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Let $(X, \mathcal{T})$ be a topological space. Suppose $A \in \mathcal{T}$. We define the interior of $A$, the closure of $A$ and the boundary of $A$ as follows:

$\operatorname{int}A=\lbrace x \in A\mid A \in \gamma(x) \rbrace$, where $\gamma(x)$ denotes the neighborhood filter of $x$.

$\operatorname{cl}A=\lbrace x \in X\mid \forall V \in \gamma(x): V \cap A \neq \emptyset \rbrace$

$\partial A= \operatorname{cl}A \cap \operatorname{cl}(X\setminus A)$

How do you prove that $\operatorname{int}(\partial A) = \emptyset$? I already have that $\operatorname{int}(\partial A)=\operatorname{int}(\operatorname{cl}A \setminus A)$, but how do you continue from this point?

Michael Hardy
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  • Then what is the boundary of $\mathbb Q\subset\mathbb R$? – drhab Feb 28 '18 at 18:28
  • @drhab $\mathbb{R}$ –  Feb 28 '18 at 18:30
  • Note that the following two lines like conspicuously different from each other in two places: $$\operatorname{cl}A=\lbrace x \in X\mid \forall V \in \gamma(x): V \cap A \neq \emptyset \rbrace$$ $$\text{cl}A=\lbrace x \in X | \forall V \in \gamma(x): V \cap A \neq \emptyset \rbrace$$ I corrected this in the question. – Michael Hardy Feb 28 '18 at 18:31
  • Sorry, I overlooked that it was about the boundary of an open set, and was about to say that $\mathbb R$ does not have empty interior. – drhab Feb 28 '18 at 18:32
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    It is a duplicate of a question about the interior of the boundary of a closed subset. Since $A^c=X-A$ is closed and $\partial A= \partial A^c$, the two questions are equivalent. – Moishe Kohan Feb 28 '18 at 18:38
  • Very strange definitions you have there. And wrong too it seems: no neighborhood filter contains a non-open set, so if A is not open, its "interior" is empty. – H.v.M. May 15 '18 at 11:48

1 Answers1

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When A is open
int $\partial$A = int (cl A $\cap$ cl $A^c$)
= int cl A $\cap$ int cl A$^c$ = int cl A - cl int A
= int cl A - cl A = empty set.

The crutial step uses the important
self dual theorem of topological duality.
cl K$^c$ = (int K)$^c$; int K$^c$ = (cl K)$^c$

William Elliot
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