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?

Your statement is not true, unless you assume more about $U$. For instance, take $X$ the affine plane with Zariski topology, $W$ a parabola and $U$ any line intersecting the parabola in two points. Then, $W\cap U$ consists of two distinct points of the line $U$, which is not irreducible at all. Their closure in $X$ also does not equal $W$.
