# 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.

• I have to ask: what sets did you try this with? Feb 26 '13 at 21:15
• 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, $\operatorname{cl}(\operatorname{int}(\mathrm{Eastwood}))$.
– user856
Feb 28 '13 at 1:16
• @Did -- Maybe all the sets OP tried were singletons? Mar 4 '13 at 8:27
• @Joseph Maybe this is a problem per se (if true).
– Did
Mar 4 '13 at 8:38

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).

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.

• 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. Feb 27 '13 at 0:13
• @ZhenLin Could you expand on this relationship?
– Did
Feb 27 '13 at 7:01
• @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. Feb 27 '13 at 8:14
• 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. Feb 27 '13 at 16:46
• Fascinating stuff! I’m going to see if I can incorporate the new information into this post. :) Feb 28 '13 at 1:02

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