Is P(A) ∪ P(B) = P(A∪B)?
Question:
Is it true that for any sets \( A \) and \( B \),
\[ P(A)\cup P(B)=P(A\cup B) \]Justify your answer.
Solution
The statement is false.
Take
\[ A=\{1\}, \quad B=\{2\} \] \[ P(A)=\{\phi,\{1\}\} \] \[ P(B)=\{\phi,\{2\}\} \] \[ P(A)\cup P(B)=\{\phi,\{1\},\{2\}\} \]Now,
\[ A\cup B=\{1,2\} \] \[ P(A\cup B)=\{\phi,\{1\},\{2\},\{1,2\}\} \]Since
\[ P(A)\cup P(B)\neq P(A\cup B) \]therefore the statement is not true in general.