2
$\begingroup$

Let $X$ be a directed-complete partial order, or even a complete lattice. A subset $S\subseteq X$ is called Scott-closed if and only if it is:

  • Downward-closed: $y\in S$ and $x\le y$ implies $x\in S$;
  • Closed under directed suprema: if $D\subseteq S$ is directed, then $\sup D\in S$.

Scott-closed sets are closed under taking finite unions and arbitary intersections, and the topology they define (as closed sets) is called the Scott topology.

Recall that an ideal on $X$ is a nonempty subset $I\subseteq X$ which is:

  • Downward closed;
  • Closed under finite suprema, i.e. if $x,y\in F$, then $x\vee y\in F$.

Now here is my question: do the Scott-closed ideals generate the whole Scott topology on $X$? If not, what would be a counterexample?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
5
$\begingroup$

The answer is no. Consider poset consisting of infinitely many incomparable elements $a_1,a_2,\dots$ and a single element $b$ larger than them all. Then $A=\{a_1,a_2,\dots\}$ is closed in the Scott topology (note it has no directed subsets with more than one element).

On the other hand, consider the topology generated by Scott-closed ideals. If an ideal contains more than one element, then it contains $b$. Therefore, since we need to allow finite unions, the closed sets in this topology are all subsets containing $b$ and finite subsets of $A$. In particular, $A$ itself is not closed in this topology.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .