Let $(E,d)$ be a metric space and $A \subset E$ a nonempty subset of $E$. Suppose that $int(A)$ and $\partial A$ are both nonempty. Then $$d(int(A),\partial A)=0$$
Aiming for a proof, let $k>0$ and $T$ the set $T=\{d(x,y)|x\in int(A), y\in \partial A\}$. For any $y\in \partial A$, we know that $B(y,k)\cap A\neq \emptyset$. If this implies that $B(y,k)\cap int(A)\neq \emptyset$, then we choose some $x\in int(A)$ such that $d(x,y)<k$. Therefore $k$ is not a lower bound of $T$, which implies that $d(int(A),\partial A)=\inf T=0$.
However, this argument is incomplete as I can't prove that last implication, nor find a counterexample. Is this statement false?
Any help would be appreciated. Thanks in advance.
Well, it depends on distance function, if we define the distance function to be $d(x,y)=\text{inf}_{x\in A^o}||x-y||$ such that $y \in \partial{A}$. Then by definition of boundary point we get $\forall \delta>0, B_\delta(y)\cap A^o \neq \emptyset$. Hence for any sequence of points in the interior, there always exist some radius of the open Ball such that, intersection remains non-empty. (Essentially, $||x-y||<\delta ,\forall x\in A^o$)
Now from our definition of the distance function: $d(x,y)\leq||x-y||<d(x,y)+\epsilon, \forall x\in A^o$, hence we can deduce $d(x,y)<\delta$, letting $\delta \rightarrow 0$, we get $d(x,y)=0$.