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Prove: There exist sets $A$, $B$ with $\mathcal P \left({A \cup B}\right) \not= \mathcal P \left({A}\right) \cup \mathcal P \left({B}\right)$

$\mathcal P$ here is the Power Set

Update: With @Simon S and @Stefanos help, I was able to write the following:

Proof:

Let $A = \{a\}$, $B = \{b\}$, where the element $a\not = b$

$\mathcal P \left({A}\right) = \{ \varnothing, \{a\}\}$

$\mathcal P \left({B}\right) = \{ \varnothing, \{b\}\}$

$\mathcal P \left({A}\right) \cup \mathcal P \left({B}\right) = \{\varnothing,\{a\},\{b\}\}$

$A \cup B = \{a,b\}$

$\mathcal P \left({A \cup B}\right) = \mathcal P \left({\{a,b\}}\right) = \{\varnothing, \{a\}, \{b\}, \{a, b\}\}$

We can see that $\mathcal P \left({A \cup B}\right) \ne \mathcal P \left({A}\right) \cup \mathcal P \left({B}\right)$

Does that look right?

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    $\begingroup$ What is $P$ here, the power set, the measure/probability? +/union? $\endgroup$ – Simon S Nov 2 '14 at 22:11
  • $\begingroup$ I guess "P" stands for power-set of a given set $\endgroup$ – Marm Nov 2 '14 at 22:13
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    $\begingroup$ Yeah, it is ok. $\endgroup$ – Jimmy R. Nov 2 '14 at 22:23
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If $A = \{ 0 \}$ and $B = \{ 1 \}$, then $P(A) = \left\{ \emptyset, \{ 0 \} \right\}$ and $P(B) = \left\{ \emptyset, \{ 1 \} \right\}$ while

$$P( A \cup B ) = \left\{ \emptyset, \{ 0 \}, \{ 1 \}, \{ 0, 1 \} \right\} \ \neq \ P(A) \cup P(B)$$

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    $\begingroup$ This works if $0\neq 1$. $\endgroup$ – Jonas Meyer Nov 2 '14 at 22:20
  • $\begingroup$ This is the same as what the OP did, no? $\endgroup$ – GFauxPas Nov 2 '14 at 22:33
  • $\begingroup$ @GFauxPas: Yes, but after Simon S & Stefanos helped me. $\endgroup$ – lucidgold Nov 2 '14 at 22:34
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    $\begingroup$ For what it's worth, I did post this before Stefanos or the OP lucidgold edited the post with a solution. $\endgroup$ – Simon S Nov 2 '14 at 22:35
  • $\begingroup$ Wouldn't this be the case for any two sets s.t. $A \cap B \neq \emptyset$? $\endgroup$ – Daniel Goldman Nov 2 '14 at 22:43
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Take two elements $a,c$ such that $a \in A$ but $a \notin B$ and $c \in B$, but $c \notin A$. Then $$\{a,c\}\in \mathcal P(A\cup B)$$ but $$\{a,c\}\notin \mathcal P(A) \quad \text{ and } \quad \{a,c\}\notin \mathcal P(B)$$ which implies that $$\{a,c\}\notin \mathcal P(A)\cup\mathcal P(B)$$

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