Intersection of infinitely many non-empty open sets in an irreducible topological space As we already know from general topological spaces, the infinite intersection of open sets does not need to be open anymore.
But is this still the case if our topological space $X$ is irreducible? In this case every open set $U \subset X$ is dense, meaning $\bar{U} = X$. Or in other words: Open sets in irreducible spaces are huge. I am also wondering if the infinite intersection of open sets can become empty. I doubt that but I can't really prove it.
Could someone help me please? Thank you.
 A: In a $T_1$ space, any subset of $X$ is an intersection of open sets. 
So any infinite cofinite space is a counterexample. 
A: You may as well consider the intersection of all nonempty open sets.  If this intersection is nonempty, that means there is a point which is in every nonempty open set of $X$.  Such a point is called a generic point of $X$, and need not exist.  A space is called sober if every irreducible closed subset of it has a unique generic point.
For an explicit example, suppose $X$ is an infinite set with the cofinite topology.  Then $X$ is irreducible, but the intersection of all nonempty open subsets of $X$ is empty, since the complement of every point is open.
A: Let $\tau$ be a $T_1$ topology on a set $X,$ with no isolated points. That is, $\{p\}\not \in \tau$ for all $p\in X.$ (E.g. let $X=\mathbb R$ and let $\tau$ be the usual topology on $\mathbb R.$).
Let $DO(\tau)$ be  the set of $\tau$-dense members of $\tau$. 
Then $\tau_{DO}=\{\phi\}\cup DO(\tau)$ is a $T_1$ topology on $X,$ and any $2$ non-empty members of $\tau_{DO}$ have non-empty intersection, so every non-empty $\tau_{DO}$-open set is $\tau_{DO}$-dense. 
Now $X$ \ $\{p\} \in DO(\tau)$ for each $p\in X.$ So if $X$ has more than $1$ point, then $F=\{X \backslash  \{p\}:p\in X\}$ is a non-empty family of non-empty members of $\tau_{DO}$ and $\cap F=\phi.$
Footnote: As already noted in the answer by Henno Brandsma, it suffices to observe that $\tau_{DO}$ is a $T_1$ topology, because in any $T_1$ space with more than $1$ point, any $Y\subset X$ is the intersection of a non-empty family of non-empty open sets.
