Tietze Extension Theorem

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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:

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

converges as a geometric series.

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.


Sources

This article incorporates material from Tietze extension theorem on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.