Tietze Extension Theorem/Lemma

From ProofWiki
Jump to navigation Jump to search

Lemma

Let $T = \left({S, \tau}\right)$ 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 $\left|{f \left({x}\right)}\right| \le 1$.


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

The sets $f^{−1} \left({\left({−\infty \,.\,.\, −\dfrac 1 3}\right]}\right)$ and $f^{−1} \left({\left[{\dfrac 1 3 \,.\,.\, \infty}\right)}\right)$ are disjoint and closed in $A$.

Since $A$ is closed, they are also closed in $S$.


We have that $S$ is normal.

Then by:

Urysohn's Lemma

and:

the fact that $\left[{0 \,.\,.\, 1}\right]$ is homeomorphic to $\left[{−\dfrac 1 3 \,.\,.\, \dfrac 1 3}\right]$

there exists a continuous mapping $g: S \to \left[{−\dfrac 1 3 \,.\,.\, \dfrac 1 3}\right]$ such that:

$g \left({f^{-1} \left({ \left({-\infty \,.\,.\, -\dfrac 1 3}\right]}\right)}\right) = -\dfrac 1 3$

and

$g \left({f^{-1} \left({ \left[{\dfrac 1 3 \,.\,.\, \infty}\right)}\right)}\right) = \dfrac 1 3$

Thus:

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


Now if $−1 \le f \left({x}\right) \le −\dfrac 1 3$, then:

$g \left({x}\right) = −\dfrac 1 3$

and thus:

$\left|{f \left({x}\right) − g \left({x}\right)}\right| \le \dfrac 2 3$

Similarly if $\dfrac 1 3 \le f \left({x}\right) \le 1$, then:

$g \left({x}\right) = \dfrac 1 3$

and thus:

$\left|{f \left({x}\right) − g \left({x}\right)}\right| \le \dfrac 2 3$

Finally, for $\left|{f \left({x}\right)}\right| \le \dfrac 1 3$ we have that:

$\left|{g \left({x}\right)}\right| \le \dfrac 1 3$

and so:

$\left|{f \left({x}\right) − g \left({x}\right)}\right| \le \dfrac 2 3$

Hence, for all $x \in S$:

$\left|{f \left({x}\right) − g \left({x}\right)}\right| \le \dfrac 2 3$

$\blacksquare$



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.