# Tietze Extension Theorem/Lemma

## 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.