Is the boundary of a boundary a subset of the boundary? The definition of a boundary of a set $S$ in a topological space $X$ is $\text{comp}\{\text{Int}(S) \cup \text{Ext}(S)\}$ (complement of the interior union exterior).
The definition for interior is the set of all interior points.
The definition for interior point of $S$ is if there exists an open neighborhood $N$ of that point such that $N \subset S$.
The definition of neighborhood of a point $x$ is a subset of the topological space $X$ that contains an open set such that $x$ is in that open set.
The definition for exterior of $S$ is $\text{Int}(\text{Comp}(S))$.
So $\text{Bdry}(\text{Bdry}(S)) = \text{comp}\{\text{Int}(\text{comp}\{\text{Int}(S) \cup \text{Ext}(S)\}) \cup \text{Ext}(\text{comp}\{\text{Int}(S) \cup \text{Ext}(S)\})\}$ which I can't make heads or tails of. Is there an easier approach to checking of the boundary of the boundary is a subset of the boundary using these definitions?
 A: Let $X$ be a space, and let $A\subseteq X$. Then $\operatorname{bdry}A=(\operatorname{cl}A)\setminus\operatorname{int}A$, so $\operatorname{bdry}A$ is a closed set. Thus,
$$\operatorname{bdry}\operatorname{bdry}A=(\operatorname{cl}\operatorname{bdry}A)\setminus\operatorname{int}\operatorname{bdry}A=(\operatorname{bdry}A)\setminus\operatorname{int}\operatorname{bdry}A\subseteq\operatorname{bdry}A\;.$$
That is, the boundary of the boundary of $A$ is always a subset of the boundary of $A$.
It’s also true, by the way, that
$$\operatorname{bdry}\operatorname{bdry}\operatorname{bdry}A=\operatorname{bdry}\operatorname{bdry}A\;;$$
you’ll find a sketch of the proof in the first part of my answer to this question.
A: The boundary of $S$ is closed in $X$, because it is the intersection of two closed subsets: $$\overline{S} \cap (X \backslash \mathrm{Int}(S)).$$ Therefore, it contains its own boundary.
A: Yes; using your definition, the question you need to answer is:

Can the boundary of the boundary of $S$ intersect $\text{Int}(S)$ or $\text{Ext}(S)$?

because if the answer is "no", then it must be a subset of the boundary, as the only points not contained in the boundary are those in the interior or exterior.
To sketch part of a proof, notice that the $U=\text{Int}(S)\cup\text{Ext}(S)$ is an open set and is, by definition, disjoint from the boundary. In particular, this means it is in the complement of the boundary and is open, implying that every point contained therein, having $U$ as a neighborhood, is in the exterior of the boundary. $U$ is therefore disjoint from in the boundary of the boundary, which implies that the boundary of the boundary is a subset of the boundary.
