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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 diagrams – Henry Oct 21 '11 at 0:41
@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. – Asaf Karagila Oct 21 '11 at 0:50
@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. – Austin Mohr Oct 21 '11 at 1:00
@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. – Asaf Karagila Oct 21 '11 at 1:02
@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). – Austin Mohr Oct 21 '11 at 1:18
up vote 1 down vote accepted

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 $x\in X\cap Y$. – Asaf Karagila Oct 21 '11 at 1:16
sorry, I was looking at another problem. But thanks, now I'm headed in the right direction – johnnymath Oct 21 '11 at 1:23

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