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The problem given is this:

$\bigcap_{i \in I}(A_i \cup B_i)$ and $(\bigcap_{i \in I}A_i) \cup(\bigcap_{i \in I}B_i)$

I am asked if they are the same. Here is the reasoning I used:

for the first one: $\forall x(\forall i \in I(x \in A_i \vee x \in B_i))$

for the second one: $\forall x(\forall i \in I(x \in A_i) \vee \forall i \in I(x \in B_i))$

And using this for two families of sets:

$A_2 = \{2, 4\}, A_3 = \{3, 6\}$

$B_2 = \{2, 3\}, B_3 = \{3, 4\}$

I can't get the proper results which are:

$\bigcap_{i \in I}(A_i \cup B_i)$ = $\{3, 4\}$

$(\bigcap_{i \in I}A_i) \cup(\bigcap_{i \in I}B_i)$ = $\{3\}$

What I get is that they both contain all elements and are both equal, where am I going wrong with this?

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Why do you think you're going wrong and where? The very first equality is false and you proved it with your example below! – DonAntonio Feb 23 '14 at 20:43
@DonAntonio the proper results are given to me next to the problem, I'm not actually reaching them on my own :( that's what I meant – johan1234 Feb 23 '14 at 20:44
Ok @johan, apparently you want to know why the outcome of the example. Read the answer. – DonAntonio Feb 23 '14 at 20:47

Let's see:

$$(A_2\cup B_2)\cap (A_3\cup B_3)=\{2,3,4\}\cap\{3,4,6\}=\{3,4\}$$

$$(A_2\cap A_3)\cup(B_2\cap B_3)=\emptyset\cup\{3\}=\{3\}$$

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Like your counterexpample proves, the sets aren't the same. You only have this: $$\bigcap_{i\in I}(A_i\cup B_i)\supseteq\left(\bigcap_{i\in I} A_i\right)\cup\left(\bigcap_{i\in I}B_i\right)$$

But if $\left(\bigcup_{i\in I} A_i\right)\cap \left(\bigcup_{i\in I} B_i\right)=\emptyset$, then the other inclusion is also true.

This is because for every $x\in \bigcap_{i\in I}(A_i\cup B_i)$, $x$ is ether in every $A_i$ or every $B_i$.

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Well as first you forgot the intersection You could reason it like this for the first set $\forall x((x \in A_1 \lor x \in B_1)\land (x \in A_2 \lor x \in B_2)\land\dots\land (x \in A_i \lor x \in B_i))$
For the second set $\forall x((x \in A_1 \land x \in A_2 \land \dots\land x\in A_i)\lor (x \in B_1 \land x \in B_2 \land \dots\land x\in B_i)\\ p_k=\tau(x \in A_k)\\q_k =\tau(x \in B_k)\\k \in I\\((p_1\lor q_1)\land(p_2\lor q_2)\land\dots\land(p_i\lor q_i))\iff (p_1\land p_2\land\dots\land p_i)\lor(q_1\land q_2\land\dots\land q_i)$ Transforming this to logic now it's pretty easy to prove that they're not the same,taking $p_1=\top \land q_2,q_3,\dots q_i=\top$ First one is true while the second one is false so it's not a tautology Not really sure is this is what you wanted.

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