903: Retract of Hausdorff Topological Space Is Closed

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description/proof of that retract of Hausdorff topological space is closed

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Proof

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Starting Context

• The reader knows a definition of retract of topological space.
• The reader knows a definition of closed set.
• The reader admits the local criterion for openness.

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Target Context

• The reader will have a description and a proof of the proposition that any retract on any Hausdorff topological space is closed on the space.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$T$: $\in \{\text{ the topological spaces }\}$
$S$: $\in \{\text{ the retracts of }T\}$
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Statements:
$S\in \{\text{ the closed subsets of }T\}$
//

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2: Natural Language Description

For any topological space, $T$, any retract, $S\subseteq T$, is closed on $T$.

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3: Proof

Whole Strategy: use the local criterion for openness for $T\setminus S$; Step 1: take a retraction, $r:T\to S$; Step 2: for each $p\in T\setminus S$, take an open neighborhood of $p$, $U\subseteq T$, such that $U\subseteq T\setminus S$.

Step 1:

There is a retraction, $r:T\to S$, which is continuous, such that $r{|}_S=id$.

Step 2:

Let $p\in T\setminus S$ be any.

$r(p)\ne p$, so, there are some open neighborhoods of $p$ and $r(p)$, $U_p\subseteq T$ and $U_{r(p)}\subseteq T$, such that $U_p\cap U_{r(p)}=\mathrm{\varnothing }$, because $T$ is Hausdorff.

Let $U:=r^{-1}(U_{r(p)}\cap S)\cap U_p$.

$p\in U$, because $r(p)\in U_{r(p)}\cap S$, so, $p\in r^{-1}(U_{r(p)}\cap S)$.

$U$ is open on $T$, because $U_{r(p)}\cap S$ is open on $S$ and $r$ is continuous.

$U\subseteq (T\setminus S)$, because $r^{-1}(U_{r(p)}\cap S)$ consists of $U_{r(p)}\cap S$ and some points outside $S$, because as $r{|}_S$ is the identity map, any point in $S\setminus (U_{r(p)}\cap S)$ is not mapped into $U_{r(p)}\cap S$, and as $U_p\cap U_{r(p)}=\mathrm{\varnothing }$, $U$ consists only of points outside $S$.

So, $T\setminus S$ is open on $T$, by the local criterion for openness. So, $S$ is closed on $T$.

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References

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