Checking my proof for the following power set problem.

Question

I am unsure if my proof for this is correct. Need verification, please!

Show that the condition A = B is not a necessary one for $P(A) \cup P(B) = P(A \cup B)$ by finding examples of sets A and B such that $P(A) \cup P(B) = P(A \cup B)$ but $A \ne B.$

Let A = {1,2} , B={1}

$P(A)$ = {{1},{2},{1,2}, $\varnothing$}

$P(B)$ = {$\varnothing$ , {1}}

$P(A)\cup P(B)$ = {$\varnothing$ , {1}, {2}, {1,2}}

$A \cup B$ = {1,2}

$P(A\cup B)$ = {$\varnothing$, {1}, {2}, {1,2}}.

Thus, $P(A\cup B)$ = $P(A) \cup P(B).$ However, $A \ne B$.

Answer 1

If you take $A$ nonempty and $B = \varnothing$, then you have it all.