- 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?