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Question
Is it true that for any sets A and \[B, P \left( A \right) \cup P \left( B \right) = P \left( A \cup B \right)\]? Justify your answer.
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Solution
\[\text{ False } . \]
\[\text{ Let } X \in P\left( A \right) \cup P\left( B \right)\]
\[ \Rightarrow X \in P\left( A \right) or X \in P\left( B \right)\]
\[ \Rightarrow X \subset A or X \subset B\]
\[ \Rightarrow X \subset \left( A \cup B \right)\]
\[ \Rightarrow X \in P\left( A \cap B \right) \]
\[ \therefore P\left( A \right) \cup P\left( B \right) \subset P\left( A \cup B \right) . . . \left( 1 \right)\]
\[\text{ Again }, \text{ let } X \in P\left( A \cup B \right)\]
\[\text{ But } X \not\in P\left( A \right) \text{ or } x \not\in P\left( B \right) \left[ \text{ For example let } A = \left\{ 2, 5 \right\} \text{ and } B = \left\{ 1, 3, 4 \right\} \text{and take } X = \left\{ 1, 2, 3, 4 \right\} \right]\]
\[So, X \not\in P\left( A \right) \cup P\left( B \right)\]
\[\text{ Thus }, P\left( A \cup B \right) \text{ is not necessarily a subset of }P\left( A \right) \cup P\left( B \right) .\]
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