Uniformly continuous map that separates $2$ closed subsets with positive distance of a metric space.

Question

I'm trying to build a continuous map that separates $2$ disjoint closed subsets of a metric space. Moreover, if the distance between them is positive, then the map is uniformly continuous.

Let $(E, d)$ be a metric space and $A,B$ closed subsets of $E$ with positive distance, i.e., $$ d(A,B) := \inf \{d(x,y) \mid x \in A, y\in B\} >0. $$ Let $$ d(x, A) := \inf_{y\in A} d(x,y) \quad \text{and} \quad d(x, B) := \inf_{y\in B} d(x,y) \quad \forall x \in E. $$ We adopt the convention that $d(x, \emptyset) = +\infty$. We define a continuous map $f:E \to [0, 1]$ by $$ f (x) := \frac{d(x, A)}{d(x, A) + d(x, B)} \quad \forall x \in E. $$ Then $f$ is uniformly continuous.

Could you verify if my attempt is fine?

I post my proof separately as below answer. This allows me to subsequently remove this question from unanswered list.

Answer 1

It follows that $d(x, A)=0 \iff x \in A$ and $d(x, B)=0 \iff x \in B$. Also, $f^{-1}(0) = A$ and $f^{-1}(1) = B$. Let $$ \alpha := d(A,B). $$

For $x \in E$, we have \begin{align} d(x, A) + d(x, B) &= \inf_{y\in A} d(x,y) + \inf_{y'\in B} d(x,y') \\ &= \inf_{y\in A} \inf_{y' \in B} d(x,y) + \inf_{y\in A} \inf_{y'\in B} d(x,y') \\ &= \inf_{y\in A} \inf_{y' \in B} (d(x,y) + d(x,y')) \\ &\ge \inf_{y\in A} \inf_{y' \in B} d(y,y') \\ &\ge \alpha. \end{align}

Also, \begin{align} &|d(x, A) - d(x', A)| \\ = &\left |\inf_{y\in A} d(x,y) - \inf_{y\in A} d(x',y) \right | \\ \le & \left |\inf_{y\in A} (d(x,y) -d(x', y)) + \inf_{y\in A} (d(x,y)-d(x',y)) + (d(x',y)-d(x,y)) \right | \\ \le & \inf_{y\in A} |d(x,y) -d(x', y)| + \inf_{y\in A} |d(x,y)-d(x',y))| + |d(x',y)-d(x,y)| \\ \le & 3d(x,x'). \end{align}

Similarly, $|d(x, B) - d(x', B)| \le 3d(x,x')$

Notice that $$ |ab-xy| = | (a-x)b + x(b-y)| \le | (a-x)b| + |x(b-y)| \quad \forall a,b,x,y \in \mathbb R. $$

Finally, \begin{align} |f(x)-f(x')| &= \left | \frac{d(x, A)}{d(x, A) + d(x, B)} - \frac{d(x', A)}{d(x', A) + d(x', B)} \right | \\ &= \frac{|d(x, A) d(x', B) - d(x', A) d(x, B)|}{(d(x, A) + d(x, B)) (d(x', A) + d(x', B))} \\ &\le \frac{|d(x, A) d(x', B) - d(x', A) d(x, B)|}{\alpha (d(x', A) + d(x', B))} \\ &\le \frac{d(x', B) |d(x, A) - d(x', A)|}{\alpha (d(x', A) + d(x', B))} + \frac{d(x', A) |d(x', B) - d(x, B)|}{\alpha (d(x', A) + d(x', B))} \\ &\le \frac{3d(x,x') d(x', B)}{\alpha (d(x', A) + d(x', B))} + \frac{3d(x,x') d(x', A)}{\alpha (d(x', A) + d(x', B))} \\ &= \frac{3}{\alpha} d(x,x'). \end{align}

This completes the proof.