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I came to point where I suppose for case 1 that $A\subseteq B$ and conclusion is trivial. For case 2 I suppose that $A\not\subseteq B$ and try to prove $B\subseteq A$, but that gets me nowhere. Any pointers here are most welcome.

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2 Answers

up vote 2 down vote accepted

Here's a proof without contrapositive, if you prefer.

Suppose that $P(A) \cup P(B) = P(A \cup B)$. Then $A \cup B \in P(A \cup B) = P(A) \cup P(B)$. This means that $A \cup B$ is an element of either $P(A)$ or $P(B)$. Thus, either $B \subseteq A \cup B \subseteq A$ or $A \subseteq A \cup B \subseteq B$.

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This is great!!! Thnx – LavaScornedOven Nov 28 '12 at 15:15
+1, thanks for providing a better answer! – Matt N. Nov 28 '12 at 15:15
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I still think @MattN. your idea with intermediary set is great, but Manny is closer to what I had in mind. I saw technique with intermediary sets in one proof long time ago, so it seems it is a well known approach. I thank both of you for help. – LavaScornedOven Nov 28 '12 at 17:52
up vote 5 down vote

Assume that neither $A \subseteq B$ nor $B \subseteq A$. Then there are elements $x \in A \setminus B$ and $y \in B \setminus A$.

Then the set $\{x,y\}$ is in $P(A \cup B)$ but not in $P(A) \cup P(B)$.

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This is basically proof by contradiction? Is there any room for direct proof? – LavaScornedOven Nov 28 '12 at 14:17
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@Vedran I think it's called contraposition. It means that if you want to show $A \implies B$ you can instead show $\lnot B \implies \lnot A$. – Matt N. Nov 28 '12 at 14:19
@Vedran As for a direct proof: Possibly. But the contraposition is short and smooth. – Matt N. Nov 28 '12 at 14:20
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Oh, you're right, although for this theorem it could be looked at both ways. Thanks anyway for quick reply. BTW, I like the step with creating additional set very much. :) – LavaScornedOven Nov 28 '12 at 14:29

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