TITLE: Basic question: closure of irreducible sets
QUESTION [8 upvotes]: 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?

REPLY [2 votes]: A space $X$ is irreducible iff $U\cap V\ne\varnothing$ whenever $U$ and $V$ are non-empty open subsets of $X$. Thus, $X$ is irreducible iff every non-empty open subset of $X$ is dense in $X$.
Now let $\langle X,\tau\rangle$ be any space, let $W$ be a non-empty, irreducible, closed subset of $X$, and suppose that $U\in\tau$ is such that $U\cap W\ne\varnothing$. (Although you didn’t say so, it’s clear that you intended $U$ to be open.) Since $W$ is irreducible, $U\cap W$ is dense in $W$, and since furthermore $W$ is closed in $X$, $\operatorname{cl}_X(U\cap W)=\operatorname{cl}_W(U\cap W)=W$, as you said.
Now let $V_0$ and $V_1$ be non-empty open subsets of $U\cap W$. Then there are $G_0,G_1\in\tau$ such that $V_0=G_0\cap(U\cap W)$ and $V_1=G_1\cap(U\cap W)$, so $V_0$ and $V_1$ are non-empty open subsets of $W$ as well, and hence $V_0\cap V_1\ne\varnothing$. Thus, $U\cap W$ is irreducible.