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?