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Let $A$ and $B$ be finite sets and $\mathbb{P}(A)$ the power set of $A$. Then $\mathbb{P}(A \cup B) \neq \mathbb{P}(A) \cup \mathbb{P}(B)$ for some $A$ and $B$.

Let $A$ be the set $\{1,2,3\}$ and $B$ the set $\{4,5,6\}$. Then $|A \cup B| = |A|+|B| = 6$ since A and B are disjoint. And so $|\mathbb{P}(A \cup B)| = 2^6$. On the other hand the size of $\mathbb{P}(A) \cup \mathbb{P}(B)$ is at most $|\mathbb{P}(A)|+|\mathbb{P}(B)| = 2^3+2^3=2^4$. Since the two sets don't have the same size, they cannot equal.

Is this correct?

Adam
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    You're proof is perfectly fine. Giving a good counterexample is a very nice way to disprove something – Stefan4024 Nov 02 '16 at 20:37
  • Also, notice for example that $A\cup B\in \mathbb{P}(A\cup B)\setminus \mathbb{P}(A)\cup \mathbb{P}(B)$ - unless $A\cup B\subset A$ or $A\cup B\subset B$. – Renan Mezabarba Nov 02 '16 at 20:39
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    I thought $\mathbb{P}(A)={X:X\subset A}$... – Renan Mezabarba Nov 02 '16 at 20:43
  • @Did the implication here is that he means $\Bbb P(A)$ as being the power set of $A$. You may be thinking he meant the probability of $A$. Indeed, I am more familiar with seeing power set notated as $\mathcal{P}(A)$ and probability as $Pr(A)$, but to each their own. – JMoravitz Nov 02 '16 at 20:54
  • @JMoravitz Oops, silly me. Thanks for the comment. – Did Nov 02 '16 at 20:55

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What you said is correct but does not prove the theorem. You only proved that the equation is not in general true. The theorem is that the equation is always false. You need a more general cardinality argument.

Jacob Wakem
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    "The equation is always false" is incorrect. Take $A=B$ for example. Then $\mathcal{P}(A)=\mathcal{P}(B)=\mathcal{P}(A)\cup \mathcal{P}(B)=\mathcal{P}(A\cup B)$ – JMoravitz Nov 02 '16 at 20:55
  • @Jmoravitz Ah yes I see. Even interpreting the task literally, it can still be afforded the graciousness of being possible. – Jacob Wakem Nov 02 '16 at 20:58