If the closure operator interchanges with tacking finite intersections, is this then true for a countable intersection? Let $(A_{i})_{i\in \mathbb{N}}$ be a family of subsets of $(X,\tau)$ satisfying $\overline{\bigcap_{i=1}^{n} A_{i}} = \bigcap_{i=1}^{n} \overline{A_{i}}$ for any $n \in \mathbb{N}$. Is it true (or under what assumptions on $X$) is it true that $\overline{\bigcap_{i\in \mathbb{N}} A_{i}}= \bigcap_{i\in \mathbb{N}} \overline{ A_{i}}$.
Thanks for all replies!
 A: It's no true, infact If you take $X=[0,1]$ with euclidean topology.
We put $B=\mathbb{Q}\cap[0,1]=\{r_{n}\}_{n\in\mathbb{N}}$ like a sequence.
Now we take
$$A_{i}=B\setminus r_{i}$$
we observe
$$\overline{A_{i}}=[0,1]$$
So
$$\overline{\cap_{i=1}^{n}A_{i}}=[0,1]=\cap_{i=1}^{n}\overline{A_{i}}$$
moreover
$$\overline{\cap_{i=1}^{+\infty}A_{i}}=\emptyset$$
and
$$\cap_{i=1}^{+\infty}\overline{A_{i}}=[0,1]$$
A: In general it is not true; just take the real line with usual topology and consider the open intervals $A_n = (0,1/n)$.  The condition for finite intersections clearly holds because the intersection of the first $n$ sets is just $A_n$, and the intersection of the first $n$ closures is the closure of $A_n$. But the infinite intersection is empty, so the closure of the infinite intersection is empty, but the intersection of the closures is the point 0.  The real line is so "nice" that it may be hard to find general conditions on $X$ that make the statement hold.
A: In general you will need a very strong condition on $X$ to guarantee such a property.  Suppose your space $X$ is Hausdorff and contains at least one countable infinite subset $A$ such that $\overline{A} \neq A$.  Then you can use Hausdorff to show that $\cap \overline{A_n} \neq \emptyset$, where $A_n = A - \{a_1,a_2,\ldots, a_n\}$ and $A = \{a_n\}_{n=1}^\infty$.  Then $\overline{\cap_n A_n} = \emptyset \neq \cap_n \overline{A_n}$, but you can also use Hausdorff to show that closure is indeed preserved under finite intersections of the $A_n$.
