# 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?

• Dear user, a little nitpick: irreducible is an absolute notion. A tpological space is plain irreducible, not irreducible in...If you take a subspace of a big space, you can then ask if it is irreducible when endowed with the subspace topology. That said, I have answered your question below. – Georges Elencwajg Feb 11 '12 at 9:21
• Also, I added in an edit that $U$ is open (which you certainly meant since in the second section you say that $W\cap U$ is open in $W$). rattle shows in his answer below that the statement needn't be true if $U$ isn't open . – Georges Elencwajg Feb 11 '12 at 9:34

## 3 Answers

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 $V=W\cap U$.

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 $W$.

• +1: let me point out that rattle's answer is perfectly correct and pertinent, but that it refers to a previous version of the question in which user 6495 had forgotten to say that $U$ is open (this condition has now be added in an edit) – Georges Elencwajg Feb 11 '12 at 9:40

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.