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Let $X$, $Y$, and $Z$ be sets of real numbers.

Is it true to say that $(X\cup Y)\cap Z\subset X\cup(Y\cap Z)$?

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    Try drawing the Venn diagrams2011-10-21
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    @Henry: While Venn diagrams are nice for some basic intuition, it is certainly not a proof and you could miss exactly the counterexamples. **Especially** if the counterexample is nontrivial.2011-10-21
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    @AsafKaragila With only three sets, the standard Venn diagram displays all possible intersections, so I believe it would be an acceptable proof for this question.2011-10-21
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    @Austin: It really depends on how you draw the circles. If you draw them disjoint or not; only two meeting; etc. I know that the psychology department accepts Venn diagrams as "proofs" in statistics classes, but in the math department we deduct points for Venn diagrams. *Especially* because we give those cases that cannot be properly handled with Venn diagrams, and either diagram will not be a complete proof.2011-10-21
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    @AsafKaragila Of course it depends on how you draw the diagram, just as it depends on how you construct a written proof. All I mean to say is that, for these early set theory questions, there is a one-to-one correspondence between written proofs and Venn diagram proofs. One is no less rigorous than the other (though I will grant you that Venn diagram proofs are significantly easier to mess up).2011-10-21
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    Even questions with four or more sets can be handled by diagrams (http://upload.wikimedia.org/wikipedia/commons/a/ac/Venn%27s_four_ellipse_construction.svg), though the pictures tend to get convoluted.2011-10-21
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    Strongly related questions: http://math.stackexchange.com/q/435483/11994 and http://math.stackexchange.com/q/544071/11994.2013-11-16

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Try element chasing:

$x\in (X\cup Y)\cap Z$, then $x\in Z$ and $x\in X\cup Y$, therefore $x\in Z$ and either in $X$ or in $Y$.

  1. If $x\in X$ then $x\in X\cup(Y\cap Z)$.
  2. If $x\in Y$ then $x\in Y\cap Z$, therefore $x\in X\cup(Y\cap Z)$.

Either way we have $x\in (X\cup Y)\cap Z$ then $x\in X\cup(Y\cap Z)$, as wanted. This proof is not just for sets of real numbers but rather for sets in general.

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    @johnny: Where did $A$ and $B$ came from? Either way if $x$ is in $X$ and not in $X\Y$ then $x\in X\cap Y$.2011-10-21
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    sorry, I was looking at another problem. But thanks, now I'm headed in the right direction2011-10-21