Help in proving that the closure of intersection is contained in the intersection of the closures

Question

I was trying to prove that the closure of the intersection of two sets is contained in the intersection of both their closures, while making clear that the reverse inclusion is not always true. My approach was as follows:
Let $A$ and $B$ be subsets of a topological space $(X,\tau)$. Let $C$ be the set of all closed sets which contain $A$, likewise (but for B) for $D$, and $E$ of the set of all closed sets that do not contain $A$ nor $B$ but contain their intersection. All these sets contain additionally the empty set. Then, for every $F_\alpha \subset F$ subset of the set of all closed sets that contain the intersection of $A$ and $B$, there exist $C_\beta \subset C$,$D_\gamma \subset D$, $E_\delta \subset E$ such that $F_\alpha=C_\beta \cup D_\gamma \cup E_\delta$. Then, $$\overline{A \cap B}=\cap_{\alpha} F_\alpha=\cap_ {\beta,\gamma, \delta}(C_\beta \cup D_\gamma \cup E_\delta)=(\cap_\beta C_\beta) \cap (\cap_\gamma D_\gamma) \cap (\cap_\delta E_\delta)=\overline{A}\cap \overline{B}\cap (\cap_\delta E_\delta)$$
Now, because the $E_\delta$ sets are all the closed sets that contain the intersection, I concluded that their intersection was the closure $\overline{A \cap B}$ itself (I'm not sure of this assessment, but my reasoning was that because even though this intersection does not include the sets which contain $A$ or $B$, if it were to include those, that bigger intersection would equal this smaller restricted one as well). Then I concluded that the only way that a set $S$ can be igual to the intersection of itself with another set $T$ is if $T$ contains $S$, and therefore $\overline{A}\cap \overline{B} \supseteq \overline{A \cap B}$.
So I have two questions: is my reasoning correct? And, if so, would it be correct that the equality would hold when the topology would have the property that all the closed sets which contain the intersection contain either $A$ or $B$?
I understand that there are easier and more straightforward proofs, which I have come across, but I wanted to try it for myself to gain greater understanding of the question.
Appreciate any help!

Answer 1

Unfortunately, your argument is incorrect. Your problem arises almost immediately in the fact that your claim $F = C_\beta \cup D_\gamma \cup E_\delta$ is not true, in general, and it doesn't follow since $C_\beta, D_\gamma,$ and $E_\delta$ were arbitrary subsets of $C, D$, and $E$, respectively. To see what I mean, let's look at the following example with $X = \mathbb{R}$ equipped with the standard topology. Suppose we consider the interval $A = (1,3)$ and the interval $B = (2,4)$. We note that $A \cap B = (2,3)$. We then will define $C$ and $D$ exactly as you've defined it.

We set $C = \{K \subset X: K \ \text{is closed and } A \subset K\}$, $D = \{K \subset X: K \text{ is closed and } B \subset K\}$, and $E$ defined exactly as you have stated for $A \cap B = (2,3)$. You then let $C_\beta$ be an arbitrary subset of $C$. Okay, let's create an arbitrary subset of $C$ with our example! Note that the closed interval $[0,4]$ is going to be an element of $C$ since $A \subset [0,4]$ and $[0,4]$ is closed. Thus, we can let $C_\beta = \{[0,4]\}$, which is a subset of $C$. Then you pick an arbitrary subset $D_\gamma$ of $D$. Alright, let's do that with our example! Note that the closed interval $[1,5]$ contains $B$, so it will be an element of $D$. Hence, $D_\gamma = \{[1,5]\}$ is a subset of $D$. Finally, the closed set $[2,3]$ contains $A \cap B$, but does not contain $A$ or $B$. Hence, $[2,3] \in E$, so we can let $E_\delta = \{[2,3]\}$. The union $C_\beta \cup D_\gamma \cup E_\delta = \{[1,5], [2,3], [0,4]\}$ is certainly not equal to $F$. An easy way to see this is that $[1.99,3] \in F$, but $[1.99, 3] \notin C_\beta \cup D_\gamma \cup E_\delta$. Even if you made a mistake and intended to write $F_\alpha = C_\beta \cup D_\gamma \cup E_\delta$, this equality still wouldn't be correct, since we could have let $F_\alpha = \{[1.99,3]\}$.

Your argument is not likely to be salvageable. There's also the additional hiccup that $\overline{A \cap B}$ is a subset of $X$, but $\cap_{\alpha} F_\alpha$ is in general not a subset of $X$. $\cap_{\alpha} F_\alpha$ is a subset of the Power Set of $X$, so your very first equality $\overline{A \cap B}=\cap_{\alpha} F_\alpha$ is incorrect. You should try and convince yourself that $\cap_{\alpha} F_\alpha$, as you've defined it, and $$\bigcap_{K \text{ is closed and }A\cap B \subset K}K$$ are not the same (the latter is the correct definition of $\overline{A \cap B}$).