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:
 * Urysohn's Lemma

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 $x \in S$:
 * $\size {\map f x − \map g x} \le \dfrac 2 3$