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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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1  
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
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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

2 Answers 2

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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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