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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?

mr eyeglasses
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3 Answers3

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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.

Community
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Brian M. Scott
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  • How do you know that $\operatorname{bdry}A=(\operatorname{cl}A)\setminus\operatorname{int}A$? – mr eyeglasses Apr 01 '15 at 12:22
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    @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 01 '15 at 12:27
  • I now have trouble seeing why $X - \operatorname{ext}A=\operatorname{cl}A$. Our definition of the closure of a set $A$ is the intersection of all the closed subsets of the topological space $X$ which contain $A$ as a subset. – mr eyeglasses Apr 01 '15 at 12:30
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    @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 01 '15 at 12:35
  • Is $(\operatorname{cl}A)\setminus\operatorname{int}A$ a closed set because any closed set with its interior removed is closed? – mr eyeglasses Apr 01 '15 at 14:22
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    @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 01 '15 at 14:24
  • Okay, so I see now that $\mathrm{Cl}(A)$ is the set of points such that every neighborhood of every point intersects $A$. How can I use this result to show that $\mathrm{Cl}(A) = X - \mathrm{Ext}(A)$ if $\mathrm{Ext}(A) = \mathrm{Int}(\mathrm{Comp}(A))$? – mr eyeglasses Apr 01 '15 at 17:55
  • @ᴇʏᴇs: Have you tried? I ask because that’s a very easy proof. – Brian M. Scott Apr 01 '15 at 18:12
  • Well, I don't see any immediate relationship between every neighborhood of points in the closure intersecting with $A$, and the exterior of $A$. So using the definitions, I have to show that $\mathrm{Cl}(A) \subset X - \mathrm{Ext}(A)$, and vice versa. So for one direction, assume $x \in \mathrm{Cl}(A)$. Then we have to show $x \in X - Int(Comp(A))$. So this is a closed set, and if I can prove that $A \subset X - Int(Comp(A))$ then I think $X - Int(Comp(A))$ also contains $Cl(A)$. But not sure how to prove that part. For the other direction, I run into a similarly circular issue. – mr eyeglasses Apr 01 '15 at 18:21
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    @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 01 '15 at 18:32
  • Sorry, I thought we were trying to prove $X - U = Cl(A)$ (since it looks like $U$ in your proof is $Ext(A)$. I am a bit confused. – mr eyeglasses Apr 01 '15 at 18:40
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    @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 01 '15 at 18:41
  • In general, are we allowed to just add/subtract sets from both sides of an equation involving sets? – mr eyeglasses Apr 01 '15 at 18:43
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    @ᴇʏᴇs: If two things are equal, and you perform the same operation on both of them, the resulting objects will still be equal. – Brian M. Scott Apr 01 '15 at 18:45
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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.

user227899
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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.

Milo Brandt
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  • 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 01 '15 at 12:24
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    @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 01 '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 01:18
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    @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 01:36
  • Intuitively thinking about the boundary of the boundary, if I take the complement of the exterior (which is denoted by $U$ here), then obviously it's disjoint from $U$, but if I take the complement of the interior of the boundary (which is disjoint from $U$), then taking the complement of something disjoint from $U$ feels like it would bring us back into $U$ somehow, so taking the union of these feels like it won't be disjoint from $U$. – mr eyeglasses Apr 29 '15 at 02:00
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    @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 02:04
  • Okay, I see exactly where I am having trouble now. I know the boundary is disjoint from $U$, but it seems like a contradiction because I thought the interior of the boundary was a subset of the boundary, so the boundary can't be totally disjoint from $U \cup I$ since the interior of the boundary is in there. – mr eyeglasses Apr 29 '15 at 02:18