Problem 2

Exercise 11.2 (Urysohn's Lemma in a Metric Space)   Suppose that $X$ is a metric space with the distance function $d(\cdot,\cdot)$. For a point $x\in X$ and a subset $A$, let

$$d(x,A) := \inf\{d(x,y)\;\vert\; y\in A\}.$$ (11.8)

1. Let $A$ and $B$ be two disjoint closed subsets in $X$. Show that

   $$f(x) = \frac{d(x,A)}{d(x,A)+d(x,B)} : X \to [0,1]$$ (11.9)

   is continuous.
2. Use (a) to verify that, for a closed subsets $A$ and an open subset $U$ such that $A\subseteq U$, there always exists an open set $V$ such that $A\subseteq V \subseteq \overline{V} \subseteq U$.

Proof. The verification that $f$ is continuous and $f\vert _A=0$ and $f\vert _B=1$ is left for Exercise Fall 2019 Exercise 5.2 or Fall 2021 Exercise 2.

Now for part (b), set $B=U^c$ and $V=f^{-1}([0,1/2))$. We show that $V\subseteq U$ and $\overline{V}\subseteq U$. Let $x\in V$. Then $f(x)\in[ 0,1/2)$. Since $f\vert _B=1$, this means $x\notin B=U^c$, so that $x\in U$. Let $\{x_n\}\subseteq V$ be a sequence converging to $x$. The continuity of $f$ yields $f(x)=\lim f(x_n)\leq 1/2$, so that $x\notin B$ again indicating $x\in U$.

Finally, $A\subseteq V$ because $A=f^{-1}(0)$, so that

$$A\subseteq V\subseteq \overline{V} \subseteq U.$$ (11.10)

$\qedsymbol$

Note that Winter 2019 Exercise 2 provides an alternative construction of $V$.