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I have to prove that for any set $A\subseteq\mathbb{R}^n$, $$ \large\overline{A^{\circ}} = \overline{\overline{A^{\circ}}^{\,\circ}} $$

This is what I got so far: for any set $A$ I'm using these definitions:

Interior:
$$\exists r > 0\text{ such that }\{x \mid B_r(x) \subseteq A\}$$

Closure: $$\{x \in \mathbb{R}\mid \exists (X_n) \subseteq A \land X_n \rightarrow x\}$$

Now what I don't get is, I think the right part of the equal, because, I have the interior of $A$, that is all the points that have a ball that is included in $A$, using this I know using the definition of closure that I can pick a sequence that converges to them (using $r$ and decreasing it with $\frac{r}{n}$, $n\to\infty$ for example). But then I don't know how to take the interior of that, I mean what I'm getting at, is that the closure of the interior is the interior, and then the right part of the equation is trivial, as it is the same (m the interior of the interior is the interior, and its closure its the interior)

I think I'm missing something..

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    $\begingroup$ no, what im trying to proove is cl(int($A$)) = cl(int(cl(int($A$)))) $\endgroup$ – Aram Jul 11 '13 at 5:58
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    $\begingroup$ I've corrected the notation in the question now. $\endgroup$ – Zev Chonoles Jul 11 '13 at 6:00
  • $\begingroup$ Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see meta. $\endgroup$ – Zev Chonoles Jul 11 '13 at 6:06
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    $\begingroup$ An equivalent formulation that might be a little easier to work with: If $U$ is an open subset of $\mathbb R^n$, then $\mathrm{cl}(U)=\mathrm{cl}(\mathrm{int}(\mathrm{cl}(U)))$. $\endgroup$ – Jonas Meyer Jul 11 '13 at 6:07
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    $\begingroup$ Related: math.stackexchange.com/q/266156 $\endgroup$ – Jonas Meyer Jul 11 '13 at 6:28
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Interestingly, you can solve this problem by appealing to only a few abstract properties of interior and closure. See this article for more information. Here are three conditions satisfied by the closure operation:

  1. $\mathrm{cl}(S) \supset S$ for all $S \subset \mathbb{R}^n$ (sets get bigger when you close them)
  2. $\mathrm{cl} (\mathrm{cl} (S)) = \mathrm{cl} (S)$ for all $S \subset \mathbb{R}^n$ (the closure of a closed set is itself)
  3. If $S \subset T \subset \mathbb{R}^n$, then $\mathrm{cl}( S) \subset \mathrm{cl}( T)$ (closure preserves containment)

The situation for interiors is nearly the same:

  1. $\mathrm{int}(S) \subset S$ for all $S \subset \mathbb{R}^n$ (sets get smaller when you take their interiors)
  2. $\mathrm{int} (\mathrm{int} (S)) = \mathrm{int} (S)$ for all $S \subset \mathbb{R}^n$ (the interior of an open set is itself)
  3. If $S \subset T \subset \mathbb{R}^n$, then $\mathrm{int}( S) \subset \mathrm{int}( T)$ (taking interiors preserves containment)

Now, let $S \subset \mathbb{R}^n$ be given. We have $\mathrm{cl} (\mathrm{int}(S)) \supset \mathrm{int}(S)$. Now take the interior of both sides to get $\mathrm{int}(\mathrm{cl}(\mathrm{int}(S))) \supset \mathrm{int}(\mathrm{int}(S)) = \mathrm{int}(S)$. Taking the closure on both sides leads to $$ \mathrm{cl}(\mathrm{int}(\mathrm{cl}(\mathrm{int}(S)))) \supset \mathrm{cl} (\mathrm{int}(S))$$ which is one half what we set out to prove. I encourage you to try to deduce the reverse inclusion using similar methods beginning from $\mathrm{int}(\mathrm{cl}(\mathrm{int}(S))) \subset \mathrm{cl}(\mathrm{int}(S))$.

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  • $\begingroup$ Using the properties you listed i know that int(cl(int($S$))) $\subset$ cl(int($S$)), applying cl to both sides, using 1, and 2 of cl i get cl(int(cl(int($S$))) $\subset $ cl(int($S$)) Is this right? $\endgroup$ – Aram Jul 11 '13 at 6:55
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    $\begingroup$ I think it's an application of interior(1) to get $int(cl(int(S))) \subset cl(int(S))$, followed by an application of closure(3) and then closure(2) to arrive at $cl(int(cl(int(S)))) \subset cl(int(S))$. Sorry about doubling up on the labels. Software seemed to insist on resetting the count at 1. $\endgroup$ – Mike F Jul 11 '13 at 7:01
  • $\begingroup$ Oh yes, i used (1) of int without saying it explicitly. Guess i got confused with all the definitions, because using what you listed it was really straightforward. $\endgroup$ – Aram Jul 11 '13 at 7:04
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First, we can rephrase the question: Prove that for an open set $U\subseteq\mathbb{R}^2$, $$\overline{U}=\overline{\overline{U}^{\circ}}$$

For the direction $\subseteq$, we have $U\subseteq \overline{U}^{\circ}$ because the RHS is the maximal open set that is contained is $\overline{U}$. Since taking a closure is monotone (if $A\subseteq B$ then $\overline{A}\subseteq \overline{B}$), we get $\overline{U}\subseteq\overline{\overline{U}^{\circ}}$.

For the direction $\supseteq$, note that for any set $A$ we have $\overline{A}=\overline{\overline{A}}$. Since $\overline{U}\supseteq\overline{U}^{\circ}$ we get $\overline{U}=\overline{\overline{U}}\supseteq\overline{\overline{U}^{\circ}}$.

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  • $\begingroup$ Nice. Not one single feather ruffled. $\endgroup$ – Mike F Jul 11 '13 at 6:55
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    $\begingroup$ Thanks. Note that I only used the topological properties of closure and interior, therefore it is true in any topological space (I see you have mentioned it in your answer). $\endgroup$ – Ido Jul 11 '13 at 6:57
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I like to take $ U = \operatorname{cl}(\operatorname{int}(A))$. Then $U$ is a closed set in $\mathbb{R}^n$.

So we have to prove $ U = \operatorname{cl}(\operatorname{int}(U))$.

$\operatorname{int}(U) \subset U $

As $U$ is closed and closer is smallest closed set containing $\operatorname{int}(U)$, we say $\operatorname{cl}(\operatorname{int}(U)) \subset U$.

Take any point $a \in U$.

If $a \in \operatorname{int} (U)$, trivially $a \in \operatorname{cl}(\operatorname{int} (U))$.

If $a \in \operatorname{bd}(U)$, $a$ shall be a limit point of $\operatorname{cl}(\operatorname{int}(U))$.

In any case $a \in \operatorname{cl}(\operatorname{int}(U))$.

Thus $U = \operatorname{cl}(\operatorname{int}(A))$.

Now replace $U$ by $\operatorname{cl}(\operatorname{int}(A))$, and get the result.

Replace notations only.

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  • $\begingroup$ @ Sleziak thanks $\endgroup$ – Dutta Jul 11 '13 at 10:44

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