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I am working on a proof where I would like the following identity to hold. $$\bigcap_{w \in \bigcup_{i \in \mathbb{N}} W_i}w = \bigcup_{i \in \mathbb{N}} \bigcap_{w \in W_i} w,$$ where $w$ are sets such that for any $w,w' \in W_i$, $w \cap w' \neq \emptyset$. After playing around with it for a while, I have not been able to come up with a counterexample, and it seems plausible to me that it should hold, but I have not been able to give a rigorous proof. Can someone help me to decide whether this is true or not?

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  • $\begingroup$ I believe this is false. Look at the case where each $W_i$ contains only one $w$. $\endgroup$ – gebruiker May 25 '15 at 9:38
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As I said in the comments, the statement is false. In fact, if we do have $$\bigcap_{w \in \bigcup_{i \in \mathbb{N}} W_i}w = \bigcup_{i \in \mathbb{N}} \bigcap_{w \in W_i} w,$$ then it is necessary that $$\bigcap_{w \in W_i} w\subseteq\bigcap_{w \in \bigcup_{i \in \mathbb{N}} W_i}w,$$ for every $i\in\mathbb N$. After a close look we see that this implies $W_i\subseteq W_j$ for all $i,j$. So we must have $W_i=W_j\,, \forall i,j\in \mathbb N$. Now the statement is trivial. So this is in fact never true, exept for the trivial case.

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Let $W_1=\{\{a\},\{a,b\}\}$, $W_2=\{\{c\},\{d\}\}$, the rest arbitrary and check the difference.

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