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Let $(X,d)$ be any arbitrary metric space? Let $A$ be some subset of $X$.

Is it always true that $\mathrm{int}(\ A \setminus \mathrm{int}(A)\ ) = \emptyset$?

I believe that this statement is false, and that for some wonky situation the above will not hold.

I thought about letting $X=\mathbb{R}^{2}$ and letting $d$ be defined by $d(x,y)=\|x-y\|$ (Euclidean one).

Then maybe taking $A$ to be some disconnected set? Like $A = B\big((0,0), 1\big)\cup \{ (21,9) \}$. Then I find that:

$\mathrm{int}(A)=B\big((0,0), 1\big)$

$A \setminus \mathrm{int}(A) = A \setminus B\big((0,0), 1\big) = \{ (21,9) \}$

$\mathrm{int}(\ A \setminus \mathrm{int}(A)\ ) = \mathrm{int}(\ \{ (21,9) \}\ ) = \emptyset$

So my exmaple doesn't work unfortunately, but maybe it will work if I take $d$ to be the discrete distance or something like that? Can somebody give me a hint?

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Suppose $x\in\operatorname{int}(A\setminus\operatorname{int}(A))$. By definition of the interior, there is an open neighborhood $U$ of $x$ contained in $A\setminus\operatorname{int}(A)$. Then $U$ is also contained in $A$, so $x$ is an interior point of $A$. Hence, $x$ is not in $A\setminus\operatorname{int}(A)$ and certainly not in its interior, a contradiction.

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Note that $\operatorname{int}\bigl(A\setminus\operatorname{int}(A)\bigr)$ is open. Furthermore, it is a subset of $A\setminus\operatorname{int}(A),$ so it is a subset of $A$. However, since it is then an open subset of $A,$ it has to be a subset of $\operatorname{int}(A).$ Hence, it is a subset of $$\bigl(A\setminus\operatorname{int}(A)\bigr)\cap\operatorname{int}(A),$$ and so....

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If $U$ is an open set of $X$ and $U\subseteq A$ then $U\subseteq \mathrm{int}(A)$. So is it possible for $U\subseteq A\setminus \mathrm{int}(A)$ to be open?

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  • $\begingroup$ I don't think that $U \subseteq A \setminus \mathrm{int}(A)$ can be open - however, I am not sure if my reasoning is correct. If $u \in U$, then $u \in \mathrm{int}(U)$ (since it's open). Since $U \subseteq A$, then $u \in \mathrm{int}(A)$. Then $u \notin A \setminus \mathrm{int}(A)$. $\endgroup$ – Greg.Paul Oct 28 '15 at 13:35
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    $\begingroup$ It is possible for $U$ to be open, but only if $U$ is a very specific set. $\endgroup$ – Cameron Buie Oct 28 '15 at 14:09

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