Questions about the proof of Tietze extention theorem on Armstrong's Topology

Question

Tietze extension theorem: Any real-valued continuous function defined on a closed subset of a metric space can be extended over the whole space.

Let $X$ be a metric space, $C$ a closed subset, and $f:C\rightarrow \mathbb E^1$ a map. Assume $f$ is bounded and $|f(x)|\leq M$ on $C$.

Armstrong's proof:

He uses $A_1= \{ x\in C : f(x)\geq \frac{M}{3} \}$ and $B_1=\{x\in C : f(x)\leq -\frac{M}{3} \}$ to construct a continuous function by Lemma (2.14): $g_1:X\rightarrow \left[-\frac{M}{3}, \frac{M}{3}\right]$ with $\begin{cases} \frac{M}{3} & \text{ on } A_1 \\ -\frac{M}{3} & \text{ on } B_1 \\ \left(-\frac{M}{3},\frac{M}{3}\right) & \text{ on } X-(A_1\cup B_1) \end{cases} $

However, both $A_1$ and $B_1$ can be empty, therefore Lemma 2.14 doesn't work anymore. How to resolve this?

Lemma 2.14: If $A,B$ are disjoint closed subsets of a metric space $X$ there is a continuous real-valued function on $X$ which takes the value 1 on points $A$, -1 on points of $B$, and values strictly between $\pm 1$ on points of $X-(A\cup B)$

Answer 1

The Tietze extension theorem says that each continuous $f : C \to \mathbb R$ has a continuous extension to $X$. Extending a constant function is a trivial special case which is possible by Lemma 2.14.

However, the proof of Lemma 2.14 has to be modified if one or both of $A,B$ are empty. The problem is the value $d(x,\emptyset) = \inf \emptyset$. Often one takes $\inf \emptyset = + \infty$ (see Infimum and supremum of the empty set), but this does not work here.

In the proof of 2.14 I suggest to replace $d(x,\emptyset)$ by $1$. Then everything works nicely.

For example, if $A= \emptyset$ and $B \ne \emptyset$, then $f(x) = \dfrac{d(x,B)-1}{d(x,B)+1}$. It takes the value $1$ for all $x \in A$ (simply because there are no $x \in A$), the value $-1$ for all $x \in B$ and the value on $x \in X \setminus (A \cup B) = X \setminus B$ lies in $(-1,1)$.