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

### Definitions:

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.)

### Problem statement:

Show that a retract of a Hausdorff space is closed.

### Proof:

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.

### Question:

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 $r$. Jan 14, 2013 at 19:12
• And the set of fixed points of $r$ is necessarily closed in $X$? Jan 14, 2013 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.) Jan 15, 2013 at 11:55
• @mathreader: Since the image of $r$ is (contained in) $A$ already, $r^{-1}(V\cap A)=r^{-1}(V)$. Aug 10, 2013 at 8:48
• @OliverG I don't think $V \cap A=V$ is implied. Clearly, $LHS \subseteq RHS$. Conversely, $x \in RHS \Rightarrow r(x) \in V$. But the image of $r$ is in $A$, so $r(x) \in V \cap A$ and $x \in r^{-1}(V \cap A)$. Jul 5, 2018 at 12:55

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? Mar 6, 2013 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 $x\notin A$. Mar 6, 2013 at 7:07
• You should use another variable. Since a person may confuse the $x$ you used for the arguement, and $x$ used in the main arguement of the question @HaraldHanche-Olsen Apr 20, 2023 at 17:04
Why $$r^{-1}(V\cap A)\cap U$$ is disjoint from $$A$$?
$$\begin{array}{rcl} y\in r^{-1}(V\cap A)\cap U & \Leftrightarrow & y\in\{ z\in X: r(z)\in V\cap A \}\cap U \\ & \Leftrightarrow & y\in\{ z\in U: r(z)\in V\cap A \}. \end{array}$$ So $$y\notin A$$. If the opposite was true $$\;$$($$y\in A$$)$$\;$$ it should be $$y\in U \Rightarrow r(y)\in U$$, $$\;$$but $$r(y)\in V\cap A\subset V$$ and $$U\cap V=\varnothing$$.