# Show that a retract of a Hausdorff space is closed.

A subspace $A \subset X$ is called a retract of $X$ if there is a map $r: X \rightarrow A$ such that $r(a) = a$ for all $a \in A$. (Such a map is called a retraction.)

Proof. Let $x \notin A$ and $a =r(x) \in A$. Since $X$ is Hausdorff, $x$ and $a$ have disjoint neighborhoods $U$ and $V$, respectively. Then $r^{−1}(V \cap A) \cap U$ is a neighborhood of $x$ disjoint from $A$. (*) Hence, $A$ is closed.

I do not understand how "$r^{−1}(V \cap A) \cap U$ is a neighborhood of $x$ disjoint from $A$" implies that $A$ is closed. I would be grateful if someone could point me in te right direction.

-
Actually, I think it is more natural to prove this by noting that $A$ is the set of fixed points of $r$. – Harald Hanche-Olsen Jan 14 '13 at 19:12
And the set of fixed points of $r$ is necessarily closed in $X$? – omar Jan 14 '13 at 19:29
Yes. The set of fixed points is the inverse image of the diagonal in $X\times X$ under the continuous map $x\mapsto(x,r(x))$, and the diagonal is itself closed. (There are more direct proofs too, none of them really difficult.) – Harald Hanche-Olsen Jan 15 '13 at 11:55
Does anyone except me has the trouble understanding why $r^{-1}(V\cap A)$ is open in the first place? – mathreader Aug 10 '13 at 6:57
@mathreader: Since the image of $r$ is (contained in) $A$ already, $r^{-1}(V\cap A)=r^{-1}(V)$. – Harald Hanche-Olsen Aug 10 '13 at 8:48

The proof shows that the complement of $A$ is open.
Why is it true that $r^{−1}(V \cap A) \cap U$ is a neighborhood of $x$ disjoint from $A$, though? How do we know we can find a neighborhood U disjoint from A? – Ryker Mar 6 '13 at 4:07
@Ryker It is true because if $x$ belongs to that set, by definition $x\in U$ and $r(x)\in V$. Therefore $r(x)\ne x$, so $x\notin A$. – Harald Hanche-Olsen Mar 6 '13 at 7:07