Characterization of interior in metric spaces

Question

I stumbled upon a theorem in my topology book that didn't have a proof. It says:

For every $x \in X$ and $A \subset X$,

$x \in \text{Int} A$ iff $d(x, A^C) > 0$

Here Int denotes interior, A^C is complement of A and d is a metric on X.

If x is in interior of A, then there is an open ball contained in A that contains x. Now, I don't have any idea how to use this to prove that distance between x and X\A is positive.

The other direction seems even more difficult.

Any help would be appreciated.

Answer 1

If for some $r>0$ we have $B(x,r) \subseteq A$, then let $y$ be any point in $A^\complement$.

We know that $d(x,y) \ge r$, because otherwise $y \in B(x,r)$ and so $y \in A$, contradiction, as $y \in A^\complement$.

This holds for all $y$ in the complement so

$$d(x,A^\complement) = \inf \{d(x,y): y \in A^\complement \}\ge r$$

because $r$ is (as I showed) a lower bound for the set of distances and the inf is the greatest lower bound of the set of distances.

So if $x \in \operatorname{int}(A)$ then $d(x,A^\complement) > 0$.

The reverse also holds:

Suppose that $r = d(x,A^\complement) > 0$. Let $y \in X$ be such that $d(x,y) < r$ (so $y \in B(X,r)$, really). Then $y$ cannot be in $A^\complement$ because then $d(x,y)$ would be a number $<r$ on the right in the formula

$$d(x,A^\complement) = \inf \{d(x,z): z \in A^\complement \}$$ and so we'd have $d(x,A^\complement) \le d(x,y) < r$ which is a contradiction wih the definition of $r$. So $y \notin A^\complement$ so $y \in A$ and we have shown $$B(x,r) \subseteq A$$ as required and so $x \in \operatorname{int}(A)$.

Answer 2

For all $x\in \text{Int}A$ there is $r>0$ with $B(r,x) \subset A$. If $d(x,A^c)=0$ there exists some $y\in A^c$ s.t. $d(x,y)<r$: a contradiction. Conversely, if $d(x,y)\ge r>0$ for all $y\in A^c$ then $B(r,x)\subset A$, otherwise, there would be some $y\in A^c\cap B(r,x)$, which contradicts $d(x,y)\ge r$.