Problem 2

Exercise 5.2 (Urysohn in a metric space)  

1. Let $X$ be a locally compact Hausdorff space and $K\subset V\subset X$ where $K$ is compact and $V$ is open. State the Urysohn Lemma in terms of $K$ and $V$.
2. Let $(X,d)$ be a metric space. For a non-empty subset $A\subset X$, the function

   $$d_A(x) := \inf\{d(x,a) \;\vert\; a\in A\}$$ (5.4)

   is uniformly continuous.
3. For disjoint closed sets $A$ and $B$, define a continuous function $f : X\to[0,1]$ for which $f(A)=0$ and $f(B)=1$. Relate this function to the Urysohn Lemma.

Lemma 1 (Urysohn Lemma)   A topological space $X$ is normal if and only if for all $K \subset V$ with $K$ compact and $V$ open, there exists a continuous function $f : X \to \mathbb{R}$ such that $f(K)=0$ and $f(X\setminus V)=1$,

For the rest of the problem:

Proof. To perform part (b), the function $d(\cdot,F)$ can be argued to be uniformly continuous as follows. Let $0<d(x,y)<\epsilon$. Then

$$\vert d(x,F)-d(y,F)\vert = \left\vert \inf_f d(x,f) - \inf_f d(y,f) \right\vert=\inf_f\vert d(x,f)-d(y,f)\vert$$ (5.5)

For any $f\in F$, we have $\inf \leq \vert d(x,f)-d(y,f)\vert$. Then

$$\vert d(x,F)-d(y,F)\vert \leq \vert d(x,f)-d(y,f)\vert\leq d(x,y) < \epsilon$$ (5.6)

Therefore, $d(\cdot,F)$ is uniformly continuous.

Now to perform part (c), define

$$f(x) := \frac{d_A(x)}{d_A(x)+d_B(x)}.$$ (5.7)

The denominator is never equal to zero, so this function inherits continuity from the functions it is composed of. To see that $d(x,A)+d(x,B)\neq 0$, suppose otherwise. Then $d(x,A)=d(x,B)=0$ which implies $x\in A\cap B$, contradicting that $A$ and $B$ are disjoint. Therefore, the function is continuous, and we can look at its action on elements in $A$ or in $B$: if $x\in A$, then $d(x, A)=0$, so $f(x)=0$. If $x\in B$, then $d(x,B)=0$, so $f(x)=d(x,A)/d(x,A)=1$.

By setting $A=K$ and $B=X\setminus V$, we can prove the Urysohn Lemma in one direction. $\qedsymbol$