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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.

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  • 1
    $\begingroup$ You are correct that $A \subseteq \overline A$ and therefore $\bigcap A \subseteq \bigcap \overline A$. Now $\bigcap \overline A$ is an intersection of closed sets, hence closed, so $\bigcap \overline A$ is a closed set containing $\bigcap A$. Since by definition $\overline{\bigcap A}$ is the smallest closed set containing $\bigcap A$, we must have $\overline{\bigcap A} \subseteq \bigcap \overline A$. $\endgroup$ – Bungo Jul 6 '16 at 23:56
  • $\begingroup$ P.S. For a simple example with proper containment, take $A_1 = (0,1)$ and $A_2 = (1,2)$. Then $A_1 \cap A_2 = \emptyset$, so also $\overline{A_1 \cap A_2} = \emptyset$. On the other hand, $\overline{A_1} \cap \overline{A_2} = [0,1] \cap [1,2] = \{1\}$. $\endgroup$ – Bungo Jul 7 '16 at 0:01
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    $\begingroup$ The intersection is contained in A, so the closure of the former is contained in the closure of the latter. Since this holds for any A, it holds for their intersection. $\endgroup$ – Steve D Jul 7 '16 at 0:38

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