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?

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.
