Interior cap for many sets Let $(X,T)$ a topological space and $A_1, A_2, A_3, \ldots$ an infinite sequence of subsets of $X$ such that for any $i \in \mathbb N$ we have $Int(A_i)=O$ I mean they all have the same interior set. I would like to enquire whether $Int( \cap A_i ) = O$ or not. Is this true or not, and how can I prove that?

This is true, and holds for any non-empty family of sets $A_i, i \in I$:

$\operatorname{Int}(\bigcap_{i \in I} A_i) \subseteq \operatorname{Int} A_{i_0} = O$ for any $i_0 \in I$.

On the other hand, we know that $O \subseteq A_i$ for all $i$, so $O \subseteq \bigcap_{i \in I} A_i$, and so $O \subseteq \operatorname{Int}(\bigcap_{i \in I} A_i)$, as $O$ is open, and the interior is the largest open set inside a set.

You can also deduce it from this dual fact, by considering $B_i = X \setminus A_i$ and noting that $\operatorname{Cl}(B_i) = X \setminus \operatorname{Int}(A_i) = X \setminus O$, and $\bigcup_i B_i = X \setminus \bigcap_i A_i$ by de Morgan.