<p>An example where $A(x)$ is empty is the set of real numbers $\mathbb{R}$ with the usual ordering. It has no compact elements at all.</p>

<p>If we assume the second condition then $A(x)$ cannot be empty: if $A(x) = \emptyset$ then by the second condition $x$ is the empty join, therefore the least element of $L$, which is compact, therefore $x \in A(x) = \emptyset$, a contradiction.</p>

<p>Your proposal to replace the first condition with non-emptyness does not work. Consider the poset $L$ which consists of two copies of $\mathbb{N}$ and $\infty$, where we write $\iota_1(n)$ and $\iota_2(n)$ for the two copies of $n$, ordered by:</p>

<ul>
<li>$\iota_1(m) \leq \iota_1(n) \iff m \leq n$</li>
<li>$\iota_2(m) \leq \iota_2(n) \iff m \leq n$</li>
<li>$x \leq \infty$ for all $x$.</li>
</ul>

<p>In words, we have two incomparable chains with a common supremum. All elements are compact except $\infty$. Now:</p>

<ol>
<li><p>${\downarrow}x \cap K(L) \neq \emptyset$, obviously.</p></li>
<li><p>$x = \bigvee ({\downarrow}x \cap K(L))$, obviously.</p></li>
<li><p>The set ${\downarrow}\infty \cap K(L) = \mathbb{N} + \mathbb{N}$ is not directed.</p></li>
</ol>

CC BY-SA 3.0
2014-02-24T18:15:26.140Z
Cool.  Great example!
Uday Reddy
CC BY-SA 3.0
2014-02-24T14:51:55.427Z
2014-02-24T14:51:55.427Z
Andrej Bauer
<p>The definition of "algebraic poset" in <a href="http://books.google.com/books?id=xPF9Hb7DPLgC&amp;printsec=frontcover&amp;dq=continuous%20lattices%20and%20domains&amp;hl=en&amp;sa=X&amp;ei=EpEKU8X8IY7e7AbTu4C4CA&amp;ved=0CEMQ6AEwAA">Continuous Lattices and Domains</a>, Definition I-4.2, says that, for all $x \in L$, </p>

<ul>
<li>the set $A(x) = {\downarrow} x \cap K(L)$ should be a directed set, and</li>
<li>$x = \bigsqcup ({\downarrow} x \cap K(L)$.</li>
</ul>

<p>Here $L$ is a poset, $K(L)$ is the set of compact elements of $L$, and ${\downarrow} x$ means $\{y \mid y \sqsubseteq x\}$.</p>

<p>I was a bit surprised by the first condition.  It is an easy argument to show that, if $k_1$ and $k_2$ are in $A(x)$ then $k_1 \sqcup k_2$ is also in $A(x)$.  So, all nonempty finite subsets of $A(x)$ have upper bounds in it.  The only question is whether the empty subset has an upper bound in it, i.e., whether $A(x)$ is nonempty in the first place.  So,</p>

<ul>
<li>Is it ok to replace the first condition with $A(x)$ is nonempty?</li>
<li>What is an example of a situation where $A(x)$ is empty?</li>
</ul>

<hr>

<p>Note added: How is $k_1 \sqcup k_2$ in A(x)?  First, since $k_1 \sqsubseteq x$ and $k_2 \sqsubseteq x$, we have $k_1 \sqcup k_2 \sqsubseteq x$.  Second, $k_1$ and $k_2$ are compact.  So, any directed set that goes "beyond" them must "pass" them.  Suppose a directed set $u$ also goes beyond $k_1 \sqcup k_2$, i.e., $k_1 \sqcup k_2 \sqsubseteq \bigsqcup u$.  Since it has gone beyond $k_1$ and $k_2$, it must have passed them, i.e., there are elements $y_1, y_2 \in u$ such that $k_1 \sqsubseteq y_1$ and $k_2 \sqsubseteq y_2$.  Since $u$ is a directed set, it must have an upper bound for $y_1$ and $y_2$, say $y$.  Now, $k_1 \sqcup k_2 \sqsubseteq y \in d$.  This shows that $k_1 \sqcup k_2$ is compact.  The two pieces together say $k_1 \sqcup k_2 \in A(x)$.</p>

CC BY-SA 3.0
2014-02-24T09:12:39.390Z
You say: “if k1 and k2 are in A(x) then k1⊔k2 is also in A(x)” — how do you prove this?
Artem Pelenitsyn
CC BY-SA 3.0
2014-02-24T09:27:12.783Z
@ArtemPelenitsyn: I have added my argument to the question.
Uday Reddy
CC BY-SA 3.0
2014-02-24T09:40:59.940Z
Please correct me if I got this wrong, but: in your note you assume that k1⊔k2 exists in L. But L is only a poset, not a directed set, so you can't do that.
Artem Pelenitsyn
CC BY-SA 3.0
2014-02-24T09:56:45.183Z
I also found the fact that second condition is sufficient in bounded complete cpo here: http://homepages.inf.ed.ac.uk/libkin/papers/alcpo.pdf (p. 1)
Artem Pelenitsyn
CC BY-SA 3.0
2014-02-24T18:11:38.497Z
@ArtemPelenitsyn.  Great, thanks very much.  Be wary of the hidden assumption!
Uday Reddy
<domain-theory>
Is this an equivalent condition for algebraic posets?
Uday Reddy
# Is this an equivalent condition for algebraic posets?

(Asked by: username_0 on 2014-02-24)

The definition of "algebraic poset" in [Continuous Lattices and Domains](http://books.google.com/books?id=xPF9Hb7DPLgC&printsec=frontcover&dq=continuous%20lattices%20and%20domains&hl=en&sa=X&ei=EpEKU8X8IY7e7AbTu4C4CA&ved=0CEMQ6AEwAA), Definition I-4.2, says that, for all $x \in L$,

* the set $A(x) = {\downarrow} x \cap K(L)$ should be a directed set, and
* $x = \bigsqcup ({\downarrow} x \cap K(L)$.

Here $L$ is a poset, $K(L)$ is the set of compact elements of $L$, and ${\downarrow} x$ means $\{y \mid y \sqsubseteq x\}$.

I was a bit surprised by the first condition. It is an easy argument to show that, if $k\_1$ and $k\_2$ are in $A(x)$ then $k\_1 \sqcup k\_2$ is also in $A(x)$. So, all nonempty finite subsets of $A(x)$ have upper bounds in it. The only question is whether the empty subset has an upper bound in it, i.e., whether $A(x)$ is nonempty in the first place. So,

* Is it ok to replace the first condition with $A(x)$ is nonempty?
* What is an example of a situation where $A(x)$ is empty?

---

Note added: How is $k\_1 \sqcup k\_2$ in A(x)? First, since $k\_1 \sqsubseteq x$ and $k\_2 \sqsubseteq x$, we have $k\_1 \sqcup k\_2 \sqsubseteq x$. Second, $k\_1$ and $k\_2$ are compact. So, any directed set that goes "beyond" them must "pass" them. Suppose a directed set $u$ also goes beyond $k\_1 \sqcup k\_2$, i.e., $k\_1 \sqcup k\_2 \sqsubseteq \bigsqcup u$. Since it has gone beyond $k\_1$ and $k\_2$, it must have passed them, i.e., there are elements $y\_1, y\_2 \in u$ such that $k\_1 \sqsubseteq y\_1$ and $k\_2 \sqsubseteq y\_2$. Since $u$ is a directed set, it must have an upper bound for $y\_1$ and $y\_2$, say $y$. Now, $k\_1 \sqcup k\_2 \sqsubseteq y \in d$. This shows that $k\_1 \sqcup k\_2$ is compact. The two pieces together say $k\_1 \sqcup k\_2 \in A(x)$.

**Tags:** domain-theory
> **Comments:**
> **username_1**: You say: “if k1 and k2 are in A(x) then k1⊔k2 is also in A(x)” — how do you prove this?  
> **username_0**: @ArtemPelenitsyn: I have added my argument to the question.  
> **username_1**: Please correct me if I got this wrong, but: in your note you assume that k1⊔k2 exists in L. But L is only a poset, not a directed set, so you can't do that.  
> **username_1**: I also found the fact that second condition is sufficient in bounded complete cpo here: http://homepages.inf.ed.ac.uk/libkin/papers/alcpo.pdf (p. 1)  
> **username_0**: @ArtemPelenitsyn.  Great, thanks very much.  Be wary of the hidden assumption!  

---

## 1 answers

### Answer by username_2 on 2014-02-24. Score: 12 | Accepted answer

An example where $A(x)$ is empty is the set of real numbers $\mathbb{R}$ with the usual ordering. It has no compact elements at all.

If we assume the second condition then $A(x)$ cannot be empty: if $A(x) = \emptyset$ then by the second condition $x$ is the empty join, therefore the least element of $L$, which is compact, therefore $x \in A(x) = \emptyset$, a contradiction.

Your proposal to replace the first condition with non-emptyness does not work. Consider the poset $L$ which consists of two copies of $\mathbb{N}$ and $\infty$, where we write $\iota\_1(n)$ and $\iota\_2(n)$ for the two copies of $n$, ordered by:

* $\iota\_1(m) \leq \iota\_1(n) \iff m \leq n$
* $\iota\_2(m) \leq \iota\_2(n) \iff m \leq n$
* $x \leq \infty$ for all $x$.

In words, we have two incomparable chains with a common supremum. All elements are compact except $\infty$. Now:

1. ${\downarrow}x \cap K(L) \neq \emptyset$, obviously.
2. $x = \bigvee ({\downarrow}x \cap K(L))$, obviously.
3. The set ${\downarrow}\infty \cap K(L) = \mathbb{N} + \mathbb{N}$ is not directed.

> **Comments:**
> **username_0**: Cool.  Great example!  

---

