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Is it true that for any sets $A$ and $B$, $P(A) \cup P(B)=P(A \cup B)$ ?
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No, Let $A=\{a\}, B=\{b\}$ and $A \cup B=\{a, b\}$ $\therefore P(A)=\{\phi,\{a\}\}, P(B)=\{\phi,\{b\}\}$
and $P(A \cup B)=\{\phi,\{a\},\{b\},\{a, b\}\} \quad \ldots (i)$
and $P(A) \cup P(B)=\{\phi,\{a\},\{b\}\} \quad \ldots (ii)$
From (i) and (ii), we have, $\quad P(A \cup B) \neq P(A) \cup P(B)$
and $P(A \cup B)=\{\phi,\{a\},\{b\},\{a, b\}\} \quad \ldots (i)$
and $P(A) \cup P(B)=\{\phi,\{a\},\{b\}\} \quad \ldots (ii)$
From (i) and (ii), we have, $\quad P(A \cup B) \neq P(A) \cup P(B)$
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