329: For Regular Topological Space, Collapsed Topological Space by Closed Subset Is Hausdorff

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A description/proof of that for regular topological space, collapsed topological space by closed subset is Hausdorff

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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: Description
• 2: Proof

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Starting Context

• The reader knows a definition of regular topological space.
• The reader knows a definition of closed set.
• The reader knows a definition of collapsed topological space by subset.
• The reader knows a definition of Hausdorff topological space.

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Target Context

• The reader will have a description and a proof of the proposition that for any regular topological space, the collapsed topological space by any closed subset is Hausdorff.

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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: Description

For any regular topological space, $T$, and any closed subset, $S\subseteq T$, the collapsed topological space, $T/S$, is Hausdorff.

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2: Proof

Let $f$ be the quotient map, $f:T\to T/S$, and $p_1,p_2\in T/S$ be any points on $T/S$ such that $p_1\ne p_2$.

If $p_1,p_2\in T\setminus S$, as $T$ is regular and so Hausdorff, there are some open sets, $U_1,U_2\subseteq T$, around $p_1,p_2$ such that $U_1\cap U_2=\mathrm{\varnothing }$. $U_i^{\prime }:=U_i\cap (T\setminus S)$ is open on $T$ and contains $p_i$, and $U_1^{\prime }\cap U_2^{\prime }=\mathrm{\varnothing }$. $f(U_i^{\prime })=U_i^{\prime }$ is open on $T/S$, because $f^{-1}(f(U_i^{\prime }))=U_i^{\prime }$ is open on $T$. $f(U_i^{\prime })$ contains $p_i$. $f(U_1^{\prime })\cap f(U_2^{\prime })=U_1^{\prime }\cap U_2^{\prime }=\mathrm{\varnothing }$.

If $p_1\in T\setminus S$ and $p_2=S$, as $T$ is regular, there are some open sets, $U_1,U_2\subseteq T$, around $p_1,S$ such that $U_1\cap U_2=\mathrm{\varnothing }$. $f(U_i)$ is on $T/S$, because $f^{-1}(f(U_i))=U_i$ is open on $T$. $f(U_i)$ contains $p_i$. $f(U_1)\cap f(U_2)=\mathrm{\varnothing }$.

If $p_1=S$ and $p_2\in T\setminus S$, the situation is symmetric with the above case.

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References

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