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So there is a similiar question in the archives which I looked at after I attempted my proof: Proving that for any sets $A,B,C$, and $D$, if $(A\times B)\cap (C\times D)=\emptyset $, then $A \cap C = \emptyset $ or $B \cap D = \emptyset $

But it is not exactly the same, so I wanted to write my proof out from start to finish to see if my thought process was correct.

1) First I just experimented with some sets to see if anything came about: let $A = \{1\}$, $B = \{2\}$, $C = \{3\}$, $D = \{4\}$:

$(A\times B) \cap (C\times D)$ side:

$$(1,2)\cap (3,4) = (\emptyset,\emptyset)$$ [not sure if i could state this, but it is what I said in my solution]

$(A\cap C)\ \times (B\cap D)$ side: $$\emptyset\ \times \emptyset = (\emptyset, \emptyset)$$

Ok so I established what appears to be equality, so now I have to prove it.

let $(x,y) \in (A\ X\ B) \cap (C\ X\ D)$ --> $(x\in A \cap y\in B) \cap (x\in C \cap y\in D)$ --> $(x\in A \cap x\in C) \cap (y\in B \cap y\in D)$ --> $x\in (A\cap C)\ X\ y\in (B\cap D)$ --> $ (x,y)\in (A\cap C)\ X\ (B\cap D)$ Done.

Then I would have to do the other way as well but it would amount to a similar argument.

P.S: How to get lines of my proof to line up with arrows?

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    $\begingroup$ Your experiment is not really senseful -- $A, B, C, D$ are supposed to be sets, not numbers. $\endgroup$ – hmakholm left over Monica May 5 '15 at 19:42
  • $\begingroup$ Please use \times for $\times$ $\endgroup$ – Gregory Grant May 5 '15 at 19:43
  • $\begingroup$ I treated them as singleton sets. I just don't know how to write the squigly bracket in latex $\endgroup$ – dc3rd May 5 '15 at 19:43
  • $\begingroup$ @dc3rd type " \{ " and " \} " $\endgroup$ – user265675 May 5 '15 at 19:46
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    $\begingroup$ Somewhat related and may help. $\endgroup$ – Daniel W. Farlow May 5 '15 at 19:50
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$$(x,y)\in (A\times B)\cap (C\times D) \iff \left\{\begin{array}{c}(x,y)\in A\times B \\ (x,y) \in C\times D \end{array}\right. \iff \left\{\begin{array}{c}x \in A,\ y\in B \\ x \in C,\ y\in D \end{array}\right. \iff$$

$$\iff \left\{\begin{array}{c}x \in A,\ x\in C \\ y \in B,\ y\in D \end{array}\right. \iff \left\{\begin{array}{c}x \in A\cap C \\ y \in B\cap D \end{array}\right. \iff (x,y) \in (A\cap C)\times (B\cap D)$$

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