The definition of "algebraic poset" in Continuous Lattices and Domains, 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 $k_1 \sqcup k_2 \in A(x)$.

  • $\begingroup$ You say: “if k1 and k2 are in A(x) then k1⊔k2 is also in A(x)” — how do you prove this? $\endgroup$ Feb 24, 2014 at 9:12
  • $\begingroup$ @ArtemPelenitsyn: I have added my argument to the question. $\endgroup$
    – Uday Reddy
    Feb 24, 2014 at 9:27
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    $\begingroup$ 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. $\endgroup$ Feb 24, 2014 at 9:40
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    $\begingroup$ I also found the fact that second condition is sufficient in bounded complete cpo here: homepages.inf.ed.ac.uk/libkin/papers/alcpo.pdf (p. 1) $\endgroup$ Feb 24, 2014 at 9:56
  • $\begingroup$ @ArtemPelenitsyn. Great, thanks very much. Be wary of the hidden assumption! $\endgroup$
    – Uday Reddy
    Feb 24, 2014 at 18:11

1 Answer 1


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.

  • 1
    $\begingroup$ Cool. Great example! $\endgroup$
    – Uday Reddy
    Feb 24, 2014 at 18:15

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