simple set theoretic question of subsets of $\mathbb R^d$ Let $A,B\subseteq\mathbb R^d$ with $A$ closed such that $A\subset\overline{B}$. Does there exist $B'\subset B$ such that $A=\overline{B'}$?
 A: Sorry, I misread the question initially.
The answer is no in general. For example, take $B$ to be plane minus $x$-axis, and $A$ to be $x$-axis. If $B'$ exists, it must be a subset of both $A$ and $B$, which is empty.
A: Here is a counterexample that is essentially the same as Sanchez's answer, but a dimension simpler. Let


*

*$A = \{0 \}$ and

*$B = (0,1)$,


so that $A \subseteq \overline{B}$. The only set whose closure is a singleton is the set itself, but that is not a valid choice in our instance since $A \not\subseteq B$.
A: If $cl(in(A))=A$ then there exists $B'$ as requested.
If $cl(in(A))=A$ then all points in $A$ are accumulation points.
We have $cl(A\cap B)\subset A\cap cl(B)=A$.
Let $x\in A\backslash cl(A\cap B)$, then since $x$ is an accumulation point in $A$, $\exists U \subset A\backslash cl(A\cap B)$, $\lambda(U)\ne 0$, with $\lambda$ the Lebesgue measure in $\mathbb{R}^d$.
However $\lambda(A)=\lambda(A\cap B)$, so $A\backslash cl(A\cap B)=\emptyset$.
I believe (but I cannot prove it) that if $cl(in(A))\ne A$, $B'$ exists only when $A\subset B$.
A: Take the set of all inner points of $A$. If $x$ is an inner point of $A$, then it is an inner point of $\overline B$ as well, so it is an inner point of $B$ (as the closure doesn't add inner points). So $IP(A) \subset B$ and $cl(IP(A)) = A$.
