Is it true that for any sets $\mathrm{A}$ and $\mathrm{B}, P(A) \cup P(B)=P(A \cup B) ?$ Justify your answer.
Is it true that for any sets $\mathrm{A}$ and $\mathrm{B}, P(A) \cup P(B)=P(A \cup B) ?$ Justify your answer.
False
Let $A=\{0,1\}$ and $B =\{1,2\}$
$\therefore A \cup B=\{0,1,2\}$
$P(A)=\{\varnothing,\{0\},\{1\},\{0,1\}\}$
$P(B)=\{\varnothing,\{1\},\{2\},\{1,2\}\}$
$P(A \cup B)=\{\varnothing,\{1\},\{2\},\{0,1\},\{1,2\},\{0,2\},\{0,1,2\}\}$
$P(A) \cup P(B)=\{\varnothing,\{1\},\{0,1\},\{2\},\{1,2\}\}$
$P(A) \cup P(B)=\{\varnothing,\{1\},\{0,1\},\{2\},\{1,2\}\}$
$\therefore P(A) \cup P(B) \neq P(A \cup B)$
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