# Tietze Extension Theorem

## Theorem

Let $T = \left({S, \tau}\right)$ be a topological space which is normal.

Let $A \subseteq S$ be a closed set in $T$.

Let $f: A \to \R$ be a continuous mapping from $A \subseteq S$ to the real number line under the usual (Euclidean) topology.

Then there exists a continuous extension $g: S \to \R$, i.e. such that:

$\forall s \in A: f \left({s}\right) = g \left({s}\right)$

## Proof

### Lemma

Let $f: A \to \R$ be a continuous mapping such that $\left|{f \left({x}\right)}\right| \le 1$.

Then there exists a continuous mapping $g: S \to \R$ such that:

$\forall x \in S: \left|{g \left({x}\right)}\right| \le \dfrac 1 3$
$\forall x \in A: \left|{f \left({x}\right) − g \left({x}\right)}\right| \le \dfrac 2 3$

### Proof of Main Theorem

First suppose that for any continuous mapping on a closed set there is a continuous extension.

Let $C$ and $D$ be disjoint sets which are closed in $S$.

Define $f: C \cup D \to \R$ by:

$f \left({x}\right) = \begin{cases} 0 & : x \in C \\ 1 & : x \in D \end{cases}$

Now $f$ is continuous and we can extend it to a continuous mapping $g: X \to \R$.

By Urysohn's Lemma, $S$ is normal because $g$ is a continuous mapping such that $g \left({x}\right) = 0$ for $x \in C$ and $g \left({x}\right) = 1$ for $x \in D$.

Conversely, let $S$ be normal and $A$ be closed in $S$.

By the lemma, there exists a continuous mapping $g_0: S \to \R$ such that:

$\forall x \in S: \left|{g_0 \left({x}\right)}\right| \le \dfrac 1 3$
$\forall x \in A: \left|{f \left({x}\right) − g_0 \left({x}\right)}\right| \le \dfrac 2 3$

Since $\left({f − g_0}\right): A \to \R$ is continuous, the lemma tells us there is a continuous mapping $g_1: X \to \R$ such that:

$\forall x \in S: \left|{g_1 \left({x}\right)}\right| \le \dfrac 1 3 \left({\dfrac 2 3}\right)$
$\forall x \in A: \left|{f \left({x}\right) − g_0\left({x}\right) - g_1 \left({x}\right)}\right| \le \dfrac 2 3 \left({\dfrac 2 3}\right)$

By repeated application of the lemma we can construct a sequence of continuous mappings $g_0, g_1, g_2, \ldots$ such that:

$\forall x \in S: \left|{g_n \left({x}\right)}\right| \le \dfrac 1 3 \left({\dfrac 2 3}\right)^n$
$\forall x \in A: \left|{f \left({x}\right) − g_0\left({x}\right) − g_1 \left({x}\right) - g_2 \left({x}\right) - \cdots - g_n \left({x}\right)}\right| \le \dfrac 2 3 \left({\dfrac 2 3}\right)^n$

Define:

$\displaystyle g \left({x}\right) = \sum_{n \mathop = 0}^\infty g_n \left({x}\right)$

We have that:

$\left|{g_n \left({x}\right)}\right| \le \dfrac 1 3 \left({\dfrac 2 3}\right)^n$

and:

$\displaystyle \sum_{n \mathop = 0}^\infty \dfrac 1 3 \left({\dfrac 2 3}\right)^n$

Thus $\displaystyle \sum_{n \mathop = 0}^\infty g_n \left({x}\right)$ converges absolutely and uniformly.

So $g$ is a continuous mapping defined everywhere.

Also, $\displaystyle \sum_{n \mathop = 0}^\infty \dfrac 1 3 \left({\dfrac 2 3}\right)^n = 1$ implies that $\left|{g \left({x}\right)}\right| \le 1$.

Now for $x \in A$, we have that $\displaystyle \left|{f \left({x}\right) − \sum_{n \mathop = 0}^k g_n \left({x}\right)}\right| \le \left({\dfrac 2 3}\right)^{k+1}$.

As $k \to \infty$, the right hand side $\to 0$ and so the sum goes to $g \left({x}\right)$.

Thus:

$\left|{f \left({x}\right) − g \left({x}\right)}\right| = 0$

Therefore $g$ extends $f$.

$\blacksquare$

## Source of Name

This entry was named for Heinrich Franz Friedrich Tietze.