I'm trying to prove the following problem from a book I found:
Let $X$ be a topological space and let $\mathscr{A}$ be a collection of subset of $X$. Prove
$\overline{ \bigcap \limits_{A \in \mathscr{A}} A}\subseteq \bigcap \limits_{A \in \mathscr{A}} \overline{A}$
*where the over line indicates closure.
First I'm going to show an example the them being equal, and then show an example of a proper subset. I know the examples are not needed for the proof, but I find them insightful, and these posts are like a notebook that I can refer back to later. Then I will attempt to prove this; I actually had a very difficult time with this one, and I feel like I'm missing something. In any case, please let me know If my examples and proof are valid, and of course alternative proof are very welcome.
(I) Example of $\overline{ \bigcap \limits_{A \in \mathscr{A}} A} = \bigcap \limits_{A \in \mathscr{A}} \overline{A}$
I find that with "most" collections of subsets, the two sides come up equal. Let $X$ be $\mathbb{R}$ with the standard Euclidean metric. Also let, $\mathscr{A}=\{(1,6),(3,10)\}$. So then $ \bigcap \limits_{A \in \mathscr{A}} A= (3,6)$, and $\overline{ \bigcap \limits_{A \in \mathscr{A}} A} =[3,6]$. Next the right side would be: $\bigcap \limits_{A \in \mathscr{A}} \overline{A}= [1,6] \cap [3,10]=[3,6]$. So it is possible for the two sides to be equal.
(II) $\overline{ \bigcap \limits_{A \in \mathscr{A}} A}\subset \bigcap \limits_{A \in \mathscr{A}} \overline{A}$
It took me a while to find an example of this, but I found that if the collection of subsets consist of infinite sets that converge to an empty set, there will be a proper subset. I also found this example, and it uses an empty set as well. Is it possible for this to be true without an empty set??
Let $X$ be $\mathbb{R}$ with the standard Euclidean metric. Also let, $\mathscr{A}=(0,\frac{1}{n}), n \in \mathbb{N}$. Now, $0 \notin \bigcap \limits_{A \in \mathscr{A}} A$, and any positive number can be removed with a large enough $n$, thus $\bigcap \limits_{A \in \mathscr{A}} A = \varnothing$. The empty set is both open and closed ergo, $\overline{ \bigcap \limits_{A \in \mathscr{A}} A} = \varnothing$. On the other hand, $ 0 \in \overline{A}$, therefore, $\bigcap \limits_{A \in \mathscr{A}} \overline{A}= $ {$0$}. So it is possible for the closure of intersections to be a subset of the intersections of closures.
Proof: $\overline{ \bigcap \limits_{A \in \mathscr{A}} A}\subseteq \bigcap \limits_{A \in \mathscr{A}} \overline{A}$
I considered many different ways to prove this, but they all lead to dead ends. The only thing I could think of was to build on the fact that $A \subseteq \overline{A}$ for any subset $A$. Thus, $ \bigcap \limits_{A \in \mathscr{A}} A\subseteq \bigcap \limits_{A \in \mathscr{A}} \overline{A}$. (Next is the part I feel is not solid) When $ \bigcap \limits_{A \in \mathscr{A}} A$ is closed the boundary points that are added will also be points included in $\bigcap \limits_{A \in \mathscr{A}} \overline{A}$. To summarize symbolically:
(a)$ \bigcap \limits_{A \in \mathscr{A}} A\subseteq \bigcap \limits_{A \in \mathscr{A}} \overline{A}$
(b)$ \partial(\bigcap \limits_{A \in \mathscr{A}} A) \subseteq \bigcap \limits_{A \in \mathscr{A}} \overline{A}$
Thus, (a) $\wedge$ (b) $\rightarrow \overline{ \bigcap \limits_{A \in \mathscr{A}} A}\subseteq \bigcap \limits_{A \in \mathscr{A}} \overline{A}$
QED
As stated above, (b) seems intuitively true, and I guess it must be true for the actual statement being proven to be true. But I don't find it obviously true, especially when considering the most general cases of a topological space.