# Tietze Extension Theorem

## Theorem

Let $T = \struct {S, \tau}$ 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:

$\forall s \in A: \map f s = \map g s$

## 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:

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

$\Box$

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:

$\map f x = \begin {cases} 0 & : x \in C \\ 1 & : x \in D \end {cases}$

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 $\map g x = 0$ for $x \in C$ and $\map g x = 1$ 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:

$\forall x \in S: \size {\map {g_0} x} \le \dfrac 1 3$
$\forall x \in A: \size {\map f x − \map {g_0} x} \le \dfrac 2 3$

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

$\forall x \in S: \size {\map {g_1} x} \le \dfrac 1 3 \paren {\dfrac 2 3}$
$\forall x \in A: \size {\map f x − \map {g_0} x - \map {g_1} x} \le \dfrac 2 3 \paren {\dfrac 2 3}$

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

$\forall x \in S: \size {\map {g_n} x} \le \dfrac 1 3 \paren {\dfrac 2 3}^n$
$\forall x \in A: \size {\map f x − \map {g_0} x − \map {g_1} x - \map {g_2} x - \cdots - \map {g_n} x} \le \dfrac 2 3 \paren {\dfrac 2 3}^n$

Define:

$\displaystyle \map g x = \sum_{n \mathop = 0}^\infty \map {g_n} x$

We have that:

$\size {\map {g_n} x} \le \dfrac 1 3 \paren {\dfrac 2 3}^n$

and:

$\displaystyle \sum_{n \mathop = 0}^\infty \dfrac 1 3 \paren {\dfrac 2 3}^n$

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

So $g$ is a continuous mapping defined everywhere.

Also, $\displaystyle \sum_{n \mathop = 0}^\infty \dfrac 1 3 \paren {\dfrac 2 3}^n = 1$ implies that $\size {\map g x} \le 1$.

Now for $x \in A$, we have that:

$\displaystyle \size {\map f x − \sum_{n \mathop = 0}^k \map {g_n} x} \le \paren {\dfrac 2 3}^{k + 1}$

As $k \to \infty$, the right hand side $\to 0$ and so the sum goes to $\map g x$.

Thus:

$\size {\map f x − \map g x} = 0$

Therefore $g$ extends $f$.

$\blacksquare$

## Source of Name

This entry was named for Heinrich Franz Friedrich Tietze.