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 $S$.

Since $S$ 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 is 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 $\left|{f \left({x}\right) − g \left({x}\right)}\right| \le \dfrac 2 3$ holds for all $x \in S$.