Tietze Extension Theorem/Lemma
Lemma
Let $T = \struct {S, \tau}$ be a topological space which is normal.
Let $A \subseteq S$ be closed in $T$.
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:
- $\forall x \in S: \size {\map g x} \le \dfrac 1 3$
- $\forall x \in A: \size {\map f x − \map g x} \le \dfrac 2 3$
Proof
The sets $f^{−1} \sqbrk {\hointl {−\infty} {−\dfrac 1 3} }$ and $f^{−1} \sqbrk {\hointr {\dfrac 1 3} \infty}$ are disjoint and closed in $A$.
Since $A$ is closed, they are also closed in $S$.
We have that $S$ is normal.
Then by:
and:
- the fact that $\closedint 0 1$ is homeomorphic to $\closedint {−\dfrac 1 3} {\dfrac 1 3}$
there exists a continuous mapping $g: S \to \closedint {−\dfrac 1 3} {\dfrac 1 3}$ such that:
- $g \sqbrk {f^{-1} \sqbrk {\hointl {-\infty} {-\dfrac 1 3} } } = -\dfrac 1 3$
and
- $g \sqbrk {f^{-1} \sqbrk {\hointr {\dfrac 1 3} \infty} } = \dfrac 1 3$
Thus:
- $\forall x \in S: \size {\map g x} \le \dfrac 1 3$
Now if $−1 \le \map f x \le −\dfrac 1 3$, then:
- $\map g x = −\dfrac 1 3$
and thus:
- $\size {\map f x − \map g x} \le \dfrac 2 3$
Similarly if $\dfrac 1 3 \le \map f x \le 1$, then:
- $\map g x = \dfrac 1 3$
and thus:
- $\size {\map f x − \map g x} \le \dfrac 2 3$
Finally, for $\size {\map f x} \le \dfrac 1 3$ we have that:
- $\size {\map g x} \le \dfrac 1 3$
and so:
- $\size {\map f x − \map g x} \le \dfrac 2 3$
Hence, for all $x \in S$:
- $\size {\map f x − \map g x} \le \dfrac 2 3$
$\blacksquare$
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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.