# Proving the idempotence Kuratowski interior criteria starting with the idempotence Kuratowski closure criteria - stuck at the end of proof

I am trying to show the $$4$$ Kuratowski criteria for the interior starting with the $$4$$ Kuratowski criteria for the closure.

If we are in a topological space $$(X, \tau)$$ with the closure of a set being the smallest closed containing the set. We can define the operation $$A \rightarrow \overline{A}$$ from $$\mathscr P(X)$$ to $$\mathscr P(X)$$ which fulfills the $$4$$ closure Koratowski criteria. If $$\sim$$ denotes the complement operation, then we have: $$\overline{A}=\sim (\sim A)°$$ where $$A°$$ denotes the greatest open set contained in $$A$$.

Then, one can easily show that:

• $$\emptyset = \overline {\emptyset} \implies X = X°$$
• $$\overline{A\cup B} = \overline {A} \cup \overline{B} \implies (A\cap B)° = A° \cap B°$$
• $$A \subseteq \overline{A} \implies A° \subseteq A$$

However, I am stuck on the last criterion:

Starting with $$\overline{\overline{A}} = \overline{A}$$, we have: $$\sim (\sim (\sim A)°)° = (\sim A)°$$

Letting $$B = \sim A$$ (as I did in the other criteria proofs), we have:

$$\sim (\sim B°)° = B°$$

From there, I don't know how to reach the conclusion. Maybe taking $$C = \sim B°$$, but then again, when substituting this change, I would loose the $$B°$$ which is crucial for reaching $$B°° = B°$$.

Edit: The Kuratowski interior criteria are, for two subset $$A$$ and $$B$$ of $$X$$:

• Intensivity: $$A° \subseteq A$$
• Preservation of the whole space: $$X° = X$$
• It preserves binary intersections: $$(A\cap B)° = A° \cap B°$$
• Idempotency: $$A°° = A°$$
• Googling "Kuratowski criteria for the interior" gives me nothing relevant. I suggest you detail these in your question Jun 11 at 15:17
• Btw, \circ gives a sort of interior symbol in MathJax. I wrote A^\circ to render $A^\circ$. There may even be a better option, I'm not sure Jun 11 at 15:26

It seems like you want to show $$(A^\circ)^\circ=A^\circ$$ for all $$A$$, given only the definition: $$A^\circ:=\sim\overline{(\sim A)}$$So you want to show: $$\sim\overline{(\sim\underset{A^\circ}{\underbrace{(\sim(\overline{\sim A}))}})}=\sim\overline{(\sim A)}$$Cancelling $$\sim$$ makes this equality equivalent to the equality: $$\overline{\overline{(\sim A)}}=\overline{(\sim A)}$$
Generalising this, if you have a set $$K$$ and:
• An involution $$a:K\to K$$
• An idempotent $$C:K\to K$$
Then the function $$I:K\to K,\,x\mapsto a(C(a(x)))$$ is also idempotent.
Here, we take $$C$$ to be closure and $$a$$ to be complementation, acting on $$K$$ the power set of $$X$$. We deduce the interior operation $$I$$ to be idempotent.