The Geometric structure of M\"obius spaces
Merlin Incerti-Medici
Classically, a M\"obius structure is defined on a metric space $(X,d)$ by the cross-ratio $\frac{d(w,z)d(x,y){d(w,y)d(x,z)$. With M\"obius structures appearing naturally in contexts where no natural metric is given -- e.\,g. on the boundary of hyperbolic spaces -- one may ask whether a M\"obius structure can be defined outside of the context of metric spaces. In this paper, we review a generalization of M\"obius structures due to Sergei Buyalo. We state that this generalized notion of M\"obius structures is compatible with the classical one, both geometrically and topologically and provide a more general proof on the topological compatibility. Having established this generalized notion of M\"obius spaces, we study some of its topological and geometric properties with an outlook towards metric completeness.
1706.10166.tex
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%opening
\title{The Geometric structure of M\"obius spaces}
\author{Merlin Incerti-Medici}

\begin{document}

\maketitle

\begin{abstract}
Classically, a M\"obius structure is defined on a metric space $(X,d)$ by the cross-ratio $\frac{d(w,z)d(x,y)}{d(w,y)d(x,z)}$. With M\"obius structures appearing naturally in contexts where no natural metric is given -- e.\,g. on the boundary of hyperbolic spaces -- one may ask whether a M\"obius structure can be defined outside of the context of metric spaces.

In this paper, we review a generalization of M\"obius structures due to Sergei Buyalo. We state that this generalized notion of M\"obius structures is compatible with the classical one, both geometrically and topologically and provide a more general proof on the topological compatibility. Having established this generalized notion of M\"obius spaces, we study some of its topological and geometric properties with an outlook towards metric completeness.
\end{abstract}


\tableofcontents 

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% INTRODUCTION

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\section{Introduction} \label{sec:Introduction}


Given a CAT(0) space $Y$, it is a well-known fact that we can define a boundary at infinity of $Y$, denoted $\partial_{\infty}Y$. If $Y$ is CAT(-1) or Gromov-hyperbolic, $\partial_{\infty}Y$ carries a family of metrics, called visual metrics or Bourdon metrics. These metrics are not isometric in general. However, the cross-ratios, defined by $\left( \frac{d(w,y)d(x,z)}{d(w,z)d(x,y)}, \frac{d(w,z)d(x,y)}{d(w,x)d(y,z)}, \frac{d(w,x)d(y,z)}{d(w,y)d(x,z)}\right)$ for a given metric $d$, are the same for all these metrics. This suggests that one might use the cross-ratio to introduce a natural geometric structure on $\partial_{\infty}Y$. This leads to the notion of M\"obius structures, which has been studied by various people (see \cite{FoertschSchroeder}). Thinking beyond metric spaces, one may ask whether the concept of a cross-ratio can be generalized to a setting where no metric is given, leading to a notion of generalized M\"obius structures. An important step in this direction has been done by Sergei Buyalo in \cite{Buyalo}.

Generalized M\"obius structures allow us to introduce a new category of spaces which lies between metric spaces and topological spaces. In this article, we will study the relationship between certain geometric properties of a (generalized) M\"obius space and its topology. Specifically, we will show that under certain topological conditions on the image of the cross-ratio-triple, which we will define below, the generalized M\"obius structure is induced by a quasi-metric (see Definition \ref{def:cornercondition} and Proposition \ref{prop:cornerconditionatinfinity}). After generalizing the notions of Cauchy sequences and completeness to generalized M\"obius spaces, we will show that a stronger condition on the image of the cross-ratio-triple allows us to show the existence of a unique completion of a generalized M\"obius space (see Theorem \ref{thm:Completion}).\\

The rest of the paper is organized as follows. In section 2, we review the work done in \cite{Buyalo} using an alternative notation that suits our purposes. In section 3, we introduce the topology induced by a generalized M\"obius structure and provide a new proof that this topology coincides with the metric topology if the generalized M\"obius structure is induced by a metric. In section 4, we study quasi-metrics and necessary and sufficient conditions for a generalized M\"obius structure to be induced by a quasi-metric. In section 5, we introduce a notion of completeness and prove the existence of completeness under a certain condition.

Throughout the paper we assume that $X$ is a set with at least three points.\\

The author is grateful to Viktor Schroeder for many discussions and helpful advice, and to Krzysztof Putyra for coming up with wonderful examples.





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% GENERALIZED M�BIUS STRUCTURES

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\section{Generalized M\"obius structures} \label{sec:Generalized Mobius structures}


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% MOTIVATION

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\subsection{Motivation} \label{subsec:motivation}

This section is a reorganized write-up of work done in \cite{Buyalo} adjusted for our needs in the following sections (see also \cite{PlatisSchroeder}).
Let $(X,d)$ be a metric space. An $n$-tuple $(x_1, \dots, x_n)$ in $X$ is called {\it admissible} if every point $x \in X$ appears at most twice in $x_1, \dots, x_n$. We denote the set of admissible $n$-tuples by $\mathcal{A}_n$. An $n$-tuple is called {\it non-degenerate} if any two elements in the tuple are different. We define the so-called {\it cross-ratio triple}

\[ crt : \mathcal{A}_4 \rightarrow \mathbb{R}P^2 \]
\[ (w,x,y,z) \mapsto ( d(w,x)d(y,z) : d(w,y)d(z,x) : d(w,z)d(x,y) ). \]

If $(X,d)$ is an extended metric space, i.e. there is a point $\omega \in X$ such that for all $x \in X \diagdown \{ \omega�\}, d(x, \omega) = \infty$, we define

\[ crt(\omega,x,y,z) = ( d(y,z) : d(x,z) : d(y,z) ) \]

and analogously for any permutation of $(\omega, x, y,z)$. The idea behind this is that "infinite distances cancel", i.\,e. $\frac{a \cdot d(x,\infty)}{b \cdot d(y, \infty)} = \frac{a}{b}$ for any real numbers $a,b$ where we put $\frac{a}{0} = \infty$. We will use this idea and notation throughout the paper.

Note that the image of $crt$ is contained in the set of points $(x : y : z) \in \mathbb{R}P^2$ such that all numbers in $\{ x,y,z \}$ that are non-zero, have the same sign. Moreover, since for any admissible quadruple no three points are equal, at most one of the three entries can be zero. We denote

\[ \Delta := \{ (a:b:c) \vert a,b,c > 0 \} \]

\[ \overline{\Delta} := \Delta \cup \{ (0:1:1), (1:0:1), (1:1:0) \}. \]

It is easy to see that the image of $crt$ is contained in $\overline{\Delta}$. A useful way to represent $\overline{\Delta}$ is to choose for every point in $\overline{\Delta}$ the representative $(a : b : c)$ that satisfies $a + b + c = 1$. This yields a triangle in $\mathbb{R}^3$ that is depicted in figure \ref{fig:Triangle}.\\

\begin{figure} %\label{fig:Triangle}
\begin{tikzpicture}
\draw (0,0);
\draw [dashed] (6,0) -- (9,0);
\draw [dashed] (6,0) -- (7.5, 2.6);
\draw [dashed] (9,0) -- (7.5, 2.6);
\draw [fill] (7.5,0) circle [radius=0.05];
\draw [fill] (6.75,1.3) circle [radius=0.05];
\draw [fill] (8.25,1.3) circle [radius=0.05];
\node [below] at (7.5, 0) {$(\frac{1}{2} : \frac{1}{2} : 0)$};
\node [above left] at (6.75, 1.3) {$(\frac{1}{2} : 0 : \frac{1}{2})$};
\node [above right] at (8.25, 1.3) {$(0 : \frac{1}{2} : \frac{1}{2})$};
\end{tikzpicture}
\caption{The set $\overline{\Delta} \subset \mathbb{R}P^2$ represented by the points $(a : b : c)$ that satisfy $a + b + c = 1$. The only points on the boundary that are part of $\overline{\Delta}$ are the marked points.}
\label{fig:Triangle} 
\end{figure}

The cross-ratio triple is not always well-suited for the computations one may want to do. It turns out to be useful to have two additional ways to describe the cross-ratio triple. A detailed discussion can be found in \cite{PlatisSchroeder}.

The second description of the cross-ratio triple can be derived as follows. Denote by $\Pi := \{ (\alpha, \beta, \gamma) \in \mathbb{R}^3 \vert \alpha, \beta, \gamma > 0, \alpha�\beta \gamma = 1 \}$ and by $\overline{\Pi} := \Pi \cup \{ (1,\infty,0), (0,1,\infty), (\infty,0,1) \}$ and define a map

\[ \Delta \rightarrow \Pi \]
\[ (a:b:c) \mapsto \left(\frac{b}{c}, \frac{c}{a}, \frac{a}{b} \right). \]

It is easy to see that this is a homeomorphism with inverse

\[ (\alpha, \beta, \gamma) \mapsto \left(\gamma^{\frac{1}{3}} \beta^{-\frac{1}{3}} : \alpha^{\frac{1}{3}} \gamma^{-\frac{1}{3}} : \beta^{\frac{1}{3}} \alpha^{-\frac{1}{3}}\right). \]

We can extend this homeomorphism to a bijection $F: \overline{\Delta} \rightarrow \overline{\Pi}$ by setting

\[ (0:1:1) \mapsto (1, \infty, 0) \]
\[ (1:0:1) \mapsto (0, 1, \infty) \]
\[ (1:1:0) \mapsto (\infty, 0, 1). \]

Consider now the map $F \circ crt : \mathcal{A}_4 \rightarrow \overline{\Pi}$. It is easy to see that this map satisfies

\[ F(crt(wxyz)) = \left( \frac{d(w,y)d(x,z)}{d(w,z)d(x,y)}, \frac{d(w,z)d(x,y)}{d(w,x)d(y,z)}, \frac{d(w,x)d(y,z)}{d(w,y)d(x,z)}\right), \]

which is the classical cross-ratio. This provides us with a second description of the cross-ratio. In particular, note that $F$ and its inverse are sequence continuous with respect to the classical notion of divergence to $\pm\infty$.\\

Using the classical cross-ratio, it is now very easy to get a third description of the cross-ratio triple. Denote by $L_4 := \{ (x,y,z) \in \mathbb{R}^3 \vert x+y+z = 0 \}$ and $\overline{L_4} := L_4 \cup \{ (0, \infty, -\infty), (-\infty, 0, \infty), (\infty, -\infty, 0) \}$ and consider the map

\[ \Psi : \Pi \rightarrow L_4 \]
\[ (\alpha, \beta, \gamma) \mapsto \left(\ln(\alpha), \ln(\beta), \ln(\gamma) \right). \]

Clearly, $\Psi$ is well-defined and a homeomorphism. Its inverse is given by

\[ \Phi : L_4 \rightarrow \Pi \]
\[ (x,y,z) \mapsto \left( e^{x}, e^{y}, e^{z} \right). \]

The maps $\Psi$ and $\Phi$ extend to a bijection

\[ \overline{\Phi} : \overline{L_4} \rightarrow \overline{\Pi} \]

by sending

\[ (0, \infty, -\infty) \mapsto ( 1, \infty, 0) \]
\[ (-\infty, 0, \infty) \mapsto (0, 1, \infty) \]
\[ (\infty, -\infty, 0) \mapsto (\infty, 0, 1). \]

We can now use the inverse of $\overline{\Phi}$, which we will denote by $\overline{\Psi}$ to translate the classical cross-ratio to a map $\overline{\Psi} \circ F \circ crt$. This gives us

\[ M_d : \mathcal{A}_4 \rightarrow \overline{L_4} \]

and
\begin{equation}
\begin{split}
M_d(wxyz) & := \overline{\Psi}(F(crt(wxyz)))\\
& \; = ( (w | z) + (x | y) - (w | y) - (x | z),\\
& \qquad  (w | x) + (y | z) - (w | z) - (x | y),\\
& \qquad  (w | y) + (x | z) - (w | x) - (y | z)  ),
\end{split}
\end{equation}

where $(x | y) = -\ln(d(x,y))$.

\begin{rem}
Note that the map $\overline{\Psi}$ is again sequence continuous. If we extend the logarithm to be $\ln(\infty) = \infty$ and $\ln(0) = -\infty$, then $\Psi$ extends naturally to $\overline{\Psi}$.
\end{rem}

\begin{rem} If $X = \partial_{\infty} Y$ for some CAT(-1) space $Y$ and $d = d_o$ the Bourdon metric with respect to some base point $o \in Y$, then $(x | y)$ is exactly the Gromov product of $x,y$ with respect to $o$. For more information, consult \cite{Bourdon}.
\end{rem}

Each of these three ways to encode the cross-ratio satisfies certain symmetries. Consider an admissible quadruple and denote its four entry slots by $1,2,3,4$. Given these four slots, they form an (abstract and possibly degenerate) tetrahedron. We denote the edges of the tetrahedron by its two endpoints, e.\,g. $(12)$ denotes the edge between $1$ and $2$. The cross-ratio always considers two edges of the tetrahedron that are on opposing sides (i.\,e. they share no endpoints). There are three of these constellations, namely $(12)(34), (13)(42), (14)(23)$. Consider these three constellations in this particular order.

Choose a permutation $\sigma \in \mathcal{S}_4$. If we apply $\sigma$ to the points $1,2,3,4$ this induces a permutation of the constellations $(12)(34), (13)(42), (14)(23)$. In other words, the permutation $\sigma$ induces a permutation $\varphi(\sigma) \in \mathcal{S}_3$. This defines a map

\[ \varphi : \mathcal{S}_4 \rightarrow \mathcal{S}_3. \]

It is easy to see that $\varphi$ is a group-homomorphism and its kernel is $\{(1), (12)(34), (13)(24), (14)(23)  \}$. To facilitate, we provide an evaluation table of $\varphi$.

\begin{table}[H] \centering
\caption{Evaluation table\label{tab:EvalTable}}
\vspace{0.4cm}
\begin{tabular}{|lr|} 
\hline
$\pi$ & $\varphi(\pi)$\\
\hline
(12)  & (23) \\
(34)  & (23) \\
(1324)  & (23) \\
(1423)  & (23) \\
\hline
(13)  & (13) \\
(24)  & (13) \\
(1234)  & (13) \\
(1432)  & (13) \\
\hline
(14)  & (12) \\
(23)  & (12) \\
(1243)  & (12) \\
(1342)  & (12) \\
\hline
(123)  & (123) \\
(142)  & (123) \\
(134)  & (123) \\
(243)  & (123) \\
\hline
(132)  & (132) \\
(124)  & (132) \\
(143)  & (132) \\
(234)  & (132) \\
\hline
(12)(34)  & 1 \\
(13)(24)  & 1 \\
(14)(23)  & 1 \\
1  & 1 \\
\hline
\end{tabular}
\end{table}

\begin{lem} \label{lem:basicproperties}
The map $M_d$ satisfies the following properties.
\begin{enumerate}
\item[1)] For all $P \in \mathcal{A}_4$ and all $\pi \in \mathcal{S}_4$, we have
\[ M(\pi P) = sgn(\pi)\varphi(\pi)M(P). \]
\item[2)] For $P \in \mathcal{A}_4, M(P) \in L_4$ if and only if $P$ is non-degenerate.
\item[3)] For $P = (x,x,y,z)$ we have $M(P) = (0, \infty, -\infty)$.
\item[4)] For any admissible $5$-tuple $(x,y,\omega, \alpha, \beta)$ such that $\alpha \neq \omega \neq \beta \neq \alpha, \alpha \neq x \neq \beta$ and $\alpha \neq y \neq \beta$ we have
\begin{equation} \label{eq:Property4}
\begin{split}
M_d(\alpha x \omega \beta ) + M_d(\alpha \omega y \beta) - M_d(\alpha x y \beta) & = \Biggl(  \ln\left(\frac{d(x,y)}{d(x,\omega)d(\omega, y)} \frac{d(\alpha, \omega)d(\omega, \beta)}{d(\alpha, \beta)}\right) ,\\
& \qquad \; -\ln\left(\frac{d(x,y)}{d(x,\omega)d(\omega, y)} \frac{d(\alpha, \omega)d(\omega, \beta)}{d(\alpha, \beta)}\right), 0 \Biggr)\\
\end{split}
\end{equation}
Moreover, the first component of the left-hand-side expression is well-defined if $\alpha \neq \beta \neq \omega \neq \alpha, x \neq \beta, y \neq \alpha$ and equals the first component of the right-hand-side. Analogously, the second component of the left-hand-side expression is well-defined if $\alpha \neq \beta \neq \omega \neq \alpha, x \neq \alpha, y \neq \beta$ and equals the second component of the right-hand-side.
\end{enumerate}
\end{lem}

\begin{rem}
Lemma \ref{lem:basicproperties} has equivalent formulations for $crt$ and the classical cross-ratio. We leave it as an exercise for the reader to formulate the precise statements.
\end{rem}

\begin{rem}
Equation (\ref{eq:Property4}) really consists of three equations, one for each component. Writing $M_d = (a_d, b_d, c_d)$, the equation in the last component says $c_d(\alpha x \omega \beta) + c_d(\alpha \omega y \beta) = c_d(\alpha x y \beta)$ which -- when translating back to the context of $crt$ via $F^{-1} \circ \overline{\Phi}$ -- is equivalent to the following equation

\[ \frac{d(\alpha, \omega) d(x, \beta)}{d(\alpha, x) d(\omega, \beta)} \frac{d(\alpha, y) d(\omega, \beta)}{d(\alpha, \omega)d(y, \beta)} = \frac{d(\alpha, y)d(x, \beta)}{d(\alpha, x)d(y, \beta)}. \]

This is called the cocycle-condition and it holds for one of the three cross-ratios, the one written down above. For the other two, there is an error of $\lambda$. If we want to generalize the notion of the cross-ratio-triple beyond metric spaces, this condition -- as formulated in Equation (2) -- turns out to be crucial.

In particular, the fact that we do not just have the encoded cocycle-condition in the last component of Equation (\ref{eq:Property4}) but also meaningful equalities in the first two components allows us to think of the value $\lambda$ as some sort of distance between $x$ and $y$ with $\omega, \alpha, \beta$ as scaling parameters.
\end{rem}

The form of $\lambda$ in equation (2) may make one hope that that the expression $\lambda$ (together with properties 1)-3)) has sufficient meaning to provide us with the necessary structure for M\"obius geometry, even if no metric is given. Thus, we make the following

\begin{mydef} \label{def:M�biusStructure}

Let $X$ be a set with at least four points. A map $M : \mathcal{A}_4 \rightarrow \overline{L_4}$ is called a {\it generalized M\"obius structure} if and only if it satisfies the following conditions:

\begin{enumerate}

\item[1)] For all $P \in \mathcal{A}_4$ and all $\pi \in \mathcal{S}_4$, we have

\[ M(\pi P) = sgn(\pi)\varphi(\pi)M(P). \]

\item[2)] For $P \in \mathcal{A}_4$, $M(P) \in L_4$ if and only if $P$ is non-degenerate.

\item[3)] For $P = (x,x,y,z)$, we have $M(P) = (0, \infty, -\infty)$.

\item[4)] For any admissible $5$-tuple $(x, y, \omega, \alpha, \beta)$ such that $\alpha \neq \omega \neq \beta \neq \alpha$, $\alpha \neq x \neq \beta$ and $\alpha \neq y \neq \beta$, there exists some $\lambda = \lambda(x,y,\omega,\alpha,\beta) \in \mathbb{R} \cup \{ \pm \infty \}$ such that

\[ M(\alpha x \omega \beta) + M(\alpha \omega y \beta) - M(\alpha x y \beta) = (\lambda, -\lambda, 0). \]

Moreover, when $\alpha \neq \omega \neq \beta \neq \alpha$, $x \neq \beta$ and $y \neq \alpha$, the first component of the left-hand-side expression is well-defined. Analogously, the second component of the left-hand-side expression is well-defined when $\alpha \neq \omega \neq \beta \neq \alpha$, $x \neq \alpha$ and $y \neq \beta$.

\end{enumerate}

The pair $(X,M)$ is called a {\it generalized M\"obius space}.

\end{mydef}



We know the meaning of $\lambda$ when $M = M_d$ is induced by a metric $d$ from Lemma \ref{lem:basicproperties}. It is natural to ask what the meaning of $\lambda$ is for a generalized M\"obius structure. For this, we look at Property 4) in Definition \ref{def:M�biusStructure}. If we write $M_d = (a,b,c)$, then

\[ \lambda = a(\alpha x \omega \beta) + a(\alpha \omega y \beta) - a(\alpha x y \beta) \; \text{if} \; x \neq \beta, y \neq \alpha \]
\[ \; \lambda = b(\alpha x y \beta) - b(\alpha x \omega \beta) - b(\alpha \omega y \beta) \; \text{if} \; x \neq \alpha, y \neq \beta. \]

Both sides of these two equations are well-defined by the symmetry of the first two components of equation (2) and the final part of property 4). We can use these expressions to define a family of maps which -- as we will see -- replace the metric $d$. Let $A = (\omega, \alpha, \beta)$ be a non-degenerate triple. Then we define

\[ d_A(x,y) := \begin{cases} 
0 & \text{if $x=y$},\\
e^{a(\alpha x \omega \beta) + a(\alpha \omega y \beta) - a(\alpha x y \beta)} & \text{if $x \neq \beta$ and $y \neq \alpha$},\\
e^{b(\alpha x y \beta) - b(\alpha x \omega \beta) - b(\alpha \omega y \beta)} & \text{if $x \neq \alpha$ and $y \neq \beta$}.
\end{cases}\]

Lemma \ref{lem:basicproperties} tells us that if $(X,d)$ is a metric space with induced $M_d$, then $M_d$ is a generalized M\"obius structure and the induced map $d_A$ satisfies $d_A(x,y) = \frac{d(x,y)}{d(x, \omega)d(\omega, y)}\frac{d(\alpha, \omega) d(\omega, \beta)}{d(\alpha, \beta)}$. For a generalized M\"obius structure, $d_A$ may not be a metric. Nevertheless, each $d_A$ defines a map $M_A : \mathcal{A}_4 \rightarrow \overline{L_4}$ by replacing $d$ by $d_A$ in the definition of $M_d$.

Before we list the most important properties of $d_A$ and $M_A$, we need to recall two more notions.

\begin{mydef} \label{def:Involution}
Let $X$ be a set, $d$ a semi-metric on $X$, i.\,e. a symmetric, non-negative and non-degenerate map $d: X^2 \rightarrow [0, \infty]$ which may have one point at infinity which is denoted by $\infty$. Let $o \in X \diagdown \{ \infty \}$. The {\it involution of $d$ at $o$} is defined by

\[ d_o(x,y) := \frac{d(x,y)}{d(x,o)d(o,y)}. \]

Since we said at the beginning that "infinite distances cancel", this means in particular

\[ d_o(x, \infty) = \frac{1}{d(x, o)}. \]

for all $x \in X \diagdown \{ \infty \}$ and

\[ d_o(\infty, \infty) = \frac{1}{\infty} = 0. \]

\end{mydef}

We can now formulate the most important properties of $d_A$ and $M_A$.

\begin{mythm} \label{thm:BuyaloSemiMetrics}
Let $(X,M)$ be a generalized M\"obius space, $d_A$ the map induced by $A$ for any non-degenerate triple $A$ in $X$. Let $M_A$ be the map induced by $d_A$. Then the following hold:

\begin{enumerate}

\item[1)] Every $d_A$ is a semi-metric on $X$ i.\,e. $d_A$ is symmetric, non-negative and non-degenerate.

\item[2)] For all $x \neq \omega$, $d_{(\omega, \alpha, \beta)}(x, \omega) = \infty$. Moreover, $d_{(\omega, \alpha, \beta)}(\alpha, \beta) = 1$.

\item[3)] Let $A = (\omega, \alpha, \beta)$, $A' = (\omega, \beta, \alpha)$, $A'' = (\beta, \alpha, \omega)$. Then $d_A = d_{A'}$ and $d_{A''}(x,y) = \frac{d_A(x,y)}{d_A(x, \beta) d_A(\beta, y)}$.

\item[4)] Let $(\omega, \alpha, \beta, b)$ be a non-degenerate quadruple in $X$. Then $d_{(\omega, \alpha, \beta)} = \lambda d_{(\omega, \alpha, b)}$ for some constant $\lambda > 0$.

\item[5)] For each non-degenerate triple $A$, $M_A = M$. In particular, $M_A$ is a generalized M\"obius structure.

\end{enumerate}
\end{mythm}

A proof of these properties can be found in \cite{Buyalo}.


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% ON THE TOPOLOGY OF M�BIUS SPACES

%-------------------------------------------------------------------------------------------------------------------------------------

\section{The topology of generalized M\"obius spaces}

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% CONSTRUCTING THE TOPOLOGY

%-------------------------------------------------------------------------------------------------------------------------------------

\subsection{Constructing the topology}

In the last section, we have seen that, given a generalized M\"obius structure $M$, we can define a family of semi-metrics $d_A$ whose induced generalized M\"obius structure is again $M$. So this family of semi-metrics intrinsically belongs to the generalized M\"obius structure $M$. In this section, we use this family of semi-metrics to define a topology on $X$. Furthermore, we will see that if $X$ is a metric space and $M$ is the generalized M\"obius structure induced by the metric, the topology induced by the semi-metrics is the same as the topology induced by the metric. The definition of the topology below is taken from \cite{Buyalo}.\\

Let $A = (\omega, \alpha, \beta)$ be a non-degenerate triple. For $y \in X \diagdown \{ \omega \}$ and $r > 0$, define

\[ B_{A,r}(y) := \{ x \in X \vert d_A(x,y) < r \}�\]

to be the {\it open ball} around $y$ of radius $r$ with respect to $d_A$. We take the family of all open balls for all non-degenerate triples $A$, all positive radii $r$ and all points $y \in X \diagdown \{ \omega \}$ as a subbasis to define a topology $\mathcal{T}_M$ on $X$. This is the {\it topology on $X$ induced by $M$}. From now on, whenever we speak of a generalized M\"obius space $(X,M)$ we assume it to be endowed with the topology induced by $M$, unless stated otherwise.\\


\begin{lem}
Consider $[0, \infty]$ with the topology where open neighbourhoods of $\infty$ are complements of compact sets in $[0, \infty)$ and open neighbourhoods of other points are just the standard euclidean open neighbourhoods. Let $(X,M)$ be a generalized M\"obius space, $A$ a non-degenerate triple in $X$ and $y \in X$. Then the maps $d_A( \cdot, y), d_A(y, \cdot) : X \rightarrow [0, \infty]$ are continuous with respect to $\mathcal{T}_M$.
\end{lem}

\begin{proof}
First note that if $y = \omega$, $d_A( \cdot, y) \equiv \infty$ is constant and hence continuous. If $y \neq \omega$, we start by defining the set

\[ C_{A,r}(y) := \{ x \in X \vert d_A(x,y) > r \} \]

which can be thought of as the complement of a `closed' ball (again $y \neq \omega$). Showing that these sets are open in $\mathcal{T}_M$ is the essential part of the proof.

Let $(a,b)$ be an open interval in $\mathbb{R}$ (possibly unbounded) and consider the map $f:= d_A(\cdot, y)$ for some fixed $y \neq \omega$. Then the preimage of $(a,b)$ under $f$ is equal to $B_{A,b}(y) \cap C_{A,a}(y)$. Hence, if $C_{A,a}(y)$ is open for all $a \geq 0$, it follows that $d_A$ is continuous.\\

By Theorem \ref{thm:BuyaloSemiMetrics}, we know that for any non-degenerate triple $(\omega, \alpha, y)$ and every $x \in X \diagdown \{ y, \omega \}$

\begin{equation} d_{(\omega, \alpha, y)} (x, y) d_{(y, \alpha, \omega)}(x, \omega) = 1 \end{equation}

and

\[ d_{(\omega, \alpha, \beta)}(x,y) = \lambda d_{(\omega, \alpha, o)}(x,y).\]

Therefore, we see that

\[ d_{(y, \alpha, o)}(x, \omega) = \lambda d_{(y, \alpha, \omega)}(x,\omega) = \frac{\lambda}{d_{(\omega, \alpha, y)}(x, y)} = \frac{\lambda}{\mu d_{(\omega, \alpha, \beta)}(x,y)} \]


for $y,$ $\omega,$ $\alpha,$ $\beta,$ $o$ mutually different and $\lambda,$ $\mu > 0$ depending only on $\alpha,$ $\omega,$ $y,$ $o$ and $\alpha,$ $\beta,$ $\omega,$ $y$ respectively. This immediately implies that $B_{(\omega, \alpha, \beta),r}(y) = C_{(y, \alpha, o), \frac{\lambda}{\mu r}}(\omega)$ for some $\lambda,$ $\mu > 0$ (notice that the points $\omega$ and $y$ behave nicely). Since this is true for all $\omega,$ $\alpha,$ $\beta,$ $y,$ $o$ and $r$ as above, we see that $C_{A,r}(y)$ is open for all non-degenerate triples $A$, all $r > 0$ and all $y \in X$. This implies the Lemma.
\end{proof}

\begin{rem}
Notice that the proof of the continuity of $d_A$ relies on the fact that we take the open balls of {\it all} semi-metrics $d_A$. It is not sufficient to take just one - or some - of the non-degenerate triples. Only as a collective can they define a topology that comes close to a metric topology. In particular, the involution again plays a critical role in making the entire construction work.
\end{rem}

\begin{lem}
The topological space $(X,\mathcal{T}_M)$ is Hausdorff.
\end{lem}

\begin{proof}
Let $x,y \in X$ be two different points. Choose a point $\alpha \in X \diagdown \{x,y\}$. We know that for every $z \in X$,

\[ d_{(y,\alpha, x)}(x,z) = \frac{1}{d_{(x,\alpha, y)}(y,z)}. \]

Therefore, the intersection of the two open balls $B_{(y, \alpha, x),1}(x), B_{(x,\alpha,y),1}(y)$ is empty. This completes the proof.

\end{proof}

Consider two generalized M\"obius spaces $(X,M), (X',M')$. So far, we avoided maps between generalized M\"obius spaces. While we still will not use them for a while, let us give a definition here and state a first nice property.

\begin{mydef} \label{def:MobiusEquivalence}
Let $(X,M), (X',M')$ be generalized M\"obius spaces. A bijection $f : X \rightarrow X'$ is called a {\it M\"obius equivalence} if and only if for every admissible quadruple $(w,x,y,z) \in \mathcal{A}_4$, we have

\[ M(wxyz) = M(f(w)f(x)f(y)f(z)). \]
\end{mydef}

\begin{lem} \label{lem:homeomorphism}
Let $(X,M), (X',M')$ be two generalized M\"obius spaces and $f : X \rightarrow X'$ a M\"obius equivalence. Then $f$ is a homeomorphism when we equip $X$ and $X'$ with their respective M\"obius topology.
\end{lem}

\begin{proof}
Let $A$ be a non-degenerate triple in $X$. Since $f$ is a bijection, it sends $A$ to a non-degenerate triple denoted $f(A)$ in $X'$. Looking at the definition of the semi-metric $d_A$, we immediately see that, since $f$ preserves the generalized M\"obius structure, we have for all $x,y \in X$

\[ d_A(x,y) = d_{f(A)}(f(x),f(y)). \]

Thus, the map $f$ sends an open ball $B_{A,r}(x)$ in $X$ to the open ball $B_{f(A),r}(f(x))$ in $X'$ and a subbasis of $\mathcal{T}_M$ to a subbasis of $\mathcal{T}_{M'}$. The same is true for $f^{-1}$ which proves the Lemma.

\end{proof}

%-------------------------------------------------------------------------------------------------------------------------------------

% COMPATIBILITY WITH THE METRIC TOPOLOGY

%-------------------------------------------------------------------------------------------------------------------------------------

\subsection{Compatibility with the metric topology}

The topology defined in the last section is an extension of what we know in the context of M\"obius structures on metric spaces. This is the statement of the following Theorem.

\begin{mythm} \label{thm:metrictopology}
Let $(X,d)$ be a metric space. Denote by $\mathcal{T}_d$ the topology on $X$ induced by $d$ and by $M$ the induced generalized M\"obius structure. Let $\mathcal{T}_M$ be the topology induced by $M$ and the family of semi-metrics $d_A$ induced by $M$. Then $\mathcal{T}_d = \mathcal{T}_M$.
\end{mythm}

Theorem \ref{thm:metrictopology} tells us that our constructions of $d_A$ and $\mathcal{T}_M$ do in fact align with the metric topology of a metric space. This theorem is a generalization of Buyalo's theorem for boundaries at infinity of hyperbolic spaces.


\begin{proof}%[Proof of Theorem \ref{thm:metrictopology}]
Since $X$ is a metric space, our observations from section 2, specifically equation (2) and the definition of $d_A$, tell us that $d_A(x,y) = \frac{d(x,y)}{d(x, \omega)d(\omega,y)}\frac{d(\alpha, \omega)d(\omega, \beta)}{d(\alpha, \beta)}$.

We need to show that the open balls in $d$ are open with respect to $\mathcal{T}_M$ and that the open balls with respect to the $d_A$ are open with respect to $\mathcal{T}_d$. We denote by

\[ B_{s}(y) :=  \{ x \in X \vert (x | y) > -\ln(s) \} = \{ x \in X \vert d(x,y) < s \} \]

the open ball of radius $s$ with respect to $d$ and by

\[ B_{A,s}(y) := \{ x \in X \vert (x | y)_A > -\ln(s) \} = \{ x \in X \vert d_A(x,y) < s \} \]

the open ball of radius $s$ with respect to $d_A$. These sets generate $\mathcal{T}_d$ respectively $\mathcal{T}_M$.\\

We now show that $B_{A,r}(y)$ is open with respect to $\mathcal{T}_d$ for all non-degenerate triples $A, r > 0$ and $y \in X \diagdown \{ \omega \}$. Let $x \in B_{A,r}(y)$, i.\,e. $(x | y)_A > -\ln(r)$. Choose $\epsilon > 0$ such that $(x | y)_A - \epsilon > -\ln(r)$. Since the logarithm is continuous, there exists an $\tilde{\epsilon} > 0$ such that for a positive real number $\mu$, we have $\vert \mu -1 \vert < \tilde{\epsilon} \Rightarrow \vert \ln(\mu) - 0 \vert < \epsilon$.\\

If $x = y$, note that $d_A(z,x)$ is continuous in $z$ with respect to $\mathcal{T}_d$ as long as $z$ is away from $\omega$. If $z$ converges to $x$ with respect to $\mathcal{T}_d$, then it is certainly away from $\omega$, as $x = y \neq \omega$. Therefore, there exists some $\delta > 0$ such that for $d(z,x) < \delta$, we have $d_A(z,x) < \epsilon := r$. Therefore, $B_{\delta}(y) \subset B_{A,r}(y)$.\\

Now suppose $x \neq y$. Consider the expression

\[ \frac{d_A(z,y)}{d_A(x,y)} = \frac{d(z,y)}{d(x,y)}\frac{d(x, \omega)}{d(z, \omega)}.\]

If we assume for $z$ to be close to $x$ with respect to $d$, then this expression is continuous in $z$ with respect to both topologies and converges to $1$ if $z$ converges to $x$. %This requires that d is a metric (or quasi-metric?)
In particular, if $z \rightarrow x$ with respect to $\mathcal{T}_d$, there exists some $\delta > 0$ such that for $d(z,x) < \delta$, we have $\vert \frac{d_A(z,y)}{d_A(x,y)} - 1 \vert < \tilde{\epsilon}$. Applying our continuity statement for the logarithm above, we get that

\[ \epsilon > \left| -\ln \left(\frac{d_A(z,y)}{d_A(x,y)}\right) \right| = \vert (z | y)_A - (x | y)_A \vert. \]

Therefore

\[ (z | y)_A > (x | y)_A - \epsilon > -\ln(r). \]

We conclude that for all $z \in \{ z \in X \vert d(z,x) < \delta \} = B_{\delta}(x), (z | y)_A > -\ln(r)$, i.\,e. $B_{\delta}(x) \subset B_{A, r}(y)$. Since $x \in B_{\delta}(x)$, we conclude that $B_{A,r}(y)$ is open with respect to $\mathcal{T}_d$. Hence we see that $\mathcal{T}_d$ is finer than $\mathcal{T}_M$.\\

In order to show that $\mathcal{T}_M$ is finer than $\mathcal{T}_d$, we consider the open ball $B_{r}(y)$ for $r > 0, y \in X$. Let $x \in B_{r}(y)$. Since $d$ is a metric, there exists a smaller ball around $x$ contained in $B_{r}(y)$, i.e. there exists $r' < r$ such that $B_{r'}(x) \subset B_{r}(y)$. Replacing $r'$ by $\epsilon$, it is now enough to show that for every $\epsilon > 0$, we can find $\delta > 0$ such that $B_{A,\delta}(x) \subset B_{\epsilon}(x)$.\\%Here we use again that d is a metric. 

We first claim that for any sequence $z_n \in X$ which converges to $x$ in $\mathcal{T}_M$, the sequence $d(z_n, \omega)$ does not converge to zero for every $\omega \in X \diagdown \{ x \}$. Put $A = (\omega, \alpha, x)$ for any $\alpha \in X \diagdown \{ x, \omega \}$ and consider

\[ d_A(z_n, x) = \frac{d(z_n, x)}{d(z_n, \omega)d(\omega, x)}\frac{d(\alpha, \omega)d(\omega, x)}{d(\alpha, x)} = \frac{d(z_n, x) d(\alpha, \omega)}{d(z_n, \omega)d(\alpha, x)}. \]

Since we know that $z_n \rightarrow x$ in $\mathcal{T}_M$ we have

\[ \lim_{n \rightarrow \infty} d_A(z_n, x) = 0. \]

Knowing this, $d(z_n, \omega)$ can only converge to zero if $d(z_n, x)$ does as well. However, since $d$ is a metric and $x \neq \omega$, this cannot be the case. Hence, $d(z_n, \omega)$ does not converge to zero for all $\omega \neq x$ if $z_n \rightarrow x$ in $\mathcal{T}_M$. Note that $d(z_n, \omega)$ also does not diverge to infinity. This can be seen by considering $d_A(z_n,y)$ for $A = (\omega, \alpha, x)$ and $y \in X \diagdown \{ x, \omega \}$ such that $d_A(z_n, y)$ does not converge to zero, (such $A$ and $y$ have to exist, as $z_n \rightarrow x$ in $\mathcal{T}_M$, which is Hausdorff).\\

Choose two points $\omega, \alpha \in X$ such that $A := (\omega, \alpha, x)$ is a non-degenerate triple. Let $z \in X$ be `arbitrarily close' to $x$ in $\mathcal{T}_M$. Then

\[ d_A(z, x) = \frac{d(z,x)}{d(z, \omega)d(\omega, x)}\frac{d(\alpha, \omega)d(\omega, x)}{d(\alpha, x)} = \frac{d(z,x) d(\alpha, \omega)}{d(z, \omega) d(\alpha, y)}. \]

Since we chose $z$ to be arbitrarily close to $x$ in $\mathcal{T}_M$ we can assume that $d_{A}(z,x) < \delta$. Hence, we have

\[ \delta > d_A(z,x) = \frac{d(z,x) d(\alpha, \omega)}{d(z, \omega) d(\alpha, x)}. \]

We can reformulate this as

\[ d(z,x) < \delta \frac{d(z, \omega) d(\alpha, x)}{d(\alpha, \omega)} < \delta C \frac{d(\alpha, x)}{d(\alpha, \omega)} \]

for $\delta > 0$ sufficiently small and $C > 0$ some finite constant. The constant $C$ exists as $d(z, \omega)$ does not diverge to infinity. Given any $\epsilon > 0$, we can now choose $0 < \delta < \epsilon \frac{d(\alpha, \omega)}{C d(\alpha, x)}$ and conclude that if $d_A(z,x) < \delta$, then $d(z,x) < \epsilon$. In other words, $B_{A,\delta}(x) \subseteq B_{\epsilon}(x)$. This implies that balls with respect to $d$ are open with respect to $\mathcal{T}_M$. Hence $\mathcal{T}_M$ is finer than $\mathcal{T}_d$. Since we have now shown that both topologies are finer than the other, they are the same, which concludes the proof of Theorem \ref{thm:metrictopology}.

\end{proof}


\begin{rem}
This proof easily extends to extended metric spaces which have a point at infinity. Let $\infty$ denote the point at infinity in the metric space $(X,d)$. Then, for any non-degenerate triple $A = (\infty, \alpha, \beta)$, we have $d_A = \lambda d$ for some positive number $\lambda$. This immediately implies that $\mathcal{T}_d \subseteq \mathcal{T}_M$. To prove equality, one modifies the proof provided above.
\end{rem}

Applying Lemma \ref{lem:homeomorphism} in the context of Theorem \ref{thm:metrictopology} immediately yields the following

\begin{cor}
Let $(X,d), (X',d')$ be -- possibly extended -- metric spaces, $M_d, M_{d'}$ the induced generalized M\"obius structures and $f : (X, M_d) \rightarrow X'$ a M\"obius equivalence. Then $f$ is a homeomorphism with respect to the metric topologies $\mathcal{T}_d, \mathcal{T}_{d'}$.
\end{cor}

\begin{proof}
We know from Lemma \ref{lem:homeomorphism} that $f$ is a homeomorphism with respect to the topologies $\mathcal{T}_M, \mathcal{T}_{M'}$. By Theorem \ref{thm:metrictopology}, the M\"obius topologies and the metric topologies coincide, i.e. $\mathcal{T}_M = \mathcal{T}_d$ and $\mathcal{T}_{M'} = \mathcal{T}_{d'}$. The statement follows.
\end{proof}


\begin{exam}
Consider two copies of the real line. For notation purposes, given a real number $x$, we will write it as $x^{(1)}$ if we consider it on the first real line and as $x^{(2)}$ if we consider it on the second real line. In the same spirit, we denote the first real line by $\mathbb{R}^{(1)}$ and the second one by $\mathbb{R}^{(2)}$. We now form a quotient in the following way. We say that $x^{(i)} \sim y^{(j)}$ if $x = y \neq 0$. This yields a quotient space $X := \mathbb{R}^{(1)} \coprod \mathbb{R}^{(2)} \diagup \sim$ with its quotient topology. This space is often called the real line with doubled zero. We denote the equivalence class $\{ x^{(1)}, x^{(2)} \}$ by $x$ for all $x \neq 0$ and we denote by $(a,b)$ the set of all $x \in X$ such that $a < x < b$, containing both zeroes if $a < 0 < b$.

Note that $X$ is locally homeomorphic to open intervals of $\mathbb{R}$, but it is not Hausdorff at the two zeroes $0^{(1)}, 0^{(2)}$.

\begin{figure}
\begin{tikzpicture}
\draw (0,1) -- (7,1);
\draw (7.2, 1) -- (14.2, 1);
\draw[fill] (7.1,2) circle [radius=0.04];
\draw[fill] (7.1, 0) circle [radius=0.04];
\node [above right] at (7.1,2) {$0^{(1)}$};
\node [below right] at (7.1,0) {$0^{(2)}$};
\end{tikzpicture}
\caption{The two zeroes are drawn as isolated points. However, each of them 'bridges' the gap between the two part of the real line.}
\end{figure}

We now define a semi-metric on $X$ whose open balls form a subbasis for the quotient topology. Let $x^{(i)}, y^{(j)}$ be representatives of two points in $X$. Define

\begin{equation*}
d([x^{(i)}],[y^{(j)}]) := \begin{cases} 1 & \text{$x = y = 0$ and $i \neq j$}\\
\vert x - y \vert & \text{otherwise.}
\end{cases}
\end{equation*}

Clearly, this is a symmetric, non-negative function which is equal to zero if and only if $x^{(i)} \sim y^{(j)}$. Hence it is a semi-metric. From now on, we will denote every element $[x^{(i)}] \in X$ simply by $x$, except if $x = 0$. Note that the topology generated by the open balls with respect to $d$ is exactly the quotient topology.\\

Since $X$ is not Hausdorff, $d$ cannot be a metric. Let us now look at the generalized M\"obius structure induced by $d$ and the induced M\"obius topology. Computations show that the open balls with respect to the induced $d_A$ are exactly the open balls with respect to $d$, except when $A = (0^{(i)}, \alpha, \beta)$ for $i \in \{ 1, 2 \}$. In that case, let $j \in \{ 1, 2 \}$, $i \neq j$ and we get that

\[ d_A(0^{(j)},y) = \frac{ \vert y \vert }{1 \vert y \vert}C_A = C_A = \frac{\vert \alpha \beta \vert}{\vert \alpha - \beta \vert} \]

for all $y \in X \diagdown \{ 0^{(1)}, 0^{(2)}�\}$. This implies that the M\"obius topology has two additional open sets, namely $\{ 0^{(1)} \}, \{ 0^{(2)} \}$. So we see that the M\"obius topology on $X$ is almost the same as before, but it isolates the two zeroes from the rest of the space, making the resulting topological space homeomorphic to the disjoint union of two real lines and two isolated points.\\

The space $X$ is an example of a topological space that is not Hausdorff. Its topology can be generated by a semi-metric $d$. The topology induced by the generalized M\"obius structure induced by $d$ is a different topology and the difference between the two topologies is exactly what is needed to turn $X$ in a Hausdorff space. This demonstrates very nicely how the generalized M\"obius structure forces a specific topology on its underlying set which excludes certain oddities that can arise from arbitrary semi-metrics. In particular, the generalized M\"obius structure ensures that its topology is Hausdorff. Notice that, in this example, the generalized M\"obius structure removes the discontinuity the semi-metric had with respect to its own topology, i.\,e. $d$ is continuous in both variables with respect to the M\"obius topology, but not with respect to the topology it itself induces.

\end{exam}



%-------------------------------------------------------------------------------------------------------------------------------------

% M�BIUS STRUCTURES, THEIR TOPOLOGY AND QUASI-METRICS

%-------------------------------------------------------------------------------------------------------------------------------------

\section{Generalized M\"obius structures, their topology and quasi-metrics}

In Lemma \ref{lem:basicproperties} in Section \ref{subsec:motivation} we stated the four basic properties of M\"obius structures induced by metrics which we then used to define generalized M\"obius structures. The first of these properties was essentially a geometric-combinatorial condition on the sides of a tetraeder. The second and third condition were about degenerate quadruples. The fourth condition was a geometric statement that considered three tetraeder that are in a specific position to each other.

So far, we used these properties to show that generalized M\"obius structures are always induced by semi-metrics, that they carry a natural topology and both of these are compatible with what a metric gives us in the classical context of M\"obius structures on metric spaces. Going forward, it turns out that there is a fifth condition one can impose on generalized M\"obius structures, which becomes more and more important in order to generalize several notions of metric spaces to generalized M\"obius spaces. In this section, we will introduce this condition and show that this condition is closely connected to the question whether a generalized M\"obius structure can be induced by a `nice' quasi-metric or not.
%-------------------------------------------------------------------------------------------------------------------------------------

% THE (CORNER)-CONDITION Find a better name for this condition.

%-------------------------------------------------------------------------------------------------------------------------------------

\subsection{The (corner)-condition}

We start by recalling the definition of a quasi-metric.

\begin{mydef} \label{def:quasimetric}
Let $X$ be a set and $K \geq 1$. A $K$-quasi-metric is a map $d : X \times X \rightarrow [0,\infty)$ satisfying

\begin{enumerate}

\item $d(x,y) = 0 \Leftrightarrow x = y$

\item $\forall x,y \in X : d(x,y) = d(y,x)$

\item $\forall x,y,z \in X : d(x,z) \leq K\max(d(x,y),d(y,z))$.

\end{enumerate}

An extended quasi-metric is a map $d : X \times X \rightarrow [0, \infty]$ satisfying the same conditions as above, except that there is exactly one point $\omega \in X$ such that for all $x \in X \diagdown \{ \omega \}$, $d(x, \omega) = \infty$. Furthermore, for all $x,y \in X \diagdown \{ \omega�\}$, $d(x,y) < \infty$.

\end{mydef}

In order to formulate this fifth condition, it is convenient to think of a generalized M\"obius structure as the map $crt : \mathcal{A}^4 \rightarrow \mathbb{R}P^2$ defined by $crt := \overline{\Phi} \circ F^{-1} \circ M$ where $F$ and $\overline{\Phi}$ are the maps defined in section \ref{subsec:motivation}. The map $crt$ fully encodes the generalized M\"obius structure $M$ and vice-versa, which is why we will also refer to $(X, crt)$ as a generalized M\"obius structure from now on. Recall that the image of $crt$ is required to be contained in $\overline{\Delta} = \{ (a : b : c) \vert x,y,z > 0 \} \cup \{ (0:1:1), (1:0:1), (1:1:0) \}$. If we consider the plane $\{ (x,y,z) \in \mathbb{R}^3 \vert x + y + z = 1 \}$, then every element of $\overline{\Delta}$ has a unique representative in the plane. Specifically, the representatives of $\overline{\Delta}$ are the following set:

\[ D := \{ (x,y,z) \vert x + y + z = 1, x,y,z > 0 \} \cup \left\{ \left(\frac{1}{2},\frac{1}{2},0\right), \left(\frac{1}{2}, 0, \frac{1}{2}\right), \left(0,\frac{1}{2}, \frac{1}{2}\right) \right\} \]

The subset $D$ is the interior of a $2$-simplex in $\mathbb{R}^3$ together with three points on the boundary. By condition 2 and 3 in the definition of a generalized M\"obius structure, we know that the points on the boundary are obtained exactly by the degenerate, admissible quadruples. Consider now the three points $(1:0:0), (0:1:0), (0:0:1)$ which correspond to the corners of the $2$-simplex $D$. We will refer to these three points as the {\it corner points} of $\overline{\Delta}$. By definition, the corner points cannot be obtained by $crt$; however, it is possible that they can be approximated within $\Ima(crt)$. The next lemma tells us that this cannot happen when the generalized M\"obius structure is induced by a metric or quasi-metric.

\begin{lem} \label{lem:cornercondition}
Let $X$ be a set, $d$ a quasi-metric on $X$ and $crt$ the generalized M\"obius structure induced by $d$. Then there exist open neighbourhoods of the corner points in $\mathbb{R}P^2$ such that the image of $crt$ doesn't intersect these neighbourhoods.
\end{lem}

I have been informed of this lemma by Viktor Schroeder, although I don't know of any proof in the literature. 
Lemma \ref{lem:cornercondition} motivates the following definition.

\begin{mydef} \label{def:cornercondition}
Let $X$ be a set, $crt$ a generalized M\"obius structure on $X$. We say that $crt$ or $(X, crt)$ satisfies the {\it (corner)-condition} if there exist open neighbourhoods of the three corner points, such that the image of $crt$ doesn't intersect these neighbourhoods.
\end{mydef}

\begin{figure}
\begin{tikzpicture}
\draw (0,0);
\draw [dashed] (6,0) -- (9,0);
\draw [dashed] (6,0) -- (7.5, 2.6);
\draw [dashed] (9,0) -- (7.5, 2.6);
\draw [fill] (6,0) circle [radius=0.05];
\draw [fill] (9,0) circle [radius=0.05];
\draw [fill] (7.5,2.6) circle [radius=0.05];
\draw [fill] (6.75,1.3) circle [radius=0.03];
\draw [fill] (8.25,1.3) circle [radius=0.03];
\draw [fill] (7.5,0) circle [radius=0.03];
\draw (6,0) circle [radius=0.5];
\draw (9,0) circle [radius=0.5];
\draw (7.5,2.6) circle [radius=0.5];
\node [below left] at (5.7,-0.3) {$(1:0:0)$};
\node [below right] at (9.3,-0.3) {$(0:1:0)$};
\node [above] at (7.5,3) {$(0:0:1)$};
\node at (7.5,1.2) {$\overline{\Delta}$};
\end{tikzpicture}
\caption{A generalized M\"obius structure $crt$ satisfies the (corner)-condition if and only if we can find open neighbourhoods as depicted above, such that the image of $crt$ in $\overline{\Delta}$ doesn't intersect these neighbourhoods.}
\end{figure}


\begin{proof}%[Proof of Lemma \ref{lem:cornercondition}]
Let $d$ be a $K$-quasi-metric on the space $X$ and $crt$ the induced generalized M\"obius structure. Let $(w,x,y,z)$ be an admissible quadruple. We want to show that $crt(wxyz)$ cannot be close to any of the three corner points. We will show this for the corner point $(0:0:1)$. The others work analogously.

In order for the point $crt(wxyz)$ to be close to $(0:0:1)$, the ratio between the first and third component has to be small, as does the ratio between the second and the third component. We will show that this cannot happen. To prove this, we need to make several case distinctions. We leave it to the reader to check that all cases can be handled analogously by simply permuting the roles and properties of $w,x,y,z$.

Let $\epsilon > 0$ (possibly quite large). Consider $crt(wxyz) = ( d(w,x)d(y,z) : d(w,y)d(x,z) : d(w,z)d(x,y))$ and suppose

\[ d(w,x)d(y,z) < \epsilon \]
\[ d(w,y)d(x,z) < \epsilon. \]

We want to bound $d(w,z)d(x,y)$ in terms of $\epsilon$, proving that the ratios $\frac{d(w,x)d(y,z)}{d(w,z)d(x,y)}, \frac{d(w,y)d(x,z)}{d(w,z)d(x,y)}$ cannot become too small. Assume without loss of generality that

\[ d(w,x) \leq d(y,z) \]

\[ d(w,y) \leq d(x,z) \]

and thus

\[ d(w,x) < \sqrt{\epsilon} \]

\[ d(w,y) < \sqrt{\epsilon}. \]

Since $d$ is a $K$-quasi-metric, we have

\[ d(x,y) \leq K \max(d(w,x), d(w,y)). \]

Without loss of generality, $d(w,x) \geq d(w,y)$ and hence

\[ d(x,y) \leq K d(w,x). \]

Further, we have

\[ d(z,w) \leq K \max(d(z,y), d(y,w)). \]

Combining these two inequalities, we conclude

\begin{equation*}
\begin{split}
d(x,y)d(z,w) & \leq K d(w,x)d(z,w)\\
& < K \frac{\epsilon}{d(y,z)}d(z,w)\\
& \leq \begin{cases} K^2 \epsilon & \text{if $d(z,y) \geq d(y,w)$}\\
K^2 d(w,x)d(y,w) \leq K^2 \epsilon & \text{if $d(z,y) < d(y,w)$}.
\end{cases}
\end{split}
\end{equation*}

Therefore, we see that $d(x,y)d(z,w) < K^2 \epsilon$. We now use this to show that $crt$ stays away from the degenerate values. Consider the triple

\[ (a,b,c) := ( d(w,x)d(y,z), d(w,y)d(x,z), d(w,z)d(x,y) ) \in \mathbb{R}^3.\]

The argument above shows that
\[ c \leq K^2\max(a,b). \]

By permuting the roles of $a,b,c$ we get

\[ c \leq K^2\max(a,b) \]
\[ b \leq K^2\max(a,c) \]
\[ a \leq K^2\max(b,c). \]

These inequalities are preserved when we multiply the triple $(a,b,c)$ with a scalar. Thus, if $c \neq 0$ we can rescale $(a, b, c)$ such that $a = \frac{d(w,x)d(y,z)}{d(w,z)d(x,y)}, b = \frac{d(w,y)d(x,z)}{d(w,z)d(x,y)}, c = 1$. Then the inequalities above yield

\[ 1 \leq K^2\max(a,b) \]

or equivalently

\[ \max(a,b) \geq \frac{1}{K^2}. \]

Projecting $(a,b,c)$ to projective space, this yields exactly that $(a:b:c) \notin \{ (a : b : 1) \in \mathbb{R}P^2 \vert a < \frac{1}{K^2}, b < \frac{1}{K^2} \}$, which is an open neighbourhood of $(0:0:1)$ in $\mathbb{R}P^2$. Since $(a:b:c) = crt(wxyz)$, this implies that $\Ima(crt)$ doesn't intersect with the $\frac{1}{K^2}$-neighbourhood of $(0:0:1)$. Analogously, the other two inequalities imply that $\Ima(crt)$ doesn't intersect the $\frac{1}{K^2}$-neighbourhood of $(1:0:0)$ and $(0:1:0)$, which completes the proof of the Lemma.

\end{proof}


Lemma \ref{lem:cornercondition} tells us that classical M\"obius structures satisfy the (corner)-condition. Hence there is a chance that it is a sensible condition to demand from a generalized M\"obius structure. In the following sections we will show that this is indeed the case.

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% QUASI-METRICS WITH A POINT AT INFINITY

%-------------------------------------------------------------------------------------------------------------------------------------

\subsection{Quasi-metrics with a point at infinity}

Let $crt$ be a generalized M\"obius structure. In section \ref{sec:Generalized Mobius structures}, we have constructed an infinite family of semi-metrics $\{ d_A \}_A$ out of $crt$. Each of these semi-metrics has one point of $X$ at infinity. It turns out that the (corner)-condition has a nice consequence for such semi-metrics.

\begin{prop} \label{prop:cornerconditionatinfinity}
Let $d$ be a semi-metric on the set $X$ such that $d$ has a point at infinity. Then $d$ is a quasi-metric if and only if its induced generalized M\"obius structure satisfies the (corner)-condition.
\end{prop}

\begin{proof}
Lemma \ref{lem:cornercondition} from the last section immediately implies one direction of the proof. Suppose now $crt$ satisfies the (corner)-condition. We want to show that $d$ is a quasi-metric.

Denote the point at infinity with respect to $d$ by $\omega$. Let $x,y,z \in X$. If two of the points are the same, or if one of the three points equals $\omega$, then the inequality for quasi-metrics is immediately satisfied. So assume $x,y,z$ are mutually different and different from $\omega$. Then $(x,y,z,\omega)$ is a non-degenerate quadruple and we can look at the cross-ratio-triple

\[ crt(xyz\omega) = (d(x,y) : d(x,z) : d(y,z)). \]

Recall the special definition of $crt$ when one of the points in the quadruple lies at infinity. Since $crt$ satisfies the (corner)-condition, we know that there is an open neighbourhood of $(1:0:0)$ such that $crt(xyz\omega)$ doesn't lie within that neighbourhood. Note that the following collection of subsets forms a basis of the neighbourhoods of $(1:0:0)$.

\[ N_{\epsilon} := \{�(1:b:c) \vert b,c \in (-\epsilon, \epsilon) \}, \; \epsilon > 0 \]

Hence there exists an $\epsilon > 0$ such that $crt(xyz\omega) \neq N_{\epsilon}$. This means nothing else than

\[ \max \left( \frac{d(x,z)}{d(x,y)}, \frac{d(y,z)}{d(x,y)} \right) \geq \epsilon \]

or equivalently

\[ \frac{1}{\epsilon}\max(d(x,z), d(z,y)) \geq d(x,y). \]

Thus, $d$ is a $\frac{1}{\epsilon}$-quasi-metric.

\end{proof}

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% BOUNDED QUASI-METRICS

%-------------------------------------------------------------------------------------------------------------------------------------

\subsection{Bounded Quasi-metrics}

Proposition \ref{prop:cornerconditionatinfinity} from the last section tells us that for a given generalized M\"obius structure, the (corner)-condition is the same as demanding that all semi-metrics that have a point at infinity and induce this generalized M\"obius structure are quasi-metrics. This result in particular applies to the semi-metrics $d_A$ induced by a given generalized M\"obius structure $crt$. Since every point $\omega \in X$ gets sent to infinity by infinitely many of the $d_A$, one may wonder how special this point at infinity really is and whether we might in fact find a bounded quasi-metric that also induces the generalized M\"obius structure $crt$. This can be done by using the following well-known construction which can be found for example in \cite{BuyaloSchroeder}.

Let $d$ be a $K$-quasi-metric on $X$ with a point at infinity. Denote the point at infinity by $\omega$. We now add an auxiliary point $z$ to the set $X$ and extend $d$ to the space $X \cup \{ \zeta \}$ in the following way: Choose a base point $\zeta_0 \in X$. Now define

\[ d(\zeta,x) = d(x,\zeta) := d(x,\zeta_0) + 1 \]

for all $x \in X$. This is obviously still a semi-metric (we don't care if its a quasi-metric). Given the space $(X \cup \{ \zeta \}, d)$, we now take the involution at the point $\zeta$. This gives us a new semi-metric

\[ \tilde{d}(x,y) = \frac{d(x,y)}{d(x,\zeta)d(\zeta,y)} \]

Restricting $\tilde{d}$ to $X$ yields a new semi-metric $\tilde{d}$ on $X$.

By Proposition 5.3.6 from \cite{BuyaloSchroeder}, $\tilde{d}$ is a $K'$-quasi-metric for some $K' \geq K^2$. Furthermore,

\begin{equation*}
\begin{split}
\tilde{d}(x,y) & = \frac{d(x,y)}{(d(x,\zeta_0) + 1)(d(y,�\zeta_0) + 1)}\\
& \leq K \frac{\max(d(x, \zeta_0),d(y,\zeta_0))}{(d(x, \zeta_0) + 1)(d(y, \zeta_0) + 1)}\\
& < K
\end{split}
\end{equation*}

and thus, $\tilde{d}$ is a bounded quasi-metric on $X$ inducing $M$. We summarize the results from the last two subsections in the following Theorem.


\begin{mythm} \label{thm:boundedquasimetrics}
Let $(X,crt)$ be a generalized M\"obius space which satisfies the (corner)-condition. Then all $d_A$ are quasi-metrics with a uniform constant $K \geq 1$. Further, the construction above yields a bounded quasi-metric on $X$ whose induced generalized M\"obius structure equals $crt$.
\end{mythm}

\begin{rem}
Instead of using the result from \cite{BuyaloSchroeder}, it can be shown that $\tilde{d}$ is a $2K$-quasi-metric, which is a better constant than the one provided by \cite{BuyaloSchroeder} when $K > 2$.
\end{rem}



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% COMPLETENESS OF M�BIUS SPACES

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\section{Completeness of generalized M\"obius spaces} \label{sec:CompletenessofMobiusSpaces}

In this section, we revisit the notion of completeness for metric spaces and study it from a M\"obius point of view. It turns out that the notion of Cauchy sequences has a natural counterpart on generalized M\"obius spaces which almost coincides with the notion of Cauchy sequences on metric spaces. We will use this to introduce a notion of completeness of generalized M\"obius spaces and show how it relates to metric completeness when considering a metric space with its generalized M\"obius structure. Furthermore, we will provide a condition under which a generalized M\"obius space admits a naturally unique minimal completion.

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% CAUCHY SEQUENCES IN M�BIUS SPACES

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\subsection{Cauchy sequences in generalized M\"obius spaces} \label{sec:CauchysequencesinMobiusSpaces}

Let $(X,d)$ be a metric space, $M$ its induced generalized M\"obius structure. We recall that a Cauchy sequence -- in its usual sense on a metric space -- is a sequence $(x_n)_n$ in $X$ such that for all $\epsilon > 0$ there exists a natural number $N_{\epsilon}$ such that for all $m,n \geq N_{\epsilon}$, we have $d(x_m,x_n) < \epsilon$. Our goal is to generalize this notion to generalized M\"obius spaces. It may be tempting to simply generalize the statement above to quasi- and semi-metrics and use that as a definition, but since a generalized M\"obius structure can be induced by many different semi-metrics, a definition relying only on the generalized M\"obius structure itself is more desirable.\\

Before we formulate the key insight, we need some notation. Given the generalized M\"obius structure $M$, we denote its three components $M(wxyz) = (a(wxyz),b(wxyz),c(wxyz))$. Let $d$ be a semi-metric that induces $M$. If $d$ has a point at infinity, we denote that point by $\omega$. Further, consider a sequence $(x_{n,m})_{n,m \in \mathbb{N}}$ in $X$. We say that $\lim \limits_{n,m \rightarrow \infty} x_{n,m} = y$, if and only if for all $\epsilon > 0$ there exists an $N_{\epsilon}$ such that for all $n,m \geq N_{\epsilon}$, we have $d(x_{n,m}, y) < \epsilon$.

In what follows below, we will often consider a sequence $(x_n)_n$ and a pair of points $y, z \in X \diagdown \{ \omega \}$ such that $y \neq z$ and $d(x_n, y)$ and $d(x_n, z)$ both don't converge to zero. Given a sequence $(x_n)_n$, we will refer such a pair $y,z$ as {\it good pairs}.

We can now characterize Cauchy sequences in terms of the generalized M\"obius structure.

\begin{lem} \label{lem:CauchySequences}
Let $(X,d)$ be a (non-extended) metric space, $(x_n)_{n \in \mathbb{N}}$ a sequence in $X$. The following are equivalent:
\begin{enumerate}

\item[1)] The sequence $(x_n)_n$ is either a Cauchy sequence, or $d(x_n, y) \xrightarrow{n \rightarrow \infty} \infty$ for all $y \in X$.

\item[2)] There exists a good pair $y, z \in X$, such that $\lim \limits_{n,m \rightarrow \infty} crt(x_n, x_m, y, z) = (0:1:1)$.

\item[3)] There exists a good pair $y, z \in X$, $\lim \limits_{n,m \rightarrow \infty} c(x_n, x_m, y, z) = -\infty$.

\end{enumerate}

Further, if 1 holds, then 2 and 3 hold for all good pairs $y, z \in X$. In addition, 2 holds for a good pair $y,z$ if and only if 3 holds for the same good pair $y,z$.
\end{lem}

%There is no proof if this lemma in BeyrerSchroeder. Should I put the full proof in the appendix?


The equivalence of 1) and 2) is stated in Lemma 2.2 of \cite{BeyrerSchroeder}. Furthermore, it is easy to see from the proof that 1) implies 2) for every good pair. We are left to prove ``2) $\Rightarrow$ 3)'' and ``3) $\Rightarrow$ 1)''. For this, we require an auxiliary result. Since it is our goal to generalize Cauchy sequences beyond the realm of metric spaces, we will formulate this result in a more general context.

\begin{lem} \label{lem:AuxiliaryforCauchy}
Let $(X,M)$ be a generalized M\"obius structure satisfying the (corner)-condition and $d$ a quasi-metric that induces $M$. Let $(x_n)_n$ be a sequence in $X$ and suppose there exists a good pair $y,z \in X$ such that $c(x_n, x_m, y, z) \xrightarrow{n,m \rightarrow \infty} -\infty$. Then one of the following two statements holds:

\begin{enumerate}

\item[a)] For every $x \in X \diagdown \{ \omega \}$, there exists some $B_x > 0$, such that $d(x_n,x) < B_x$ for all $n$. Furthermore, $d(x_n,x_m) \xrightarrow{n,m \rightarrow \infty} 0$. We say that $x_n$ is bounded.

\item[b)] For every $x \in X \diagdown \{ \omega \}$, we have $d(x_n, x) \xrightarrow{n \rightarrow \infty} \infty$. We say that $x_n$ diverges to infinity and write $x_n \rightarrow \infty$.

\end{enumerate}
\end{lem}

Lemma \ref{lem:AuxiliaryforCauchy} is a generalization of the statement 3) $\Rightarrow$ 1) in Lemma \ref{lem:CauchySequences}.

\begin{rem}
Lemma \ref{lem:CauchySequences} and Lemma \ref{lem:AuxiliaryforCauchy} also hold for extended metric spaces. One can prove 1) $\Rightarrow$ 2) for the case $y = \omega$ separately (and by symmetry, the same proof works for $z = \omega$). The proof of 2) $\Rightarrow$ 3) that we see below immediately generalizes to extended metric spaces. For 3) $\Rightarrow$ 1), we can use the fact that by Lemma \ref{lem:AuxiliaryforCauchy}, this statement also holds for quasi-metrics. If $y = \omega$ for a given quasi-metric, we can perform involution of $d$ at any point $x \in X \diagdown \{ y,z \}$. This provides us with a quasi-metric that induces the same generalized M\"obius structure, but neither $y$ nor $z$ lies at infinity.
\end{rem}

\begin{proof}[Proof of Lemma \ref{lem:AuxiliaryforCauchy}]

Let $(x_n)_n$ be a sequence in the generalized M\"obius space $(X,M)$, let $d$ be a quasi-metric that induces $M$ and let $y,z$ be a good pair such that $c(x_n,x_m,y,z) \xrightarrow{n,m \rightarrow \infty} -\infty$. Let $y',z'$ be another good pair. By definition of the generalized M\"obius structure induced by $d$, we can write

\[c(x_n, x_m, y, z) = (x_n | y) + (x_m | z) - (x_n | x_m) - (y | z) = \ln \left( \frac{d(x_n,x_m)d(y,z)}{d(x_n,y)d(x_m,z)} \right). \]

Using this equality, the statement $c(x_n, x_m,y,z) \xrightarrow{n,m \rightarrow \infty} -\infty$ becomes equivalent to

\begin{equation} \label{eq:coefficientc}
\frac{d(x_n,x_m)d(y,z)}{d(x_n,y)d(x_m,z)} \xrightarrow{n,m \rightarrow \infty} 0.
\end{equation}\\

We will distinguish between two cases, which will turn out to be exactly the distinction between Case a) and Case b). Suppose there exists some $x \in X \diagdown \{ \omega \}$ and some constant $B > 0$ such that $d(x_n, x) < B$ for all $n$. We want to show that we are in Case a).

Since $d$ is a quasi-metric, we have that for all $x' \in X \diagdown \{ \omega \}$,

\[ d(x_n, x') \leq K \max(d(x_n, x), d(x,x')) \leq K \max(B, d(x,x')).\]

Therefore, we see that $d(x_n,x')$ is bounded for all $x' \in X \diagdown \{ \omega \}$. In particular, $d(x_n,y), d(x_n,z)$ are both bounded by some constant $B > 0$.

We are left to show that $d(x_n,x_m) \xrightarrow{n,m \rightarrow \infty} 0$. For this we look at the convergence condition (\ref{eq:coefficientc}). Using the fact that $x_n$ is bounded yields

\begin{equation*}
\begin{split}
\frac{d(x_n,x_m)d(y,z)}{d(x_n,y)d(x_m,z)} & \geq d(x_n,x_m) \frac{d(y,z)}{B^2}.
\end{split}
\end{equation*}

Since the left-hand-side of this equation goes to zero by assumption, the right-hand-side has to go to zero as well. Hence we see that $d(x_n,x_m) \xrightarrow{n,m \rightarrow \infty} 0$.\\

We are left to show that we end up in Case b), whenever there is no $x \in X \diagdown \{ \omega \}$ such that $d(x_n,x)$ is bounded. Suppose $d(x_n,x)$ is unbounded for all $x \in X \diagdown \{ \omega \}$. Then there exists a subsequence $(x_{n_i})_i$ of $(x_n)_n$ such that $d(x_{n_i},x) \rightarrow \infty$ for one (and hence all, since $d$ is a quasi-metric) $x \in X \diagdown \{ \omega \}$.

Now suppose by contradiction that $d(x_n,x)$ does not converge to infinity for one and hence all $x \in X \diagdown \{ \omega \}$. Then we find another subsequence $(x_{m_j})_j$ of $(x_n)_n$, which is bounded. In particular, we find a constant $B > 0$ such that

\[ d(x_{m_j}, y) \leq B \]

and

\[ d(x_{m_j}, z) \leq B \]

for all $j$. From our treatment of Case a), we know that for this subsequence, $d(x_{m_j}, x_{m_{j'}}) \xrightarrow{n,m \rightarrow \infty} 0$. In particular, we find a number $J$ such that for all $j,j' \geq J$, we have

\[ d(x_{m_j},x_{m_{j'}}) < 1. \]

Now, we estimate the distance between these two subsequences. For this, we need to to take $x_{m_J}$ as an auxiliary point. Since $x_{n_i}$ diverges to infinity, there is a number $I$ such that $d(x_{m_J}, x_{n_i}) > \max(K, K \cdot B)$ for all $i \geq I$. Now we use the fact that $d$ is a quasi-metric to get that for all $i \geq I, j \geq J$ we have

\[ \max(K, K \cdot B) \leq d(x_{m_J}, x_{n_i}) \leq K \max( d(x_{m_J}, x_{m_{j}}), d(x_{m_{j}}, x_{n_i}) ) = K d(x_{m_{j}}, x_{n_i}), \]

where the last equality follows from the fact that $d(x_{m_J}, x_{m_j}) < 1$ for all $j \geq J$. Now consider for $i \geq I, j \geq J$

\begin{equation*}
\begin{split}
\frac{d(x_{m_j},x_{n_i})d(y,z)}{d(x_{m_j},y)d(x_{n_i},z)} & \geq \frac{d(x_{m_j},x_{n_i})d(y,z)}{B d(x_{n_i},z)}\\
& \geq \frac{d(x_{m_j},x_{n_i})d(y,z)}{B K \max( d(x_{n_i}, x_{m_j}), d(x_{m_j},z) )}\\
& = \frac{d(x_{m_j},x_{n_i})d(y,z)}{B K d(x_{n_i}, x_{m_j}) }\\
& = \frac{d(y,z)}{BK},
\end{split}
\end{equation*}

where in the second-to-last step we use the fact that $d(x_{n_i}, x_{m_j}) \geq \max(1,B) \geq B \geq d(x_{m_j},z)$ for all $i \geq I, j \geq J$. This inequality shows that $\frac{d(x_{m_j},x_{n_i})d(y,z)}{d(x_{m_j},y)d(x_{n_i},z)}$ is bounded from below by a positive constant. But by assumption, $\frac{d(x_{m_j},x_{n_i})d(y,z)}{d(x_{m_j},y)d(x_{n_i},z)}$ converges to zero, a contradiction. We see that, if a subsequence $(x_{n_i})_i$ diverges to infinity, the sequence $(x_n)_n$ has to diverge to infinity as well. Thus, we are in Case b), which completes the proof.

\end{proof}

\begin{proof}[Proof of Lemma \ref{lem:CauchySequences}]
Let (X,d) be a non-extended metric space, $(x_n)_n$ a sequence in $X$ and $y,z \in X$ such that $\lim \limits_{n \rightarrow \infty} x_n \neq y,z$.\\

1) $\Rightarrow$ 2): Instead of proving just 1) $\Rightarrow$ 2), which follows directly from \cite{BeyrerSchroeder}, we will also prove the second part of the Lemma, i.e. we will prove that $\lim\limits_{n,m \rightarrow \infty} crt(x_n, x_m, y, z) = (0:1:1)$ for all good pairs $y,z$.\\

Step 1: We start by proving that for every Cauchy sequence, we have

\[ \lim_{n,m \rightarrow \infty} crt(x_n,x_m, y,z) = (0:1:1). \]

Suppose $(x_n)$ is a Cauchy sequence. Let $\epsilon > 0$. Then we know there exists some $N_{\epsilon} \in \mathbb{N}$ such that for all $n, m \geq N_{\epsilon}$, we have $d(x_n, x_m) < \epsilon$. Since $y,z$ is a good pair, we can choose $\epsilon$ sufficiently small such that there is an $N_{\epsilon}$ such that additionally, $d(x_n,y),d(x_n,z) > \epsilon^{\frac{1}{4}}$ for all $n \geq N_{\epsilon}$. Therefore, we get

\begin{equation*}
\begin{split}
\frac{d(x_n,x_m)d(y,z)}{d(x_n,y)d(x_m,z)} & < \frac{\epsilon d(y,z)}{d(x_n,y)d(x_m,z)}\\
& < \frac{\epsilon}{\sqrt{\epsilon}}d(y,z)\\
& = \sqrt{\epsilon}d(y,z).
\end{split}
\end{equation*}

Thus we see that $\frac{d(x_n,x_m)d(y,z)}{d(x_n,y)d(x_m,z)} \xrightarrow{n,m \rightarrow \infty} 0$. For symmetry reasons, we immediately see that also $\frac{d(x_n,x_m)d(y,z)}{d(x_n,z)d(x_m,y)} \xrightarrow{n,m \rightarrow \infty} 0$.

We are left to show that $\frac{d(x_n,y)d(x_m,z)}{d(x_n,z)d(x_m,y)} \xrightarrow{n,m \rightarrow \infty} 1$ in order to prove that $crt(x_n,x_m,y,z) \xrightarrow{n,m \rightarrow \infty} (0:1:1)$. Let $n,m \geq N_{\epsilon}$ for $N_{\epsilon}$ as above. Then

\begin{equation*}
\begin{split}
\frac{d(x_n,y)d(x_m,z)}{d(x_n,z)d(x_m,y)} & \leq \frac{d(x_n,y)(d(x_n,z) + d(x_n,x_m))}{d(x_n,z)(d(x_n,y) - d(x_n,x_m))}\\
& < \frac{d(x_n,y)d(x_n,z) + \epsilon d(x_n,y)}{d(x_n,y)d(x_n,z) - \epsilon d(x_n,z)}\\
& \leq \frac{1 + \frac{\epsilon}{d(x_n,z)}}{1 - \frac{\epsilon}{d(x_n,y)}}\\
& \leq \frac{1 + \epsilon^{\frac{3}{4}}}{1 - \epsilon^{\frac{3}{4}}}.
\end{split}
\end{equation*}

We can now introduce $\tilde{\epsilon} > 0$ and choose $\epsilon > 0$ such that

\[ \frac{1 + \epsilon^{\frac{3}{4}}}{1 - \epsilon^{\frac{3}{4}}} < 1 + \tilde{\epsilon} \]

and

\[ \frac{1 - \epsilon^{\frac{3}{4}}}{1 + \epsilon^{\frac{3}{4}}} > 1 - \tilde{\epsilon}. \]

An analogous computation to the one above yields a bound from below, such that we get that for all $\tilde{\epsilon} > 0$, there exists an $N_{\epsilon}$ such that for all $n,m \geq N_{\epsilon}$, we have

\[ 1 - \tilde{\epsilon} < \frac{d(x_n,y)d(x_m,z)}{d(x_n,z)d(x_m,y)} < 1 + \tilde{\epsilon}. \]

It follows that $\frac{d(x_n,y)d(x_m,z)}{d(x_n,z)d(x_m,y)} \xrightarrow{n,m \rightarrow \infty} 1$ and hence $crt(x_n,x_m,y,z) \xrightarrow{n,m \rightarrow \infty} (0:1:1)$. Note that we relied on the triangle-inequality for this part of the proof.\\

Step 2: We show that if $(x_n)$ diverges to infinity, we get

\[ \lim_{n,m \rightarrow \infty} crt(x_n,x_m, y,z) = (0:1:1). \]

Suppose that $d(x_n, x) \rightarrow \infty$ for all $x \in X$ as $n$ goes to infinity (except for the point $x \in X$ that may lie at infinity). Let $C > 0$ be a (large) constant. Choose $y,z \in X$. Then we find a natural number $N_{C}$ such that for all $n \geq N_{C}, d(x_n,y), d(x_n,z) > C$. Hence we see that for all $n, m \geq N_{C}$,

\begin{equation*}
\begin{split}
\frac{d(x_n,x_m)d(y,z)}{d(x_n,y)d(x_m,z)} & \leq \frac{(d(x_n, y) + d(y, x_m))d(y,z)}{d(x_n,y)d(x_m,z)}\\
& = \frac{d(y,z)}{d(x_m,z)} + \frac{d(x_m,y)d(y,z)}{d(x_n,y)d(x_m,z)}\\
& \leq \frac{d(y,z)}{d(x_m,z)} + \frac{(d(x_m,z) + d(z,y))d(y,z)}{d(x_n,y)d(x_m,z)}\\
& = \frac{d(y,z)}{d(x_m,z)} + \frac{d(y,z)}{d(x_n,y)} + \frac{d(y,z)^2}{d(x_n,y)d(x_m,z)}\\
& < d(y,z) ( \frac{1}{C} + \frac{1}{C} + \frac{d(y,z)}{C^2}).
\end{split}
\end{equation*}

As we increase $C$, we see that $\frac{d(x_n,x_m)d(y,z)}{d(x_n,y)d(x_m,z)}$ converges to zero. Since we chose $y,z$ arbitrarily, we can interchange them. This leaves us with the task to show that $\frac{d(x_n,y)d(x_m,z)}{d(x_n,z)d(x_m,y)} \xrightarrow{n,m \rightarrow \infty} 1$. For this, we do the following estimate.

\begin{equation*}
\begin{split}
\frac{d(x_n,y)d(x_m,z)}{d(x_n,z)d(x_m,y)} & \leq \frac{(d(x_n,z) + d(y,z))(d(x_m,y) + d(y,z))}{d(x_n,z)d(x_m,y)}\\
& = 1 + \frac{d(y,z)}{d(x_n,z)} + \frac{d(y,z)}{d(x_m,y)} + \frac{d(y,z)^2}{d(x_n,z)d(x_m,y)}\\
& < 1 + \frac{d(y,z)}{C} + \frac{d(y,z)}{C} + \frac{d(y,z)^2}{C^2}
\end{split}
\end{equation*}

In the same way, we have

\begin{equation*}
\begin{split}
\frac{d(x_n,y)d(x_m,z)}{d(x_n,z)d(x_m,y)} & \geq \frac{(d(x_n,z) - d(y,z))(d(x_m,y) - d(y,z))}{d(x_n,z)d(x_m,y)}\\
& = 1 - \frac{d(y,z)}{d(x_n,z)} - \frac{d(y,z)}{d(x_m,y)} + \frac{d(y,z)^2}{d(x_n,z)d(x_m,y)}\\
& > 1 - \frac{d(y,z)}{C} - \frac{d(y,z)}{C}.
\end{split}
\end{equation*}

From these two estimates, we conclude that $\frac{d(x_n,y)d(x_m,z)}{d(x_m,y)d(x_n,z)} \xrightarrow{n,m \rightarrow \infty} 1$. This concludes the proof of step 2 and the proof that 1) $\Rightarrow$ 2).\\

2) $\Rightarrow$ 3): Recall that $crt$ and $M$ are related by a homeomorphism $\overline{\Psi} \circ F : \overline{\Delta} \rightarrow \overline{L_4}$, which sends $(0:1:1)$ to $(0, \infty, -\infty)$. Thus statement 2) is equivalent to the statement that there exist some $y,z \in X$ such that $\lim\limits_{n \rightarrow \infty} x_n \neq y,z$ and we have $\lim \limits_{n,m \rightarrow \infty} M(x_n, x_m, y, z) = (0, \infty, -\infty)$. From this, one can immediately see that 2) $\Rightarrow$ 3). In particular, if 2) holds for a given pair $y,z$ then 3) holds for the same pair $y,z$.\\

3) $\Rightarrow$ 1): This is a special case of Lemma \ref{lem:AuxiliaryforCauchy}. Since we have seen that 1) $\Rightarrow$ 2) for all good pairs $y,z$ we also see that, if 3) holds for a good pair $y,z$ then 2) holds for the same good pair $y,z$. This concludes the proof of Lemma \ref{lem:CauchySequences}




\end{proof}



Among other things, Lemma \ref{lem:CauchySequences} tells us that for metric spaces, we only need to find one good pair $y, z$ that satisfies property 2 or 3 to get the same property for all good pairs $y, z$ that aren't the limit of $(x_n)_n$. It would be good to have the same property in any generalized M\"obius space that isn't necessarily induced by a metric. Then we could define a sequence in a generalized M\"obius space to be a Cauchy sequence if for one good pair $y,z$ and hence all nice pairs, we have $crt(x_n,x_m,y,z) \rightarrow (0:1:1)$, which would be much easier to check in practice than if we had to check all good pairs. The next lemma tells us, that this is actually true in the case of property 3.

\begin{lem} \label{lem:Conditionc}
Let $(X,M)$ be a generalized M\"obius space, such that $M$ satisfies the (corner)-condition. Let $(x_n)_n$ be a sequence in $X$. Suppose there is a good pair $y, z$ such that

\[c(x_n,x_m,y,z) \xrightarrow{n,m \rightarrow \infty} -\infty.\]

Then the same holds for all good pairs $y', z' \in X$.
\end{lem}

\begin{proof}

By Lemma \ref{lem:AuxiliaryforCauchy}, we know that $(x_n)$ is either bounded or diverges to infinity. Let $y',z'$ be a good pair. As we have seen in the proofs of Lemma \ref{lem:CauchySequences} and \ref{lem:AuxiliaryforCauchy}, we get the right convergence of $c(x_n,x_m,y',z')$ if $\frac{d(x_n,x_m)d(y',z')}{d(x_n,y')d(x_m,z')}$ converges to zero.\\

Case 1: Suppose $(x_n)_n$ is bounded. Since $y',z'$ is a good pair, $(x_n)_n$ doesn't converge to $y'$. Therefore, we find some $\epsilon > 0$ and a subsequence $(x_{n_i})_i$ such that $d(x_{n_i},y') \geq \epsilon$ for all $i$. From Lemma \ref{lem:AuxiliaryforCauchy}, we know that $d(x_n, x_m) \xrightarrow{n,m \rightarrow \infty} 0$ and we find a number $N$ such that for all $n,m \geq N, d(x_n,x_m) < \frac{\epsilon}{2K}$. Thus, we have for all $n \geq N$

\begin{equation*}
\begin{split}
\epsilon \leq d(x_{n_i}, y') \leq K \max( d(x_{n_i}, x_n), d(x_n, y') ).
\end{split}
\end{equation*}

Since $K d(x_{n_i},x_n) \leq \frac{\epsilon}{2} < \epsilon$, we see that

\[ \frac{\epsilon}{K} \leq \frac{1}{K} d(x_{n_i},y') \leq d(x_n,y') \]

for $n \geq N$. This implies that the sequence $(x_n)_n$ stays away from $y'$ for large $n$, specifically, $d(x_n, y') \geq \frac{\epsilon}{K}$ for $n \geq N$. The same is true for $(x_n)_n$ and $z'$ and some other $\tilde{\epsilon} > 0$. Hence, we have

\begin{equation*}
\begin{split}
\frac{d(x_n,x_m)d(y',z')}{d(x_n,y')d(x_m,z')} \leq K^2\frac{d(x_n,x_m)d(y',z')}{\epsilon \tilde{\epsilon}} \xrightarrow{n,m \rightarrow \infty} 0.
\end{split}
\end{equation*}

We see that $\frac{d(x_n,x_m)d(y',z')}{d(x_n,y')d(x_m,z')}$ converges to zero and hence $c(x_n, x_m, y', z') \rightarrow -\infty$.\\

Case 2: Suppose $x_n$ diverges to infinity. We can find a number $N$ such that for all $n \geq N, d(x_n, y')\geq d(y',z')$ and $d(x_n,z') \geq d(y',z')$. Then we have

\begin{equation*}
\begin{split}
\frac{d(x_n,x_m)d(y',z')}{d(x_n,y')d(x_m,z')} & \leq \frac{K \max(d(x_n, y'), d(y',x_m)) d(y',z')}{d(x_n,y')d(x_m,z')}\\
& \leq \frac{K^2 \max( d(x_n,y'), d(y', z'), d(z',x_m) ) d(y',z')}{d(x_n,y')d(x_m,z')}\\
& = K^2 \frac{d(y',z')}{\min(d(x_n,y'),d(x_m,z'))} \rightarrow 0.
\end{split}
\end{equation*}

Hence, we see that also in this case, $\frac{d(x_n,x_m)d(y',z')}{d(x_n,y')d(x_m,z')}$ converges to zero and, therefore, $c(x_n, x_m, y', z') \rightarrow -\infty$. This completes the proof.

\end{proof}

One might hope that an analogous statement for condition true holds. However, the following example illustrates that Lemma \ref{lem:AuxiliaryforCauchy} and \ref{lem:Conditionc} are the best that we can hope for.


\begin{exam} \label{exam:ExampleCauchy}
Consider the circle, represented as $S^1 = \mathbb{R} \diagup 4\mathbb{Z}$. We will mostly use representatives in $[-2, 4]$ to represent points on the circle. Consider the space $X := S^1 \diagdown \{ [0] \}$ and define a map $d : X \times X \rightarrow [0,\infty)$ by

\begin{equation*}
d([x],[y]) := \begin{cases} \vert x - y \vert & (x,y) \in (0,2]^2 \cup [1,3]^2 \cup [2,4)^2 \cup ([-1,1] \diagdown \{ 0 \})^2\\
2 \vert x - y \vert & (x,y) \in \left( (0,1) \times (2,3) \right) \cup \left( (2,3) \times (0,1) \right) \cup \left( (1,2) \times (3,4) \right) \cup \left( (3,4) \times (1,2) \right).
\end{cases}
\end{equation*}

Notice the use of different representatives depending on the case. Geometrically, $(X,d)$ can be thought of as follows. Think of $X$ as a subset of the circle of circumference $4$ with the shortest path metric. This circle can be embedded into $\mathbb{R}^2$ such that it is centered at the origin, i.e. it is the boundary of a disk centered at the origin.

We can now consider the intersection of the circle with each quarter of $\mathbb{R}^2$. We call them the 'upper-right', 'upper-left', 'lower-left', 'lower-right' segment of $S^1$, based on their position in the standard coordinate system of $\mathbb{R}^2$.




The distance $d(x,y)$ between two points $x,y$ is now defined to be the same as on $S^1$ if $x,y$ lie on the same segment of $S^1$ or if they lie on segments that are neighbours of each other. If $x,y$ lie on segments of $S^1$ that lie opposite to each other, then $d(x,y)$ is exactly twice the length of the path from $x$ to $y$ that passes through the point $(0,-1)$.\\

It is not very hard to see that $d$ is a $12$-quasi-metric. Thus, we get a generalized M\"obius space $(X, M_d)$ which satisfies the (corner)-condition. Consider now the following sequence in $X$:

\[ x_n = \left[\frac{1}{n}(-1)^n\right]. \]

One can show that there is a good pair for $(x_n)_n$ that satisfies convergence condition 3), but not convergence condition 2). Furthermore, one can even find another good pair for $(x_n)_n$ that satisfies both convergence condition 2) and 3).



The issue at hand is that even if we understand the convergence behaviour of $\frac{d(x_n,x_m)d(y,z)}{d(x_n,y)d(x_m,z)}$, we cannot control the convergence behaviour of $\frac{d(x_n,y)d(x_m,z)}{d(x_n,z)d(x_m,y)}$ if $d$ is not a metric. So we have found a quasi-metric -- and thus a generalized M\"obius structure $M_d$ that satisfies the (corner)-condition -- for which the statement "3) $\Rightarrow$ 2)" that we have proven for metrics in Lemma \ref{lem:CauchySequences}, does not hold.\\


This example illustrates the relationship between the different possible conditions one could use to define Cauchy sequences in a generalized M\"obius space. If condition 2 holds for one good pair $y, z$, this does not imply that condition 2 holds for all good pairs, unless we work with a metric space. In the same way, if condition 3 holds for all good pairs, this doesn't imply the same for condition 2. However, from Lemma \ref{lem:Conditionc} we know that, if condition 3 holds for one good pair, it holds for all of them. 


\end{exam}


Our study of example \ref{exam:ExampleCauchy} leads us to the following definition of Cauchy sequences in a generalized M\"obius space.


\begin{mydef} \label{def:CauchySequence}
Let $(X,M)$ be a generalized M\"obius space that satisfies the (corner)-condition. A sequence $(x_n)_n$ in $X$ is called a {\it Cauchy sequence} if and only if for one (and hence all) good pairs $y,z$ in $X$, we have

\[ c(x_n,x_m,y,z) \xrightarrow{n,m \rightarrow \infty} -\infty. \]
\end{mydef}


\begin{mydef} \label{def:Completeness}
A generalized M\"obius space $(X,M)$ is called {\it complete} if and only if all Cauchy sequences in $(X,M)$ converge.
\end{mydef}

Using the previous lemmata, the following results are easy to see.

\begin{prop} \label{prop:MobiusInvarianceofCauchySequencesandCompleteness}

Let $(X,M), (X',M')$ be two generalized M\"obius spaces that satisfy the (corner)-condition, $f : X \rightarrow X'$ a M\"obius equivalence between them.

\begin{enumerate}

\item[1)] Let $(x_n)_n$ be a sequence in $X$. Then $(x_n)_n$ is a Cauchy sequence in $(X,M)$ if and only if $(f(x_n))_n$ is a Cauchy sequence in $(X',M')$.

\item[2)] The generalized M\"obius space $(X,M)$ is complete if and only if $(X',M')$ is.

\end{enumerate}
\end{prop}

\begin{proof}
Proof of 1): The sequence $(x_n)_n$ is a Cauchy sequence if and only if for some good pair $y,z$ in $X$, we have

\[ c(x_n,x_m,y,z) \rightarrow -\infty. \]

Since $f$ is a M\"obius equivalence, this implies

\[ c'(f(x_n),f(x_m), f(y),f(z)) = c(x_n,x_m,y,z) \rightarrow -\infty. \]

Since $f$ is a homeomorphism by Lemma \ref{lem:homeomorphism} and $y,z$ is a good pair, so is $f(y),f(z)$ for $(f(x_n))_n$. Thus, $(f(x_n))_n$ is a Cauchy sequence in $(X',M')$.\\

Proof of 2): Suppose $(X,M)$ is complete and let $(x'_n)_n$ be a Cauchy sequence in $(X',M')$. By the first part of the Proposition, $(f^{-1}(x'_n))_n$ is a Cauchy sequence in $(X,M)$ which converges to some $x \in X$ by completeness. Since $f$ is a homeomorphism, $(x'_n)_n$ has to converge to $f(x)$. This implies completeness.

\end{proof}

\begin{prop}
Let $(X,d)$ be a (possibly extended) metric space and denote the induced generalized M\"obius structure by $M$. Equivalent are:

\begin{enumerate}

\item[1)] $(X,M)$ is complete as a generalized M\"obius space

\item[2)] $(X,d)$ is complete as a metric space and is either bounded or has a point at infinity.

\end{enumerate}
\end{prop}

\begin{proof}
1) $\Rightarrow$ 2): Suppose, $(X,M)$ is complete as a generalized M\"obius space and let $(x_n)_n$ be a Cauchy sequence in the metric sense. By Lemma \ref{lem:CauchySequences}, $(x_n)_n$ is also a Cauchy sequence in the M\"obius sense. Hence, $(x_n)$ has to converge in M\"obius topology. Since the M\"obius topology is the same as the metric topology on a metric space (see Theorem \ref{thm:metrictopology}), $(x_n)_n$ converges in metric topology and $(X,d)$ is complete in the metric sense.

2) $\Rightarrow$ 1): Suppose $(X,d)$ is complete as a metric space and let $(x_n)_n$ be a Cauchy sequence in the M\"obius sense. By Lemma \ref{lem:CauchySequences}, $(x_n)_n$ is either a Cauchy sequence in the metric sense, or it diverges to infinity. If it is a Cauchy sequence in the metric sense, it converges in metric topology (and thus in M\"obius topology) by metric completeness. If $x_n$ diverges to infinity, the metric space cannot be bounded. Hence, it has a point at infinity by assumption and $x_n$ converges to the point at infinity in metric and M\"obius topology.

\end{proof}


%-------------------------------------------------------------------------------------------------------------------------------------

% THE COMPLETION OF A M�BIUS SPACE

%-------------------------------------------------------------------------------------------------------------------------------------


\subsection{The completion of a generalized M\"obius space} \label{sec:TheCompletionofaMobiusSpace}

Now that we have a notion of Cauchy sequences and a notion of completeness for generalized M\"obius spaces, an obvious question is whether every generalized M\"obius space has a naturally unique completion, as metric spaces do.

Certainly, if we take a metric space $(X,d)$ and consider the induced generalized M\"obius structure $M$, the metric completion $(\overline{X},\overline{d})$ is either complete with respect to the induced generalized M\"obius structure $\overline{M}$, which is just an extension of $M$, or one has to add one point at infinity to make it complete in the M\"obius sense. Adding a point at infinity doesn't change that $X$ is dense in its completion and it is easy to see that uniqueness up to isometry for the metric case implies uniqueness up to M\"obius equivalence (even up to isometry) in M\"obius sense.\\

We want to see whether we can create a completion even beyond the metric case. For this, we will do the same construction that is used to construct the metric completion. Let $(X,crt)$ be a generalized M\"obius space. (We will talk about the necessary extra condition later.) Define the set

\[ \overline{X} := \{ (x_n)_n \vert (x_n) \text{ a Cauchy sequence in } (X,crt)  \} \diagup \sim \]

where $(x_n) \sim (x'_n)$ if and only if for every pair $y \neq z$ in $X$ that is a good pair for both $(x_n)$ and $(x'_n)$, we have

\[ c(x_n,x'_n,y,z) \rightarrow -\infty. \]

There is a canonical embedding of $X$ into $\overline{X}$ by sending $x$ to the constant sequence $x_n = x$. This is clearly a Cauchy sequence and the map $x \mapsto [(x)_n]$ is injective, since two different constant sequences are not equivalent in the sense defined above.\\

The next step is to extend the generalized M\"obius structure $crt$ to $\overline{X}$. What we would like to do is, to define

\[ \overline{crt}([(w_n)],[(x_n)],[(y_n)],[(z_n)]) := \lim_{n \rightarrow \infty} crt(w_n,x_n,y_n,z_n). \]

There are two questions that arise immediately when stating this definition. Does the limit on the right-hand-side exist and is it independent of the choice of representative of a point $[(w_n)] \in \overline{X}$? In general, the answer to these two questions is no. The reason for that has already appeared in example \ref{exam:ExampleCauchy}, namely that, if $d(x_n,x_m) \rightarrow 0$, we cannot make sure that $d(x_n,y)$ converges for all $y \in X$. From this, we could derive sequences with $crt(x_{2n},x_{2n+1},y,z) \rightarrow (0:9:100)$ and $crt(x_{2n}, x_{2n+2},y,z) \rightarrow (0:1:1)$, i.e. sequences such that $\lim \limits_{n,m \rightarrow \infty}crt(x_n,x_m,y,z)$ does not exist. This example is a special case that will appear in the definition of $\overline{crt}$ given above and makes this construction not well-defined in general.

As mentioned in example \ref{exam:ExampleCauchy}, the problem at hand is that we cannot control the behaviour of $\frac{d(x_n,y)d(x_m,z)}{d(x_m,y)d(x_n,z)}$ for a Cauchy sequence $(x_n)$. If we knew that $crt(w_n,x_n,y_n,z_n)$ could only converge to points in $\mathbb{R}P^2$ that are allowed to be obtained by a generalized M\"obius structure, then we could resolve this problem (as we will see below). Because of this, we now introduce an new condition on our generalized M\"obius structure $crt$.

\begin{mydef} \label{mydef:symmetrycondition}
A generalized M\"obius structure $crt$ or a generalized M\"obius space $(X,crt)$ satisfies the (symmetry)-condition if and only if

\[ \overline{\Ima(crt)} \subseteq \overline{\Delta} = \{ (a:b:c) \vert a,b,c > 0 \} \cup \{ (0:1:1), (1:0:1), (1:1:0) \} \]

where $\overline{\Ima(crt)}$ denotes the closure of the image of $crt$ in $\mathbb{R}P^2$.
\end{mydef}

\begin{figure} 
\begin{tikzpicture}
\draw (0,0);
\draw [dashed] (6,0) -- (9,0);
\draw [dashed] (6,0) -- (7.5, 2.6);
\draw [dashed] (9,0) -- (7.5, 2.6);
\draw [fill] (6.75,1.3) circle [radius=0.05];
\draw [fill] (8.25,1.3) circle [radius=0.05];
\draw [fill] (7.5,0) circle [radius=0.05];
\draw [thick] (6.75,1.3) to [out=240, in=200] (6.8,0.7) to [out=20, in=120] (7.125,0.65) to [out=300, in=30] (7.01,0.343) to [out=220, in=180] (7.5,0) to [out=0, in=320] (7.99,0.343) to [out=140, in=240] (7.875, 0.65) to [out=60, in=160] (8.2,0.7) to [out=340, in=300] (8.25,1.3) to [out=120, in=80] (7.706, 1.555) to [out=260, in=0] (7.5, 1.299) to [out=180, in=280] (7.292, 1.552) to [out=100, in=60] (6.75,1.3);
\node [below left] at (5.7,-0.3) {$(1:0:0)$};
\node [below right] at (9.3,-0.3) {$(0:1:0)$};
\node [above] at (7.5,3) {$(0:0:1)$};
\node at (7.5,1.0) {$\Ima(crt)$};
\end{tikzpicture}
\caption{A generalized M\"obius structure $crt$ satisfies the (symmetry)-condition if and only if no point at the boundary of $\overline{\Delta}$ can be approximated by a sequence of points in $\Ima(\Delta)$ except for $(\frac{1}{2} : \frac{1}{2} : 0), (\frac{1}{2} : 0 : \frac{1}{2})$ and $(0 : \frac{1}{2} : \frac{1}{2})$. In other words, the image doesn't touch the boundary at any other than those three points.}
\end{figure}

The (symmetry)-condition tells us that any sequence of cross-ratio-triples $crt(w_n,x_n,y_n,z_n)$ can only accumulate at points inside the triangle $\Delta$ or the three distinct points on the boundary of the triangle that are assumed by degenerate quadruples. It tells us that, if we have convergence of $crt(w_n,x_n,y_n,z_n)$ it certainly doesn't converge towards anything bad. It turns out, this is enough to prove actual convergence.

Now suppose $(X,crt)$ satisfies the (symmetry)-condition. Let $(x_n)$ be a Cauchy sequence. By symmetry of $x_n,x_m$ we see that $\frac{d(x_n,x_m)d(y,z)}{d(x_n,y)d(x_m,z)}$ and $\frac{d(x_n,x_m)d(y,z)}{d(x_m,y)d(x_n,z)}$ both converge to zero as $n,m$ tend to infinity. Therefore, the sequence $crt(x_n,x_m,y,z)$ can be written in the form $(a_n : b_n : c_n)$ with all three entries being non-negative, where we scale $a_n,b_n,c_n$ such that $a_n+b_n+c_n = 2$. By the convergence statements above, $a_n$ has to converge to zero. Since $crt$ satisfies the (symmetry)-condition, the only point $(0:b:c)$ that can be approximated arbitrarily well in $\Ima(crt)$ is $(0:1:1)$. Therefore, $crt(x_n,x_m,y,z) \rightarrow (0:1:1)$.

We want to consider one particular case of the conclusion above. Suppose, $d$ is a quasi-metric induced by $crt$ and suppose, $d$ has a point at infinity, denoted by $\infty$. Assume, the Cauchy sequence $x_n$ is bounded with respect to $d$, i.e. $x_n$ does not converge to $\infty$. Then we find a good pair $y, \infty$ and by the conclusion above, we see that

\[ crt(x_n,x_m,y,\infty) \xrightarrow{n,m \rightarrow \infty} (0:1:1). \]

This implies however

\begin{equation} \label{eq:ConvergenceCauchy}
\frac{d(x_n,y)}{d(x_m,y)} \xrightarrow{n,m \rightarrow \infty} 1.
\end{equation}

We can now use this to prove that $d(x_n,y)$ converges for every Cauchy sequence $(x_n)$ and any $y \in X$. If $(x_n)$ converges to $y$, then $d(x_n,y) \rightarrow 0$ by definition. If $(x_n)$ diverges to infinity with respect to $d$, then $d(x_n,y) \rightarrow \infty$. If $(x_n)$ is bounded with respect to $d$ then $0 \leq d(x_n,y) \leq B$ and hence - by compactness - has a convergent subsequence $d(x_{n_i},y)$. Applying equation (\ref{eq:ConvergenceCauchy}) in the case $m = n_i$ gives us

\[ \frac{d(x_n,y)}{d(x_{n_i},y)} \xrightarrow{n,i \rightarrow \infty} 1. \] 

Since $d(x_{n_i},y)$ converges, this implies that the limit of $d(x_n,y)$ exists and

\[ \lim_{n \rightarrow \infty} d(x_n,y) = \lim_{i \rightarrow \infty} d(x_{n_i},y). \]


This proves the following

\begin{prop} \label{prop:Convergence}
Let $(X,crt)$ be a generalized M\"obius structure satisfying the (symmetry)-condition. Let $(x_n)$ be a Cauchy sequence in $X$, $y \in X$ and $d$ a quasi-metric that induces $M$ that has a point at infinity (e.g. $d = d_A$). Then $d(x_n,y)$ converges, possibly to infinity.
\end{prop}

Recall that every sequence $(x_{n,m})$ in $\mathbb{R}$ parametrized by $\mathbb{N}^2$ with the property that $\lim_{n \rightarrow \infty} x_{n,m}$ exists for every $m$, $\lim_{m \rightarrow \infty} x_{n,m}$ exists for every $n$ and $\lim_{n,m \rightarrow \infty} x_{n,m}$ exists, satisfies

\[ \lim_{n \rightarrow \infty} \lim_{m \rightarrow \infty} x_{n,m} = \lim_{m \rightarrow \infty} \lim_{n \rightarrow \infty} x_{n,m} = \lim_{n,m \rightarrow \infty} x_{n,m}. \]

Repeating the argument from above, one can show that for any two Cauchy-sequences $(x_n)_n, (y_m)_m$, $d(x_n, y_{m_i})$ converges for a certain subsequence of $y_m$ and with the same argument again, one can show that $d(x_n, y_m)$ converges. Hence, $crt(w_n,x_n,y_n,z_n)$ converges as well and, by the (symmetry)-condition, it converges to a point in $\overline{\Ima(crt)} \subseteq \overline{\Delta}$.

We are left to show that $\lim \limits_{n \rightarrow \infty} crt(w_n,x_n,y_n,z_n) = \lim \limits_{n \rightarrow \infty} crt(w'_n,x'_n,y'_n,z'_n)$ for $(w_n) \sim (w'_n), (x_n) \sim (x'_n), (y_n) \sim (y'_n), (z_n) \sim (z'_n)$. Again, we will prove the statement for $d(x_n,y)$ and a quasi-metric $d$ that induces $crt$ and has a point at infinity. Repeating this argument then implies as above that the statement for $crt(w_n,x_n,y_n,z_n)$ holds.

So let $d$ be a quasi-metric that induces $crt$ and has a point at infinity, denoted by $\infty$. Let $(x_n) \sim (x'_n)$. Since $c(x_n,x'_n,y,z) \rightarrow -\infty$ for all good pairs, it is easy to see that either $d(x_n,x'_n) \rightarrow 0$ or $x_n, x'_n$ both diverge to infinity.

Let $y \in X$. If $(x_n)$ converges to $\infty$, then $x'_n$ has to diverge to infinity as well as we noted above and hence $d(x_n,y) = d(x'_n,y)$ for all $y \in X$.

Now suppose, $(x_n)$ doesn't diverge to $\infty$, hence it has to be bounded by Lemma \ref{lem:AuxiliaryforCauchy} and $d(x_n, x'_n) \xrightarrow{n \rightarrow \infty} 0$. By Proposition \ref{prop:Convergence}, $d(x_n,y)$ and $d(x'_n,y)$ both converge. Suppose $d(x_n,y) \xrightarrow{n \rightarrow \infty} 0$. Then $d(x'_n,y) \leq K \max(d(x'_n,x_n), d(x_n,y)) \leq K \epsilon$ for $n$ sufficiently large. Thus, $\lim \limits_{n \rightarrow \infty} d(x'_n,y) = 0 = \lim \limits_{n \rightarrow \infty} d(x_n,y)$.

Finally, suppose, $d(x_n, y) \rightarrow r$ for some positive real number $0 < r < \infty$. Then, by swapping $x_n$ and $x'_n$ in the argument above, $d(x'_n,y)$ doesn't converge to zero. Therefore and because $(x_n),(x'_n)$ are both bounded, $y, \infty$ is a good pair for both sequences. Since the two sequences are equivalent by assumption, we have

\[ c(x_n,x'_n,y, \infty) \rightarrow -\infty. \]

Using the (symmetry)-condition and the same argument as in the proof of Proposition \ref{prop:Convergence}, we get

\[ crt(x_n,x'_n,y,\infty) \rightarrow (0:1:1). \]

In other words,

\[ crt(x_n,x'_n,y,\infty) = \frac{d(x_n,y)}{d(x'_n,y)} \rightarrow 1 \]

and therefore

\[ \lim_{n \rightarrow \infty} d(x_n,y) = \lim_{n \rightarrow \infty} d(x'_n,y). \]

By the same argument as above, we see that $\lim_{n \rightarrow \infty} d(x_n,y_n) = \lim_{n�\rightarrow \infty} d(x'_n,y_n)$ for all Cauchy sequences $(x_n) \sim (x'_n), (y_n)$. Thus, $\lim_{n \rightarrow \infty} crt(w_n,x_n,y_n,z_n) = \lim_{n \rightarrow \infty} crt(w'_n,x'_n,y'_n,z'_n)$ and therefore, $\overline{crt}$ is well-defined.\\

Given a generalized M\"obius space $(X, crt)$ that satisfies the (symmetry)-condition, we have constructed a new generalized M\"obius space $(\overline{X}, \overline{crt})$. We also have a canonical map of $X$ into $\overline{X}$ that preserves the generalized M\"obius structure (hence it is also a topological embedding).\\

We are left to show that $\overline{X}$ is complete and that $\overline{X}$ is unique.

\begin{mythm} \label{thm:Completion}
Let $(X,crt)$ be a generalized M\"obius space that satisfies the (symmetry)-condition. Then, the generalized M\"obius space $(\overline{X}, \overline{crt})$ is complete and the image of the embedding $X \hookrightarrow \overline{X}$ is dense in $\overline{X}$.\\

Furthermore, if $(Y, crt')$ is a complete generalized M\"obius space, such that there exists a M\"obius embedding $i : X \hookrightarrow Y$ such that $i(X)$ is dense in $Y$, then there exists a unique M\"obius equivalence $f : \overline{X} \rightarrow Y$ such that $i = f \circ i_X$ where $i_X$ denotes the canonical inclusion of $X$ into $\overline{X}$.
\end{mythm}


\begin{proof}%[Proof of Theorem \ref{thm:Completion}]
We start by proving completeness. Let $\xi^m = [(x^{(m)}_n)_n] \in \overline{X}$ be such that $(\xi^m)_m$ is a Cauchy sequence in $\overline{X}$. We will often identify $\xi^m$ with the representative $x^{(m)}_n$. Choose a quasi-metric $d$ on $X$ that induces $crt$ and let $\overline{d}$ be the extension to $\overline{X}$. Clearly, $\overline{d}$ induces $\overline{crt}$. By Lemma \ref{lem:AuxiliaryforCauchy}, $(\xi^m)_m$ either diverges to infinity, or it is bounded with respect to $\overline{d}$.

We analyze the point at infinity in $\overline{X}$ with respect to $\overline{d}$. Let it be represented by a Cauchy sequence $(z_n)$ in $X$. Then $\overline{d}((z_n),(y_n)) = \infty$ for all Cauchy sequences $(y_n)$ in $X$ that are not equivalent to $(z_n)$. This means that

\[ \infty = \overline{d}((z_n),(y_n)) = \lim_{n \rightarrow \infty} d(z_n,y_n),�\]

which is the same as saying that $(z_n)$ diverges to infinity. So the point at infinity with respect to $\overline{d}$ is the equivalence class of all sequences in $X$ that diverge to infinity with respect to $d$.\\

Before we study the convergence of our sequence $(\xi^m)_m$, we need to take a look at the topology of $\overline{X}$. By Definition \ref{def:Completeness}, we need to show completeness with respect to the M\"obius topology of $(\overline{X}, \overline{crt})$. Convergence to a point $p \in \overline{X}$ with respect to the M\"obius topology is the same as convergence with respect to all quasi-metrics $d_A$ induced by $\overline{X}$ that do not have $p$ at infinity. Recall that, if a generalized M\"obius structure $crt$ is induced by a quasi-metric $d$, then the induced semi-metrics $d_A$ are quasi-metrics by Proposition \ref{prop:cornerconditionatinfinity} and have the form

\[ d_A(x,y) = \frac{d(x,y)}{d(x, \omega)d(\omega,y)} \frac{d(\alpha, \omega)d(\omega, \beta)}{d(\alpha, \beta)}. \]

It is easy to see that $d_A(x,y)$ goes to zero if and only if $d(x,y)$ goes to zero, or $d(x, \omega)$ goes to infinity, or $d(\omega,y)$ goes to infinity. Therefore, if $d(x_n,x)$ goes to zero, this implies that $d_A(x_n,x)$ goes to zero for all $d_A$ that don't have $x$ at infinity. This implies that $x_n$ converges to $x$ in M\"obius topology. Thus it is enough to check convergence with respect to $\overline{d}$ to prove convergence in M\"obius topology, as long as $\xi^m$ doesn't diverge to infinity. Also note that divergence to infinity immediately implies convergence to the point at infinity in M\"obius topology (if there is a point at infinity).\\

Back to the sequence $(\xi^m)_m$. We first treat the case when the sequence diverges to infinity. If $(\xi^m)_m$ diverges to infinity, we need to show that there is a point at infinity with respect to $\overline{d}$. If $d$ is bounded, then so is $\overline{d}$ and therefore, the sequence $(\xi^m)_m$ can only diverge to infinity, if there is a point at infinity in $X$ with respect to $d$ and hence a point at infinity in $\overline{X}$ with respect to $\overline{d}$. Thus we see that, if $(\xi^m)_m$ diverges to infinity, there is a point in $\overline{X}$ that lies at infinity with respect to $\overline{d}$ and the sequence converges to that point.\\

Now suppose, $(\xi^m)_m$ is bounded. We need to find a Cauchy sequence $(x_l)_l$ in $X$ such that $(\xi^m)_m$ converges to that sequence in M\"obius topology as $m$ tends to infinity. Let $(y_n)$ be a Cauchy sequence in $X$ that doesn't diverge to infinity with respect to $d$. Since $(\xi^m)_m$ is bounded, we find some constant $B > 0$ such that $\overline{d}(\xi^m, (y_n)) < B$ for all $m \in \mathbb{N}$. Therefore, for some other constant $B > 0$ and every $m$ we find some natural number $N_m$ such that for all $n \geq N_m$, we have

\[d(x^{(m)}_n, y_n) \leq B.\]

This implies that $\xi^m = [ (x^{(m)}_n)_n ]$ is represented by a bounded Cauchy sequence for every $m$. By Lemma \ref{lem:AuxiliaryforCauchy},

\[ d(x^{(m)}_n, x^{(m)}_{n'}) \xrightarrow{n,n' \rightarrow \infty} 0. \]

Thus, for every fixed $m$ and every $\epsilon > 0$, we find a natural number $N_m$, such that for all $n, n' \geq N_m$, we have

\begin{equation} \label{eq:estimatealongn}
d(x^{(m)}_n, x^{(m)}_{n'}) < \epsilon.
\end{equation}

Since $(\xi^m)_m$ is a bounded Cauchy sequence by assumption, we also find for every $\epsilon > 0$ a natural number $M$ such that for all $m, m' \geq M$,

\begin{equation} \label{eq:estimatealongm}
\overline{d}(\xi^m, \xi^{m'}) < \epsilon.
\end{equation}

We now use the following technical lemma.

\begin{lem}\label{lem:technicallemma}

There exists a sequence $(x_l)_l$ in $X$, satisfying the following properties:

\begin{enumerate}

\item $x_l = x^{(m_l)}_{n_l}$

\item The sequences $m_l, n_l$ are increasing.

\item For every $l \in \mathbb{N}$ and all $n \geq n_l$, $d(x^{(m_l)}_{n_l}, x^{(m_l)}_n) < \frac{1}{lK}$.

\item For every $l \in \mathbb{N}$ and all $m,m' \geq m_l, \overline{d}( \xi^m, \xi^{m'} ) \leq \frac{1}{2lK}$.

\item For all $l \leq l' \in \mathbb{N}$ and all $n \geq n_{l'}$, $d(x^{(m_l)}_{n}, x^{(m_{l'})}_{n}) < \frac{1}{lK}$.

\end{enumerate}

\end{lem}

We first show how the Lemma completes the proof of Theorem \ref{thm:Completion}. Given such a sequence $(x_l)_l$, one immediately sees that for all $l$ and all $l' \geq l$, we have

\[ d(x_l, x_{l'}) = d(x^{(m_l)}_{n_l}, x^{(m_{l'})}_{n_{l'}}) \leq K \max ( d( x^{(m_l)}_{n_l}, x^{(m_l)}_{n_{l'}} ), d( x^{(m_l)}_{n_{l'}}, x^{(m_{l'})}_{n_{l'}} ) ) \leq K \frac{1}{lK} = \frac{1}{l}. \]

This implies that $x_l$ is bounded and a Cauchy sequence. Furthermore, for $m \geq m_l$

\[ \overline{d}(\xi^m, (x_l)_l) = \lim_{l \rightarrow \infty} d( x^{(m)}_l, x_l) = \lim_{l \rightarrow \infty} d( x^{(m)}_l, x^{(m_l)}_{n_l}) \leq \lim_{l \rightarrow \infty} K \frac{1}{ \min(l, n_l) K} =  0. \]

Therefore, for any other point $(y_l)_l \in \overline{X}$, we have $\overline{d}(\xi^m, (y_l)_l) > 0$ for $m$ sufficiently large. This implies that for all $(y_l)_l, (z_l)_l \in \overline{X} \diagdown \{ (x_l)_l \}$, we have $\overline{crt}(\xi^m, (x_l)_l, (y_l)_l, (z_l)_l) \xrightarrow{m \rightarrow \infty} (0:1:1)$. Since we assume that $(\xi^m)_m$ does not diverge to infinity, $(x_l)_l \neq \infty$ and we can choose $(z_l)_l = \infty$ (add a point at infinity if necessary). Then, denoting $y := (y_l)_l, \infty = (\infty)_l$, this limit gets the form

\[ \overline{crt}(\xi^m, (x_l)_l, y, \infty) \xrightarrow{m \rightarrow \infty} = (0:1:1). \]

By the definition of $\overline{crt}$ this implies

\[ \frac{\overline{d}(\xi^m, y)}{\overline{d}((x_l)_l,y)} \xrightarrow{m \rightarrow \infty} 1. \]

In other words,

\[ \lim_{m \rightarrow \infty} \overline{d}(\xi^m, y) = \overline{d}((x_l)_l, y). \]

This implies that $\xi^m$ converges to $(x_l)$ with respect to $\overline{d}$ and hence also in $(\overline{X}, \overline{crt})$ as we have discussed at the beginning of the proof.\\

We are left to show that the completion $(\overline{X}, \overline{crt})$ is unique up to unique M\"obius equivalence. Let $(Y, crt')$ be a complete generalized M\"obius space and $i : X \hookrightarrow Y$ a M\"obius embedding, i.e. an injective map that is a M\"obius equivalence onto its image. Further, assume $i(X)$ is dense in $Y$ with its M\"obius topology. Denote the canonical inclusion of $X$ into $\overline{X}$ by $i_X$. Since $i, i_X$ are both injective, we get a bijection $f : i(X) \rightarrow i_X(X)$ which sends $i(x)$ to $i_X(x)$. Since $i, i_X$ are M\"obius equivalences onto their images, they are also homeomorphisms onto their images. Therefore, the map $f$ is a homeomorphism with respect to the subspace topology on $i(X)$ and $i_X(X)$. Since $i(X), i_X(X)$ are both dense in $Y, \overline{X}$ respectively, the homeomorphism $f$ extends to a unique continuous map $F : Y \rightarrow \overline{X}$. It is an easy exercise to show that $F$ is bijective. (Hint: Recall that the $d_A$ are continuous with respect to M\"obius topology.)

We claim that $F$ is a M\"obius equivalence. Let $(w,x,y,z)$ be a non-degenerate quadruple in $Y$ (clearly, $F$ preserves the generalized M\"obius structure on degenerate, admissible quadruples). Then we can approximate these four points by sequences $w_n, x_n, y_n,z_n$ in $i(X)$. By definition of $F$,

\[ F(w) = \lim_{n \rightarrow \infty} F(w_n) \]
\[ F(x) = \lim_{n \rightarrow \infty} F(x_n) \]
\[ F(y) = \lim_{n \rightarrow \infty} F(y_n) \]
\[ F(z) = \lim_{n \rightarrow \infty} F(z_n) \]

and hence

\begin{equation*}
\begin{split}
\overline{crt}(F(w)F(x)F(y)F(z)) & = \lim_{n \rightarrow \infty} crt(F(w_n)F(x_n)F(y_n)F(z_n))\\
& = \lim_{n \rightarrow \infty} crt(f(w_n)f(x_n)f(y_n)f(z_n))\\
& = \lim_{n \rightarrow \infty} crt'(w_n,x_n,y_n,z_n)\\
& = crt'(w,x,y,z). 
\end{split}
\end{equation*}

This shows that $F$ preserves the generalized M\"obius structure on non-degenerate quadruples. Since $F$ is bijective, it trivially preserves the generalized M\"obius structure on degenerate, admissible quadruples. Hence, $F$ is a M\"obius equivalence. Since all M\"obius equivalences are homeomorphisms, uniqueness follows from the fact that $F \vert_{i(X)} = f$ is given and the fact that $i(X)$ is dense in $Y$. This completes the proof of Theorem \ref{thm:Completion} up to the proof of Lemma \ref{lem:technicallemma}.

\end{proof}


\begin{proof}[Proof of Lemma \ref{lem:technicallemma}]

We are left to construct the sequence $x_l$. We construct $x_l$ inductively. The induction start goes as follows: By equation (\ref{eq:estimatealongm}), we find natural numbers $M_1 < M_2$, such that

\[ \forall m,m' \geq M_1,\, \overline{d}(\xi^m, \xi^{m'}) < \frac{1}{2K} \]

\[ \forall m,m' \geq M_2,\, \overline{d}(\xi^m, \xi^{m'}) < \frac{1}{4K}. \]

Now we fix $m = M_1, m' = M_2$. Using equation (\ref{eq:estimatealongn}), we find a natural number $N_1$ such that

\[ \forall n \geq N_1,\, d\left(x^{(M_1)}_n, x^{(M_2)}_n\right) < \frac{1}{K}. \]

Since $(x^{(M_1)}_n)_n$ is a Cauchy sequence in $X$, we can choose $N_1$ such that additionally,

\[ \forall n,n' \geq N_1,\, d\left(x^{(M_1)}_n, x^{(M_1)}_{n'}\right) < \frac{1}{K}. \]

Set

\[ x_1 := x^{(M_1)}_{N_1}. \]

We see that $x_1$ satisfies conditions 3 and 4 from above. Now we do the inductive construction.

Suppose, we are given points $x_1, \dots, x_l$ in $X$ satisfying properties 1-5. Since $(\xi^m)_m$ is a Cauchy sequence in $\overline{X}$, we find some $M_{l+1} > m_l$, such that

\[ \forall m,m' \geq M_{l+1}, \overline{d}( \xi^m, \xi^{m'}) < \frac{1}{2(l+1)K}. \]

Put $m_{l+1} := M_{l+1}$. Since we have chosen $M_{l+1} > m_l$, condition 2 stays satisfied for $(m_l)_l$. Further, $m_{l+1}$ clearly satisfies condition 4. Since $\xi^{m_{l+1}}$ is a Cauchy sequence, we find some natural number $N_0$, such that

\[ \forall n,n' \geq N_0, d(x^{(m_{l+1})}_n, x^{(m_{l+1})}_{n'}) < \frac{1}{(l+1)K}. \]

Thus condition 3 is satisfied, if we choose $n_{l+1} \geq N_0$. By condition 4, we know that for all $i < l+1$, we have

\[ \overline{d}( \xi^{m_i}, \xi^{m_{l+1}}) < \frac{1}{2iK}. \]

Therefore, we find some natural numbers $N_i$, such that

\[ \forall n \geq N_i, d( x^{(m_i)}_n, x^{(m_{l+1})}_n ) < \frac{1}{iK}. \]

We put $N := \max(N_0, N_1, \dots, N_l, n_l)$ and get

\[ \forall n \geq N, \forall i < l+1, d( x^{(m_i)}_n, x^{(m_{l+1})}_n) < \frac{1}{iK}. \]

Put $n_{l+1} := N$ and put

\[ x_{l+1} := x^{(m_{l+1})}_{n_{l+1}}. \]

By the definition of $N$, the sequence $(n_l)_l$ satisfies condition 2. Condition 3 is satisfied as $n_{l+1} \geq N_0$. Condition 4 is satisfied by choice of $m_{l+1}$ Finally, condition 5 is satisfied because $n_{l+1} \geq \max(N_1, \dots, N_l)$. Condition 1 is trivially satisfied and hence we have constructed a sequence with properties 1-5. We have seen before that such a sequence is a Cauchy sequence in $(X, crt)$ and $(\xi^m)_m$ converges to $(x_l)_l$ in $(\overline{X}, \overline{crt})$. Hence the Cauchy sequence $(\xi^m)_m$ converges. This implies that $(\overline{X}, \overline{crt})$ is complete.

\end{proof}


%-------------------------------------------------------------------------------------------------------------------------------------

% CONCLUDING REMARKS

%-------------------------------------------------------------------------------------------------------------------------------------

\section{Concluding remarks}

In this article, we presented the generalized notion of M\"obius spaces introduced by Buyalo. We presented the topology of a generalized M\"obius space and gave a more general proof that the M\"obius topology coincides with the metric topology, if the generalized M\"obius structure is induced by a metric. We then specialized to generalized M\"obius structures that are induced by a quasi-metric and generalized the notions of Cauchy sequences and completeness to such generalized M\"obius spaces. Finally, we showed that any generalized M\"obius space that satisfies a certain (symmetry)-condition admits a completion which is unique up to unique M\"obius equivalence.

The compatibility of all these generalized notions with the special case of a metric space and the fact that several core results carry over to the generalized notions suggests that the notion of a generalized M\"obius space indeed provides an interesting theory which may be useful in the study of spaces where generalized M\"obius structures appear naturally.

%-------------------------------------------------------------------------------------------------------------------------------------

% APPENDIX

%-------------------------------------------------------------------------------------------------------------------------------------

%\appendix


%\section{Technical Proofs}




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\end{document}

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[G]{Gromov M. Gromov, Hyperbolic groups, MSRI Publ. 8, 75-263, Springer, 1987.
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[S]{Schroederdyadic Viktor Schroeder, Quasi-metric and metric spaces, Conformal Geometry and Dynamics of the American Mathematical Society, Volume 10, 2016.