Sayed Ahmed, T. and Samir B., {\it Omitting types for first order logic with infinitary predicates}
Mathematical Logic Quaterly  {\bf 53}(6) (2007) p.564-576.

\bibitem{hv} Hodkinson Venema{ \it Canonivcal varities with no canonical axiomatizations} Tarns 357 (2005) 4579-4605
 
\bibitem{ghv} {\it Erdos graphs resolves Fine's canonicity problem} Bull 10(2) (2004) 186-208.

\bibitem{Kechris} A.S. Kechris, \emph{Classical Descriptive Set Theory}, Springer Verlag, New York, 1995.


\bibitem{sep} M. Assem,  \emph{Separating Models by Formulas and the Number of Countable Models}, submitted (2013).
\bibitem{BeckerKechris} H. Becker and A.S. Kechris, \emph{The Descriptive Set Theory of Polish Group Actions}, Cambridge University Press, 1996. CMP 97:

\end{thebibliography}

%\bibitem{hv} Barely canonical
\end{document}
1304.1149.extracted_bib
\begin{thebibliography}{100}

\bibitem{AGMNS} Andr\'eka, H., Givant, S., Mikulas, S. N\'emeti, I., Simon A., 
{\it Notions of density
that imply representability in algebraic logic.} 
Annals of Pure and Applied logic, {\bf 91}(1998) p. 93 --190. 


%\bibitem{AGMNS}

\bibitem{1} H. Andreka, M. Ferenczi, I. Nemeti (editors) {\it Cylindric-like algebras and algebraic logic} Bolyai Society, Mathematical Studies, Springer
(2013).

\bibitem{ANT} Andr\'eka, H., N\'emeti, I., Sayed Ahmed, T., 
{\it Omitting types for finite variable fragments and complete representations of algebras.}
Journal of Symbolic Logic {\bf 73}(1) (2008) p.65-89

%\bibitem{B} Becker H {\it Polish group actions, dichotomies and generalized elementary embeddings} Journal of the Americal Mathematical Soceity, 11(2) 397-449 1998.

\bibitem{CF} Casanovas, E., and Farre, R. {\it Omitting Types in Incomplete Theories.}  
Journal of Symbolic logic, Vol. 61, Number 1, p. 236-245.
March 1996. 

%\bibitem{M80} Miller, A. {\it Covering $^{\omega}2$ with $\omega_1$ disjoint closed sets.} 
%The Kleene Symposuim (proceedings, Madison, Wisconsin, 1978). Studies
%in Logic and the Foundation of Mathematics, vol 101, North-Holland,
%Amsterdam, 1980 pp. 415--421.

%\bibitem{Miller1} Miller A. {\it A characterization of the least cardinal for which 
%the Baire Category Theorem 
%fails}. Proceedings of the Americal Mathematical Society, {\bf 86}(3) (1982) p. 498--502.
%\bibitem{Bulletin} T.Sayed Ahmed {\it A sufficient and necessary condition for omitting types}
%Bulletin section of logic {\bf 34}(1)(2005)  23-28

\bibitem{DM} Daigneault, A., and Monk,J.D., 
{\it Representation Theory for Polyadic algebras}. 
Fund. Math. {\bf 52}(1963) p.151-176.




\bibitem{HMT1}L. Henkin, D. Monk, A. Tarski {\it Cylindric algebras, part 1} 1970

\bibitem{HMT2}L. Henkin, D. Monk, A. Tarski {\it Cylindric algebras, part 2} 1985
%\bibitem{Hodkinson}  Cylindric algebras are not closed under completemath lohg 3 2008 294-306

\bibitem{r} R. Hirsch {\it Relation algebra reducts of cylindric algebras and complete representations}. Journal of Symbolic Logic
(72) (2007) p. 673-703

\bibitem{Hodkinson} Hodkinson{\it Atom structures of relation and cylindric algebas} Appals(1997)p. 117-148.

\bibitem {Hirsch} Hirsch and Hodkinson  {\it  Complete representations in algebraic logic} JSL (62)(1997) p. 816-647


\bibitem{R} Hirsch R. {\it Relation algebra reducts of cylindric algebras and complete representations} Journal of Symbolic Logic (72) (2007) 673-703

\bibitem{HHbook} Hirsch, Hodkinson {\it Relation lgebras by Games}  147 Studies in Logic and Foundations of Mathematics Elsevier 
North Holland 2002

\bibitem{strong} Hirsch, Hodkinson {\it Strongly representable  atom structures of cylindric algebra} Journal of Symbolic Logic (74)(2009) 811-828

\bibitem{HHM}  Hirsch, Hodkinson and Maddux {\it On provability with finitely many varibales} Bulletin of Symbolic Logic 
(8) (2002) p. 329-347.

\bibitem {HHK} Hirsch Hodkinson and Kurucz {\it On modal logics between $K\times K\times K$ and $S5\times S5\times S5$.} 
Journal of Symbolic Logic(67) (2002) 221-234.


\bibitem{HH} Hirsch and Hodkinson {\it Completions and complete representations}
In \cite{1} p. 61-90.

\bibitem{Shelah} S. Shelah {\it Classification Theory } 1978

\bibitem{m} Maddux {\it Non finite axiomatizability results for cylindric and relation algebras} JSL, 54 (1989) 951-974.
\bibitem{N} Newelski, L. {\it Omitting types and the real line.}
Jornal of Symbolic Logic, {\bf 52}(1987),  p.1020-1026.

\bibitem{IGPL} Sayed Ahmed {\it The class of neat reducts is not elementary} Logic Journal of $IGPL$(9) (2001) 593-628

\bibitem{MLQ} Sayed Ahmed {\it A model theoretic solutin to a problem of Tarski} 
Mathematical Logic quarterly (48) (2002) 343-355.

\bibitem{FM} T. Sayed Ahmed {\it The class of 2 dimensional neat reducts of polyadic algebras is not elementary}. 
Fundementa Mathematica (172) (2002) p.61-81

\bibitem{MLQ} T. Sayed Ahmed {\it A model theoretic solution to a problem of Tarski} Mathematical Logic Quarterly (48) (2002)
p.343-355

\bibitem{Bulletin}  Sayed Ahmed,T., {\it Algebraic Logic, where does it stand today?}
Bulletin of Symbolic Logic. {\bf 11}(4)(2005), p.465-516.

\bibitem{w} Sayed Ahmed {\it  Weakly representable atom structures that are not strongly representable with an application to first order logic}
Math Logic Quarterly (2008) p. 294-306


\bibitem{OTT} Sayed Ahmed T., {\it Completions, complete representations and omiting types} In \cite{1} p. 205-222

\bibitem{OTT2} Sayed Ahmed, T. and Samir B., {\it Omitting types for first order logic with infinitary predicates}
Mathematical Logic Quaterly  {\bf 53}(6) (2007) p.564-576.


\bibitem{KS} Khaled, Sayed Ahmed {\it On complete representations of algebras of logic} 
Logic Journal of $IGPL$, (2009)   p.267-272

\bibitem{IGPl} Sayed ahmed {\it  The class of neat reducts is not elementary} Logic Journal of IGPL (9) (2001)p. 593-628

%\bibitem{MLQ} A model theoretic MLQ 48 (2002) 343-355.

\bibitem{OTT2} Sayed Ahmed, T. and Samir B., {\it Omitting types for first order logic with infinitary predicates}
Mathematical Logic Quaterly  {\bf 53}(6) (2007) p.564-576.

\bibitem{hv} Hodkinson Venema{ \it Canonivcal varities with no canonical axiomatizations} Tarns 357 (2005) 4579-4605
 
\bibitem{ghv} {\it Erdos graphs resolves Fine's canonicity problem} Bull 10(2) (2004) 186-208.

\bibitem{Kechris} A.S. Kechris, \emph{Classical Descriptive Set Theory}, Springer Verlag, New York, 1995.


\bibitem{sep} M. Assem,  \emph{Separating Models by Formulas and the Number of Countable Models}, submitted (2013).
\bibitem{BeckerKechris} H. Becker and A.S. Kechris, \emph{The Descriptive Set Theory of Polish Group Actions}, Cambridge University Press, 1996. CMP 97:

\end{thebibliography}