0
$\begingroup$

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!

$\endgroup$

1 Answer 1

1
$\begingroup$

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 $\overline{A \cap B}$).

$\endgroup$
7
  • $\begingroup$ First of all thank you for such a thorough answer. But why are those two expressions at the end different? When I write $\cap_\alpha F_\alpha$ does that not mean the intersection of all closed sets that contain the intersection? That's what I thought it meant. $alpha$ is some index that selects the closed set in question and then I intersect all of them to get the closure. Is that not how it works? $\endgroup$ Dec 15, 2022 at 12:54
  • $\begingroup$ About the first issue you pointed out, I have edited my argument in that department, would you please read that paragraph again and tell me if that difficulty has been avoided? For example, in your example, I would choose the empty set in C and D, and $[1.99,3]$ in E and your $F_\alpha$ would then be generated by that union. With those definitions of the sets C, D and E, I believe I am justified in asserting that any subset of F can be generated by the union of certain subsets of C, D and E. $\endgroup$ Dec 15, 2022 at 15:24
  • 1
    $\begingroup$ Whether you realize it or not, $\cap_\alpha F_\alpha$ is the intersection of collections of closed sets containing the intersection, not the intersection of closed sets containing the intersection. For instance, let's find two subsets of $F$. Note that both of the closed intervals $[1.5, 3.5]$ and $[2,3]$ are elements of $F$. Thus, $F_\alpha = \{[1.5, 3.5]\}$ and $F_{\alpha '} = \{[2,3]\}$ are subsets of $F$. Note that $F_\alpha \cap F_{\alpha '} = \emptyset$ $\endgroup$
    – Michael
    Dec 15, 2022 at 16:26
  • 1
    $\begingroup$ You could say something like $$\bigcap_{x \in F} x = \overline{A \cap B}$$ if you wished. If you require $F_\alpha \in F$, you don't have need for the index $\alpha$ anymore. $\endgroup$
    – Michael
    Dec 15, 2022 at 16:39
  • 1
    $\begingroup$ Thanks, this was really instructive. $\endgroup$ Dec 15, 2022 at 17:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .