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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.

tryst with freedom
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onimoni
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    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
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    And the set of fixed points of $r$ is necessarily closed in $X$? – onimoni Jan 14 '13 at 19:29
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    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 06:57
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    @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 08:48
  • How does $V \cap A = V$? If $V$ is just an open neighborhood of $r(x)$ then how can you be sure that there aren't points in $V$ that are in $X$ \ $A$? – Oliver G Feb 19 '17 at 20:30
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    @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 $x \in r^{-1}(V \cap A)$. – Bryan Shih Jul 05 '18 at 12:55
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    The set $V \cap A$ is an open set in the subspace topology for $A$. Since $r : X \to A$ is continuous, that implies that the inverse image is open. – Alan Yan Aug 01 '20 at 17:27

2 Answers2

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The proof shows that the complement of $A$ is open.

Harald Hanche-Olsen
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    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 06 '13 at 04:07
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    @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 06 '13 at 07:07
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    Thanks! I actually managed to figure it out on my own in the meantime, but I figured I'd leave the question, so that if you answered someone else might find it useful, as well. – Ryker Mar 06 '13 at 07:37
  • 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 – tryst with freedom Apr 20 '23 at 17:04
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    @TrystwithFreedom You have a point there, but I can't edit such an old comment. In any case, the reader coming here will be able to see your notice and take corrective action. – Harald Hanche-Olsen Apr 21 '23 at 07:03
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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 $U\cap V=\varnothing$.

knstans
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