# 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?

• I think you need to know something about the topology to say anything interesting about the boundary of the boundary. – Cameron Williams Apr 1 '15 at 3:31
• – Moya Apr 1 '15 at 3:35
• A cheap answer,I know,but I found it useful and so will you: en.wikipedia.org/wiki/Boundary_%28topology%29 – Mathemagician1234 Apr 1 '15 at 3:58

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.

• @ᴇʏᴇs: That’s a standard equivalence; some people take it as the definition, in fact. The definition that you’re using reduces immediately to $$(X\setminus\operatorname{ext}A) \cap (X\setminus\operatorname{int}A) = (\operatorname{cl}A)\cap(X\setminus\operatorname{int}A) = (\operatorname{cl}A)\setminus\operatorname{int}A\;.$$ – Brian M. Scott Apr 1 '15 at 12:27
• @ᴇʏᴇs: Then your first step should probably be to prove the fundamental fact that $\operatorname{cl}A$ is also the set of points $x$ such that every open nbhd of $x$ intersects $A$. There are times (and this is one of them) when this equivalent characterization of $\operatorname{cl}A$ is by far the easiest to use. – Brian M. Scott Apr 1 '15 at 12:35
• @ᴇʏᴇs: It’s more general than that: if $F$ is closed, and $U$ is open, then $F\setminus U=F\cap(X\setminus U)$ is the intersection of two closed sets and is therefore closed. – Brian M. Scott Apr 1 '15 at 14:24
• @ᴇʏᴇs: Let $U=\operatorname{int}(X\setminus A)$. Clearly $X\setminus\operatorname{cl}A$ is an open subset of $X\setminus A$, so $X\setminus\operatorname{cl}A\subseteq U$. On the other hand, if $x\in U$, then $U$ is an open nbhd of $x$ disjoint from $A$, so $x\notin\operatorname{cl}A$, and therefore $U\subseteq X\setminus\operatorname{cl}A$. Thus, $X\setminus\operatorname{cl}A=U$. – Brian M. Scott Apr 1 '15 at 18:32
• @ᴇʏᴇs: Yes, $U$ is $\operatorname{ext}A$. Proving that $X\setminus\operatorname{cl}A=U$ immediately proves that $X\setminus U=\operatorname{cl}A$: just take complements on both sides. This really should not cause you more than a moment’s hesitation. – Brian M. Scott Apr 1 '15 at 18:41

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.

Can the boundary of the boundary of $S$ intersect $\text{Int}(S)$ or $\text{Ext}(S)$?
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.
• I don't understand why the question boils down to the boundary of the boundary of $S$ intersecting $\text{Int}(S)$ or $\text{Ext}(S)$ – mr eyeglasses Apr 1 '15 at 12:24
• @ᴇʏᴇs $\text{Int}(S)\cup \text{Ext}(S)$ is the complement of the boundary; saying $A$ is a subset of $B$ is equivalent to saying a set $A$ doesn't intersect that complement of $B$. – Milo Brandt Apr 1 '15 at 21:57
• Sorry, I was reading this over and I understand that $U$ is disjoint from the boundary, but I don't understand why $U$ is disjoint from the boundary of the boundary. Is it because the boundary of boundary is a closed set? If so, isn't it possible that an open set contains a closed set? I can't visualize the boundary of a boundary, so all I can tell is that it's a closed set that may or may not be in $U$, and we want to prove that it's not in $U$, but I can't see how a set being the exterior of a boundary implies that the exterior is disjoint from the boundary of the boundary – mr eyeglasses Apr 29 '15 at 1:18
• @ᴇʏᴇs The point is that the exterior of the boundary contains (and, in fact, equals) $U$. From definition, $U$ is the complement of the boundary. So, the interior of $U$ is the interior of the complement (i.e. the exterior) of the boundary - but if $U$ is open, then $\text{int}(U)=U$, as each point has $U\subseteq U$ as an open neighborhood. Thus, $U$ is the exterior of the boundary - and the boundary of the boundary must be disjoint from the exterior of the boundary (being the complement of the union of the exterior and interior of the boundary) – Milo Brandt Apr 29 '15 at 1:36
• @ᴇʏᴇs The important bit is that we take the complement after the union. If we let $I$ be the interior of the boundary, then the boundary of the boundary is $(U\cup I)^C$ - which is a set disjoint from $U\cup I$. Since $U$ is a subset of $U\cup I$, it follows that any set disjoint from $U\cup I$ is disjoint from $U$ too - and as such, the boundary is. – Milo Brandt Apr 29 '15 at 2:04