Prove that the Interior of the Boundary is Empty Suppose X is a Metric Space
Let S $\subset X$
Prove that if S is Closed then, the Interior of the Boundary of S is Empty
Totally stuck on how to solve this.
 A: Hint: Let $\partial S$ be the boundary of $S$.  (Not sure if you're familiar with that notation.)  $x\in\partial S$ if and only if every neighborhood $U$ of $x$ contains $y\in S$ different from $x$, and a point $z$ which is not in $S$.
A: It is true in general topology that the boundary of an open set has empty interior, and the same is true for a closed set.
Lemma: A set $U$ is open iff $\partial U = \bar{U}\setminus U$.
Let $U$ be an open set. Then $\partial U$ is disjoint from $U$. Suppose for contradiction that $\partial U$ contains an non-empty open set $O$, and let $x \in O$. Then since $x \in \bar{U}$, every neighborhood of $x$ intersects $U$, and in particular $O\cap U \neq\emptyset$, a contradiction. Now for each set $A$, $\partial A = \partial (A^C)$, the boundary of every closed set has empty interior as well.
A: Note that  $int(\partial A)=int(cl(A) \setminus int(A))$, then we have $int(cl(A)\setminus int(A))=cl(A) \setminus cl(int(A))$. Since $int(A) \subset A $, thus $cl(int(A)) \subset cl(A))$. Hence $cl(A) \setminus cl(int(A)) = \emptyset $. Therefore $int(\partial A)=\emptyset$.
