How to prove that $ \text{int}(\text{cl}(A)) = \text{cl}(\text{int}(A)) $? 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.
 A: 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
A: I would like to add a little something to Arthur’s answer. In general, we only have

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.

A: 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).
