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 $B°° = B°$.

Thanks for your hints in advance.

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: $A°° = A°$
  • $\begingroup$ Googling "Kuratowski criteria for the interior" gives me nothing relevant. I suggest you detail these in your question $\endgroup$
    – FShrike
    Commented Jun 11, 2023 at 15:17
  • 1
    $\begingroup$ 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 $\endgroup$
    – FShrike
    Commented Jun 11, 2023 at 15:26

1 Answer 1


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: $$\overline{\overline{(\sim A)}}=\overline{(\sim A)}$$

Which is true.

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.


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