Is it always true that $int(\overline{A}) = int(\overline{int (\overline{A})} )?$

Question

Let $X$ be a topological space. Fix a subset $A\subseteq X.$ Recall that interior of $A$, denoted as $int(A),$ is the largest open subset of $A.$ Also, recall that closure of $A,$ denoted as $\overline{A},$ is the smallest closed set containing $A.$

Question: Is it always true that $$int(\overline{A}) = int(\overline{int (\overline{A})} )?$$

Clearly $int(\overline{A})\subseteq \overline{A}.$ It follows that $int(\overline{A}) \supset int(\overline{int (\overline{A})} ).$ However, I do not know whether the reverse inclusion holds.

If we let $A = int(\overline{A}),$ then the desired equality becomes $$A = int(\overline{A}).$$ However, this is false in general, for example, $A = (0,1)\cup (1,2).$

Answer 1

Yes, Because $int(\overline{A})\subset \overline{int(\overline{A})}$, Because the left side is open, then $int(\overline{A})=int(int(\overline{A}))\subset int(\overline{int(\overline{A})})$

Answer 2

Yes. For all A, int cl A = int cl int cl A where cl K = $\overline A$.
Proof. int cl A = int int cl A subset int cl int cl A
subset int cl cl A = int cl A.

Sets A for which A = int cl A are called regular open.
From above, one sees that regular open sets are of the form
int cl A. The example you gave shows not all open sets are
regular open. Regular open sets are open sets with their pin holes and cracks filled.