TITLE: Is it always true that $\mathrm{int}(\ A \setminus \mathrm{int}(A)\ ) = \emptyset$?
QUESTION [11 upvotes]: 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?

REPLY [4 votes]: 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....