# 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 $A=\overline{B'}$?

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I know, but this is not the question! – Andy Teich Nov 5 '12 at 19:33
I am inclined to say that this set is $A\cap B$. Well I think if you take the closure of that you end up with $A$. – tst Nov 5 '12 at 19:35
$A\cap B$ can be empty! – Andy Teich Nov 5 '12 at 19:37
ok, then I think it depends on whether the set $A$ has empty interior or not. Well actually I think this can happen when $cl(int(A))=A$. – tst Nov 5 '12 at 19:40
If $A=\{1\}$ and $B=(0,1)$ then clearly the answer is no. – tst Nov 5 '12 at 19:42

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.

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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 $cl(IP(A)) = A$.

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Closures can in fact add inner points. For example, take $$B=\{x\in\Bbb R^d:0<\lVert x\rVert<1\}.$$ – Cameron Buie Nov 5 '12 at 19:38

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 $A \not\subseteq B$.

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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 Lebesque 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 $A\subset B$.

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