Let $X$ be a metric space, $A$ a non-empty subset of $X$ and $x \in X$ a point. We define the distance $d(x, A)$ via $$ d(x,A)=\inf(\{ d(x,a)\mid a \in A\}) $$
(a) Show that $d\colon X \to \mathbb{R}$ is a continuous function and that $x\in \overline{A}$ if and only if $d(x, A) = 0$.
(b) Now let $A$ and $B$ be disjoint closed and non-empty subsets of $X$, and consider the function $f \colon X \to \mathbb{R}$ defined by $$ f(x)=\frac{d(x, A)}{d(x, A) + d(x, B)} $$ Show that $f$ is well-defined (i.e., that the denominator is never zero) and that it has the properties as in the Urysohn Lemma, i.e., it takes values in $[0, 1]$ and satisfies $f(a) = 0$ for all $a ∈ A$ and $f(b) = 1$ for all $b ∈ B$.
(c) Show that in fact $A = f^{-1}(0)$ and $B = f^{-1}(1)$
My thoughts
a) Can we say that the function is continuous if we have $x,y\in X$ so, that
$d(x,A)\leq d(x,a)\leq d(x,y) + d(y,a)$ for $a \in A$ $d(x,A)-d(x,y) \leq \inf{d(y,a)}=d(y,A)$ so $d(x,A)-d(y,A)\leq d(x,y)$
Is this right or am I missing something? And how can I show that $x\in\overline{A}$ if and only if $d(x,A)=0$? Is it just to show that we can create a ball in $A$, and show that the intersection is non-empty?
Not sure how to prove b) and c)
Can someone help with this?
a is incorrect.
Prove instead: x in $\overline A$ iff d(x,A) = 0.
Give an example when d(x,A) = 0 and x not in A.
b. If the denominator is 0, then d(x,A) = d(x,B) = 0.
Since A and B are closed, x in A and x in B.
That is why A and B are required to be disjoint.
c. x in A iff d(x,A) = 0 iff f(x) = 0.
x in B iff d(x,B) = 0 iff f(x) = 1.
d. Since you mentioned compact in your title,
prove compact subsets of a metric space are bounded.