# Intersection of elements of a union

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?

• I believe this is false. Look at the case where each $W_i$ contains only one $w$. – gebruiker May 25 '15 at 9:38

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