# Is the Scott topology generated by the ideals as the closed sets?

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?

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.