The Tietze extension theorem states that a space $X$ is normal iff every continuous function $f:A \rightarrow \mathbb{R}$, with $A$ a closed subset of $X$, can be extended to a continuous function $g:X \rightarrow \mathbb{R}$.
The proof can be found here. I came across the following proposition:
A space $X$ is normal iff every lower semi-continuous multi-valued map $F: X \rightarrow 2^\mathbb{R}$ with compact and convex images admits a continuous selection.
The statement maybe requires some additional definitions:
- A multi-valued map $F: X \rightarrow 2^Y$ is called lower semi-continuous if for every $G \subseteq Y$ open, $\left\{x \in X \mid F(x) \cap G \neq \emptyset\right\}$ is also an open set.
- If $F: X \rightarrow 2^Y$ is a lower semi-continuous multi-valued map, we call $f: X \rightarrow Y$ a continuous selection if $f$ is continuous and $f(x) \in F(x)$ holds for all $x \in X$.
I don't know if this is needed to prove the proposition, but I'll include it anyways; I've proved that $c(F)$, defined by $c(F)(x) = c(F(x))$ (where $c(A)$ denotes the convex hull of a set $A$), is lower semi-continuous if $F: X \rightarrow 2^Y$.
Any help with proving the blockquoted statement is appreciated, I don't really know how to get started. It does state however that the proof uses the Tietze extension theorem.
The implication from right to left isn’t too hard.
Suppose that $X$ has the continuous selection property. Let $F$ be an arbitrary closed subset of $X$ and $f:F\to\Bbb R$ a continuous function. Let $h:\Bbb R\to(0,1)$ be a homeomorphism, and let $g=h\circ f$. Define
$$G:X\to\wp(\Bbb R):x\mapsto\begin{cases} \{g(x)\},&\text{if }x\in F\\ [0,1],&\text{otherwise}\;. \end{cases}$$
If $U\subseteq\Bbb R$ is open,
$$\{x\in X:G(x)\cap U\ne\varnothing\}=\begin{cases} g^{-1}[U]\cup(X\setminus F),&\text{if }U\cap[0,1]\ne\varnothing\\ \varnothing,&\text{otherwise}\;, \end{cases}$$
so $G$ is lower semicontinuous and takes compact, convex values. Let $\sigma:X\to\Bbb R$ be a continuous selection for $G$; then $\sigma(x)$ is a continuous extension of $g$ to all of $X$, and $h^{-1}\circ\sigma$ is a continuous extension of $f$. It now follows from the Tietze extension theorem that $X$ is normal.
I don’t immediately see how to prove the opposite implication; I’ll give it some thought after I get some sleep.