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Let $X$ be an infinite set and let $(M_i)_{i\in X}$ be a family of sets indexed by $X$. Suppose $\bigcap_{i\geq 1} M_{a_i} = \{1\}$ whenever the sequence $a_i \in X$ contains infinitely many distinct entries.

Consider the intersection:

$$ \bigcap_{\text{distinct } a,b,c \in X} (M_a \cup M_b \cup M_c) \tag{1} $$

Can we write this expansion as an infinite union of finite intersections of the $M_i$? What is the most succinct form you can think of to write it out as a union?

I'm tripping up because of the fact that there are three unioned terms and so I'm not sure what form the final union will take.


Attempt. The intersection seen in (1) above is equal to:

$$ \bigcap_{a \in X} \bigcap_{b \in X \\ b \neq a} \bigcap_{c \in X \\ c \neq a,b} (M_a \cup M_b\cup M_c) = \bigcap_{a \in X} \left(M_a\cup \bigcap_{b \in X \\ b \neq a} \left ( M_b \cup \bigcap_{c \in X \\ c \neq a,b} M_c\right)\right) \tag {2} $$

I'm thinking the intersection (1) is trivially $= \{1\}$. Do you get the same result?

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  • $\begingroup$ Thanks @EricWofsey you've saved the post. This is definitely regarding another problem, that I need not mention and will not because it incurs automatic downvoting, which isn't fair if you ask me. It's as if saying a realm of math problems is off-topic for the site. $\endgroup$ Aug 18, 2021 at 2:36
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    $\begingroup$ I've edited to remove some more irrelevant details and clarify the presentation. $\endgroup$ Aug 18, 2021 at 2:37
  • $\begingroup$ @EricWofsey thank you. Seeing as I can write it in the last form at the bottom. It looks (since $X$ is infinite) and we assume that $\bigcap_{i\geq 1} M_{a_i} = \{1\}$ that the whole thing collapses to $\{1\}$ the element contained in every $M_{j}$. $\endgroup$ Aug 18, 2021 at 2:39

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You can indeed compute the intersection by following your attempt and observing that the innermost big intersection is just $\{1\}$, and consequently so are each of the outer intersections in turn. There is a more intuitive way to think about it, though.

Let $S$ denote your intersection. Instead of thinking about $S$, let's think about its complement. When is $x\not\in S$? Well, that just means there exist distinct $a,b,c\in X$ such that $x$ is not in any of $M_a,M_b,$ or $M_c$. In other words, this means there are at least three different $i\in X$ such that $x\not\in M_i$. So, negating this, $x\in S$ just means that $x$ is in all but at most two of the $M_i$. If you like, you can write this as $$S=\bigcup_{i,j\in X}\bigcap_{k\neq i,j}M_k.$$ Here for each fixed $i,j\in X$, the inner intersection consists of all the elements that are in every $M_k$ except possibly $M_i$ and $M_j$.

In your case, since you know any intersection of infinitely many of the $M_i$ is just $\{1\}$, all of these inner intersections will be $\{1\}$, and so $S$ will be $\{1\}$.

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