How to prove that $ \text{int}(\text{cl}(A)) = \text{cl}(\text{int}(A)) $?

Question

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.

This is true for all sets I have tried... This is a surprising statement. — Did, Feb 26 '13 at 21:17
I feel the set should be named $\mathrm{Eastwood}$. Because, you know, $\operatorname{cl}(\operatorname{int}(\mathrm{Eastwood}))$. — , Feb 28 '13 at 01:16
@RahulNarain Shame, shame, SHAME on you... :-)) (Spilled my coffee because of you.) — Did, Mar 04 '13 at 08:38

score 48 · Answer 1 · edited Apr 24 '16 at 21:33

48

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.

edited Apr 24 '16 at 21:33

Berrick Caleb Fillmore

5,269

answered Feb 26 '13 at 21:35

Haskell Curry

19,524

9

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. – Zhen Lin Feb 27 '13 at 00:13
@ZhenLin Could you expand on this relationship? – Did Feb 27 '13 at 07:01
2

@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. – Zhen Lin Feb 27 '13 at 08:14
@ZhenLin This I know. The relationship to intuitionistic propositional logic I was asking about. – Did Feb 27 '13 at 15:36
1

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. – Zhen Lin Feb 27 '13 at 16:46
@ZhenLin $\langle$ Shivers $\rangle$. Thanks. – Did Feb 27 '13 at 18:35
1

Fascinating stuff! I’m going to see if I can incorporate the new information into this post. :) – Haskell Curry Feb 28 '13 at 01:02
It appears that Greg Strabel's paper is not conveniently available anymore (the link above is broken), so I posted a copy here: http://www.mathtransit.com/cornucopia/2009_strabel.pdf – mathematrucker Apr 22 '16 at 15:55
Do you have a source proving the transitions in this diagram? It could be very useful for many questions on the topic – Dean Gurvitz Oct 27 '18 at 17:53

score 37 · Accepted Answer · edited Apr 21 '15 at 03:18

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).

score 6 · Answer 3 · edited Mar 02 '13 at 07:14

6

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

edited Mar 02 '13 at 07:14

Haskell Curry