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 $g: S \to \R$, that is, 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 $\size {\map f x} \le 1$.
Then there exists a continuous mapping $g: S \to \R$ such that:
- $\forall x \in S: \size {\map g x} \le \dfrac 1 3$
- $\forall x \in A: \size {\map f x − \map g x} \le \dfrac 2 3$
$\Box$
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:
- $\ds \map g x = \sum_{n \mathop = 0}^\infty \map {g_n} x$
We have that:
- $\size {\map {g_n} x} \le \dfrac 1 3 \paren {\dfrac 2 3}^n$
and:
- $\ds \sum_{n \mathop = 0}^\infty \dfrac 1 3 \paren {\dfrac 2 3}^n$
converges as a geometric series.
Thus $\ds \sum_{n \mathop = 0}^\infty \map {g_n} x$ converges absolutely and uniformly.
So $g$ is a continuous mapping defined everywhere.
Also, $\ds \sum_{n \mathop = 0}^\infty \dfrac 1 3 \paren {\dfrac 2 3}^n = 1$ implies that $\size {\map g x} \le 1$.
Now for $x \in A$, we have that:
- $\ds \size {\map f x − \sum_{n \mathop = 0}^k \map {g_n} x} \le \paren {\dfrac 2 3}^{k + 1}$
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
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $3$: Compactness: Global Compactness Properties
- This article incorporates material from Tietze extension theorem on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.