334: Reverse of Tietze Extension Theorem

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

A description/proof of reverse of Tietze extension theorem

---

Topics

About: topological space
About: map

---

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 continuous map.
• The reader admits the proposition that any map between topological spaces is continuous if the domain restriction of the map to each closed set of a finite closed cover is continuous.
• The reader admits the proposition that the preimages of any disjoint subsets under any map are disjoint.

---

Target Context

• The reader will have a description and a proof of the reverse of the Tietze extension theorem: any topological space is normal if any continuous map from any closed set to $\mathbb{R}$ has a continuous extension to the whole topological space.

---

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 topological space, $T$, if any continuous map from any closed set to $\mathbb{R}$, $f:C\to \mathbb{R}$, has a continuous extension, $\bar{f}:T\to \mathbb{R},\bar{f}{|}_C=f$, $T$ is normal.

---

2: Proof

Suppose that any continuous map from any closed set to $\mathbb{R}$, $f:C\to \mathbb{R}$, has a continuous extension, $\bar{f}:T\to \mathbb{R}$. Let us think of any disjoint closed sets, $C_1,C_2\subseteq T,C_1\cap C_2=\mathrm{\varnothing }$. Let us think of a continuous map, $f:C_1\cup C_2\to \mathbb{R}$, such that for any $p\in C_1$, $f(p)=-1$; for any $p\in C_2$, $f(p)=1$. $f$ is continuous, because $\{C_1,C_2\}$ is a closed cover of $C_1\cup C_2$ and $f{|}_{C_1}$ and $f{|}_{C_2}$ are continuous, by the proposition that any map between topological spaces is continuous if the domain restriction of the map to each closed set of a finite closed cover is continuous. There is a continuous extension, $\bar{f}$. Let us define $U_1:={\bar{f}}^{-1}((-\mathrm{\infty },0))$ and $U_2:={\bar{f}}^{-1}((0,\mathrm{\infty }))$, open as preimages of open sets under a continuous map. $U_1$ and $U_2$ are disjoint by the proposition that the preimages of any disjoint subsets under any map are disjoint. $C_1\subseteq U_1$ and $C_2\subseteq U_2$. So, $U_1$ and $U_2$ are disjoint neighborhoods of $C_1$ and $C_2$. As $C_1$ and $C_2$ are arbitrary, any 2 disjoint closed sets have some disjoint neighborhoods, which is the definition of being normal.

---

References

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