I'm trying to prove that for a closed set $S$, $S\subset X$ and $(X,d)$ a metric space, $\partial(\partial S)=\partial S$, where $\partial$ is the boundary of S defined as $$\partial S=\{\forall x\in X:\forall \epsilon >0 D(x,\epsilon)\cap S \neq \emptyset \:and\: D(x,\epsilon)\cap S^C = \emptyset\}=cl(S)\cap cl(S^C) $$ where $cl(S)$ is the closure of S.
Here's what I've done so far:
Let $T=\partial S$:
We know that $T$ is closed in X as it is a boundary (proof).
I know how to prove that $\partial T \subseteq T$, as since we define a closure of a set as the smallest closed set that contains $T$, thus since $T$ is closed $cl(T)=T$. We also know that $\partial T \subseteq cl(T)$ and thus $\partial T \subseteq T\implies \partial(\partial S)\subseteq \partial S$.
It's proving that $\partial(\partial S)\supseteq \partial S$ is where I'm having difficulty.
Best Answer
$\partial S = \partial (X \setminus S)$ (by definition) and the boundary of an open set is nowhere dense (e.g. this link). This means that $\partial S$ (being closed) has no interior points and so $\partial S \subseteq \partial \partial S$ is then easy to see. The other inclusion you already did (basically, the boundary is closed and a set is closed iff it contains its boundary).