328: For Normal Topological Space, Collapsed Topological Space by Closed Subset Is Normal

<The previous article in this series | The table of contents of this series | The next article in this series>

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

---

Topics

About: topological space

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

---

Starting Context

• The reader knows a definition of normal topological space.
• The reader knows a definition of closed set.
• The reader knows a definition of collapsed topological space by subset.
• The reader admits the proposition that the preimage of any closed set under any continuous map is a closed set.

---

Target Context

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

---

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.

---

Main Body

---

1: Description

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

---

2: Proof

Let $f$ be the quotient map, $f:T\to T/S$, and $C_1,C_2\in T/S$ be any closed subsets on $T/S$ such that $C_1\cap C_2=\mathrm{\varnothing }$. $f^{-1}(C_i)$ is closed on $T$ as $f$ is continuous, by the proposition that the preimage of any closed set under any continuous map is a closed set. $f^{-1}(C_1)\cap f^{-1}(C_2)=\mathrm{\varnothing }$. As $T$ is normal, there are some open sets, $U_1,U_2\subseteq T$, around $f^{-1}(C_1),f^{-1}(C_2)$ such that $U_1\cap U_2=\mathrm{\varnothing }$.

If $S\notin C_1,C_2$, $f^{-1}(C_i)\subseteq T\setminus S$, so, $U_i^{\prime }:=U_i\cap (T\setminus S)$, open on $T$, still contains $f^{-1}(C_i)$, while $U_1^{\prime }\cap U_2^{\prime }=\mathrm{\varnothing }$. $f(U_i^{\prime })$ is open on $T/S$, because $f^{-1}(f(U_i^{\prime }))=U_i^{\prime }$ is open on $T$. $f(U_i^{\prime })$ contains $C_i$. $f(U_1^{\prime })\cap f(U_2^{\prime })=U_1^{\prime }\cap U_2^{\prime }=\mathrm{\varnothing }$.

If $S\notin C_1$ and $S\in C_2$, $f^{-1}(C_1)\subseteq T\setminus S$ and $S\subseteq f^{-1}(C_2)$, and $U_1\cap S=\mathrm{\varnothing }$. $f(U_i)$ is open on $T/S$, because $f^{-1}(f(U_i))=U_i$ is open on $T$. $f(U_i)$ contains $C_i$. $f(U_1)\cap f(U_2)=\mathrm{\varnothing }$.

If $S\in C_1$ and $S\notin C_2$, the situation is symmetric with the above case.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>