Is it true that for any sets A and B,P(A) cup | vedclass

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(N/A) The statement is $False$.Let $A = \{0, 1\}$ and $B = \{1, 2\}$.Then $A \cup B = \{0, 1, 2\}$.The power set $P(A) = \{\varnothing, \{0\}, \{1\},

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(N/A) The statement is $False$.Let $A = \{0, 1\}$ and $B = \{1, 2\}$.Then $A \cup B = \{0, 1, 2\}$.The power set $P(A) = \{\varnothing, \{0\}, \{1\}, \{0, 1\}\}$.The power set $P(B) = \{\varnothing, \{1\}, \{2\}, \{1, 2\}\}$.Therefore,$P(A) \cup P(B) = \{\varnothing, \{0\}, \{1\}, \{2\}, \{0, 1\}, \{1, 2\}\}$.The power set $P(A \cup B) = \{\varnothing, \{0\}, \{1\}, \{2\}, \{0, 1\}, \{0, 2\}, \{1, 2\}, \{0, 1, 2\}\}$.Comparing the two sets,we observe that $\{0, 2\} \in P(A \cup B)$ but $\{0, 2\} 
otin P(A) \cup P(B)$.Thus,$P(A) \cup P(B) 
eq P(A \cup B)$.

Is it true that for any sets $A$ and $B$,$P(A) \cup P(B) = P(A \cup B)$? Justify your answer.

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