Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $A,B\subseteq\mathbb R^d$ with $A$ closed such that $A\subset\overline{B}$. Does there exist $B'\subset B$ such that $A=\overline{B'}$?

share|cite|improve this question
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
up vote 4 down vote accepted

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.

share|cite|improve this answer

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 $A \not\subseteq B$.

share|cite|improve this answer

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 $A\subset B$.

share|cite|improve this answer

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.