# Another question over set theory (intersection of an indexed family).

I'm working with the Tao's book, by the way amazing. And I'm not sure about my proof of the next exercise (I really appreciate some help :).) $$\left(\bigcup_{\alpha \in I } A_{\alpha}\right) \cap \left(\bigcup_{\beta \in I } B_{\beta}\right) =\bigcup_{\langle \alpha,\beta\rangle \in I \times J} \left(A_{\alpha} \cap B_{\beta} \right)$$

$z\in\left(\bigcup_{\alpha \in I } A_{\alpha}\right) \wedge z\in \left(\bigcup_{\beta \in I } B_{\beta}\right) \leftrightarrow \exists \alpha \in I.\,z\in A_{\alpha} \wedge\exists \beta \in J.\,z\in B_{\beta}$

And the next step is what I'm not sure if it's completely legal. It's a little tricky :P

$\exists \alpha \in I. \exists \beta \in J. z\in A_{\alpha} \wedge z\in B_{\beta} \leftrightarrow \exists \alpha\;\exists \beta\; ( (\alpha \in I \wedge \beta \in J) \wedge z\in A_{\alpha} \cap B_{\beta})$

Then

$\exists\alpha\; \exists \beta\, ( \langle\alpha, \beta\rangle \in I \times J) \wedge z\in A_{\alpha} \cap B_{\beta}) \leftrightarrow z\in \bigcup_{\langle\alpha,\beta\rangle \in I \times J} \left(A_{\alpha} \cap B_{\beta} \right)$

I have more questions but I don't want to bother them with that.

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Some MathJax advice: < and > mean "less than" and "greater than", and produce spacing correct for that meaning only; to make angle brackets, use \langle and \rangle. – Zev Chonoles Jul 7 '13 at 2:56
It's alright. Although, when proving such stuff, I would recommend not dropping the intermediate $\exists \alpha \in I. \exists \beta \in J. z \in A_\alpha \land z \in B_\beta$. – Daniel Fischer Jul 7 '13 at 2:56
@Zev Chonoles Thanks for the advice :). – Jose Antonio Jul 7 '13 at 3:00
@DanielFischer, thanks. My problem is to manipulate the quantifiers in this kind of operation. Do you know a good source to read about it? – Jose Antonio Jul 7 '13 at 3:03

The argument is correct, though I frankly think that it would be clearer if stated with more words and fewer symbols:

Suppose that $$x\in\left(\bigcup_{\alpha\in I}A_\alpha\right)\cap\left(\bigcup_{\beta\in J}B_\beta\right)\;;$$ then there are an $\alpha\in I$ and $\beta\in J$ such that $x\in A_\alpha$ and $x\in B_\beta$. But then $\langle\alpha,\beta\rangle\in I\times J$, and $x\in A_\alpha\cap B_\beta$, so $$x\in\bigcup_{\langle\alpha,\beta\rangle\in I\times J}(A_\alpha\cap B_\beta)\;.$$ Each of the steps in the argument is clearly reversible, so $$\left(\bigcup_{\alpha\in I}A_\alpha\right)\cap\left(\bigcup_{\beta\in J}B_\beta\right)=\bigcup_{\langle\alpha,\beta\rangle\in I\times J}(A_\alpha\cap B_\beta)\;.$$

Note that the argument can also be made purely computationally, using nothing more than distributivity:

\begin{align*} \left(\bigcup_{\alpha\in I}A_\alpha\right)\cap\left(\bigcup_{\beta\in J}B_\beta\right)&=\bigcup_{\alpha\in I}\left(A_\alpha\cap\bigcup_{\beta\in J}B_\beta\right)\\\\ &=\bigcup_{\alpha\in I}\bigcup_{\beta\in J}(A_\alpha\cap B_\beta)\\\\ &=\bigcup_{\langle\alpha,\beta\rangle\in I\times J}(A_\alpha\cap B_\beta)\;. \end{align*}

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I got lost reading your forest of symbols, so I'll share my verbose thoughts (I also renamed $A\to X, B\to Y$ for my own sake).

Pick an $x$ in $$\left(\bigcup_{\alpha\in I}X_{\alpha}\right)\bigcap \left(\bigcup_{\beta\in J}Y_{\beta}\right)$$ Then $x$ is in some $X_{\alpha}$, and it is also in some $Y_{\beta}$. Hence, there is a pair $(\alpha,\beta)$ such that $x$ is in $$X_{\alpha}\cap Y_{\beta}$$ Thus, $LHS \subset RHS$.

Now, pick a $y$ in $$\bigcup_{\alpha\in I, \ \beta\in J}\left(X_{\alpha}\cap Y_{\beta}\right)$$ Thus, $y\in X_{\alpha}\cap Y_{\beta}$ for some $\alpha, \beta$. So $y\in \cup X_{\alpha}$, and likewise for $\cup Y_{\beta}$. So $RHS \subset LHS$ and the result follows.

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