Let $X$ be a topological space, $A\subset X$ and $B=Fr(A)$.
I have two questions.
How do you prove that if $A$ is open or closed, then $B^\circ=\emptyset$?
Why can this not be proven for a set that is both open and closed?
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Asaf Karagila
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Barbara
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Just to be clear, what is your definition of boundary? – Exodd Jan 09 '17 at 18:17
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@Exodd. $ Fr(A)=\overline{A}\cap \overline{(A^c)} $ – Barbara Jan 09 '17 at 18:22
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By your definition, $Fr(X) = Fr(\emptyset) = \emptyset$, so even if $A$ is both open and closed, the interior of $B$ is empty.. – Exodd Jan 09 '17 at 18:30
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$$(Fr(A))^\circ=(\overline{A}\cap \overline{A^c})^\circ=(\overline{A})^\circ\cap (\overline{A^c})^\circ=A^\circ\cap (A^c)^\circ=A^\circ\cap (\overline{A})^c=\emptyset$$
Barbara
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1It’s not necessarily true that $\operatorname{int}\operatorname{cl}A=\operatorname{int}A$: consider the set $A=(0,1)\cup(1,2)$ in $\Bbb R$. – Brian M. Scott Jan 09 '17 at 20:07