Let $X$ be a [directed-complete partial order](https://ncatlab.org/nlab/show/dcpo), or even a [complete lattice](https://ncatlab.org/nlab/show/complete+lattice). A subset $S\subseteq X$ is called [Scott-closed](https://ncatlab.org/nlab/show/Scott+topology) 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](https://ncatlab.org/nlab/show/ideal#in_lattices_and_other_prosets) 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?



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


