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$\newcommand{\bdy}{\operatorname{bdy}}$

I'm trying to show that $\bdy(A) = \bdy(A^c)$.

I know that $\bdy(A) = \operatorname{closure} A \setminus \operatorname{int}(A) = (\operatorname{int}(A^c))^c \setminus \operatorname{int}(A)$, but I don't know where to go from there.

Any help or hints would be very much appreciated.

Michael Hardy
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therexists
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  • bdy(A^c) = closure A^c \ int(A^c)? – therexists Oct 04 '13 at 03:19
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    How do you define closure and int? $\partial A$ is often defined as the set of elements of the ambient space $X$ such that every neighborhood contains elements both of $A$ and $A^c$, and of course this is symmetric in $A$ and $A^c$ – Cocopuffs Oct 04 '13 at 03:20
  • @Cocopuffs Agree, but I don't know how to apply that to this proof. – therexists Oct 04 '13 at 03:23
  • @therexists: Cocopuffs asked a question. Please edit your question to answer hir. – dfeuer Oct 04 '13 at 03:24

4 Answers4

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If you know that $\mathrm{closure}(A)=(\mathrm{int}(A^c))^c$, then you also have $\mathrm{closure}(A^c)=(\mathrm{int}(A))^c$ because $(A^c)^c=A$. Therefore

$$\mathrm{boundary}(A)=\mathrm{closure}(A)\cap(\mathrm{int}(A))^c = \mathrm{closure}(A)\cap \mathrm{closure}(A^c).$$

The last expression is symmetric in $A$ and $A^c$.

Jonas Meyer
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A point in X is said to be a boundary point of a set A if each neighborhood of the point intersects both A and X\A. If you replace A with the complement of A in the statement, you get the same statement. So very simply both the sets have the same boundary.

Suraj
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  • This is a question about topological spaces. There are no spheres here. – José Carlos Santos Feb 04 '20 at 14:21
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    @JoséCarlosSantos: That's right, but if the word "sphere" is replaced by "set" or better "neighborhood", a cogent Answer results. – hardmath Feb 04 '20 at 15:42
  • If you edit your explanation to say "neighbourhood" instead of "open sphere", then other people will understand your explanation, and maybe upvote it. – Alex Feb 04 '20 at 15:48
  • I thought this question was on metric spaces. Apologies for the confusion! – Suraj Feb 04 '20 at 18:27
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Note that for a metric space $(X,d)$, and $A \subset X$, $bdy(A)$ is $cl(A) \cap$ $cl(X \backslash A)$. To show that $bd(A)$ is a subset of $bd(X \backslash A)$ (a similar argument can show the reverse inclusion), it really just suffices to show that $X \backslash (X \backslash A)$ is $A$.

Note that an equivalent definition of the boundary of A is $cl(A) \backslash int(A)$. If you’re using this definition, I suggest you prove that the two definitions are indeed equivalent.

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To show that the boundary of a set A is equal to the boundary of it complement , it suffices to show that there exist an element x in between the boundaries of $A$ and $A^c$ . So if $x$ belongs to boundary of $A$, This implies there exist $y$ and $z$ such that $y$ belongs to $A$ and $z$ belongs to $A^c$ . This implies bd $(A)=x=bd (A^c)$. Hence $bd (A)=bd(A^c)$ .

Hume2
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leonard_nfor
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