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How can one prove that $ \text{int}(\text{cl}(A)) = \text{cl}(\text{int}(A)) $, where $ A \subseteq \mathbb{R}^{n} $?

This is true for all sets that I have tried, but I can’t prove it formally.

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    $\begingroup$ I have to ask: what sets did you try this with? $\endgroup$ Feb 26, 2013 at 21:15
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    $\begingroup$ This is true for all sets I have tried... This is a surprising statement. $\endgroup$
    – Did
    Feb 26, 2013 at 21:17
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    $\begingroup$ I feel the set should be named $\mathrm{Eastwood}$. Because, you know, $\operatorname{cl}(\operatorname{int}(\mathrm{Eastwood}))$. $\endgroup$
    – user856
    Feb 28, 2013 at 1:16
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    $\begingroup$ @Did -- Maybe all the sets OP tried were singletons? $\endgroup$ Mar 4, 2013 at 8:27
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    $\begingroup$ @Joseph Maybe this is a problem per se (if true). $\endgroup$
    – Did
    Mar 4, 2013 at 8:38

3 Answers 3

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I would like to add a little something to Arthur’s answer. In general, we only have Interior-Closure Diagram where ‘$ \longrightarrow $’ means ‘$ \subseteq $’.

It is possible for all $ 7 $ sets displayed in the diagram above to be distinct. The following example in $ \mathbb{R} $ is given in Theorem $ 1 $ of this set of notes by Greg Strabel: $$ A := (0,1) \cup (1,2) \cup \{ 3 \} \cup ([4,5] \cap \mathbb{Q}). $$

We clearly do not require such a complicated subset of $ \mathbb{R} $ to prove that $ \text{cl}(\text{int}(A)) \neq \text{int}(\text{cl}(A)) $ in general. By letting $ A = [0,1] $, we obtain $ \text{int}(\text{cl}(A)) = (0,1) $ and $ \text{cl}(\text{int}(A)) = [0,1] $. As $ (0,1) \neq [0,1] $, we are done.

Now, using the Kuratowski Closure-Complement Theorem (also known as the Kuratowski $ 14 $-Set Theorem), one can prove the following fact.

Fascinating fact: Let $ A $ be a subset of a topological space. If we apply the interior and closure operators to $ A $ repeatedly in any order, then we will obtain at most $ 7 $ distinct sets in the end. Each of these sets will be one of the $ 7 $ sets displayed in the diagram above (this should hint to you that some of the sets in the diagram may coincide). As shown earlier, $ \mathbb{R} $ contains an example for which the maximum possible total of $ 7 $ distinct sets is achieved.

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    $\begingroup$ Doubly fascinating fact: This is closely related to the fact that $\lnot \lnot \lnot \phi \to \lnot \phi$ is a tautology in intuitionistic propositional logic. $\endgroup$
    – Zhen Lin
    Feb 27, 2013 at 0:13
  • $\begingroup$ @ZhenLin Could you expand on this relationship? $\endgroup$
    – Did
    Feb 27, 2013 at 7:01
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    $\begingroup$ @Did $\lnot$ translates to "interior of complement" in point set topology, and the crucial step in the proof of Kuratowski's theorem is to show that doing $\lnot$ three times is the same as doing it once. $\endgroup$
    – Zhen Lin
    Feb 27, 2013 at 8:14
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    $\begingroup$ There is a soundness theorem regarding valuations of intuitionistic propositional theories in lattices of open subsets of topological spaces, and $\lnot \lnot \lnot \phi \to \lnot \phi$ is something that can be proven by purely logical means. $\endgroup$
    – Zhen Lin
    Feb 27, 2013 at 16:46
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    $\begingroup$ Fascinating stuff! I’m going to see if I can incorporate the new information into this post. :) $\endgroup$ Feb 28, 2013 at 1:02
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You can’t. The set on the left-hand side is open, and the set on the right-hand side is closed, and the only subsets of $ \mathbb{R}^{n} $ that are both open and closed are $ \varnothing $ and $ \mathbb{R}^{n} $.

Added: ... and there are subsets of $ \mathbb{R}^{n} $ that have neither $ \varnothing $ nor $ \mathbb{R}^{n} $ as the closure of their interior (or the interior of their closure).

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The following interactive page lets you vary the initial set $ E $ to see (among other things) whether or not it satisfies the identity $ \text{int}(\text{cl}(E)) = \text{cl}(\text{int}(E)) $:

http://mathdl.maa.org/images/upload_library/60/bowron/k14.html

For more on this topic and others like it, this page has plenty of references:

http://www.mathtransit.com/cornucopia.php

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    $\begingroup$ Nice websites! :) $\endgroup$ Mar 2, 2013 at 22:49

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