Basic question: closure of irreducible sets Let $(X,\tau)$ be a topological space, $U \subseteq X$ a non-empty open subset of $X$ and let $W$ be an irreducible non-empty closed subset of $X$. Assuming $W \cap U \neq \emptyset$ why is this intersection irreducible in $U$ and its closure (taken in $X$) equal $W$?
Clearly $W \cap U$ is dense in $W$ because it is an open subset of $W$ and $W$ is irreducible so its closure, taken in $W$ is $W$ itself. I don't see why its closure taken in $X$ is equal $W$. Why is this? Also, why is it irreducible in $U$?
EDIT: OK $W=cl_{X}(W)=cl_{X}(W \cap U)$ I think. Why is it irreducible in $U$ though?

To say that $W\neq \emptyset$ is  irreducible means that it is not the union of two closed subsets both $\neq W$.
Passing to complements, it means that two non-empty open subsets of $W$ have non-empty intersection.
But then every open subset $V\subset W$ has the same property (since opens of $V$ are in particular opens of $W$) and is thus irreducible.
This applies in particular to $V=W\cap U$.
