Is this proof claiming that P(A union B) does not equal P(A) union P(B) correct? Let $A$ and $B$ be finite sets and $\mathbb{P}(A)$ the power set of $A$. Then $\mathbb{P}(A \cup B) \neq \mathbb{P}(A) \cup \mathbb{P}(B)$ for some $A$ and $B$.

Let $A$ be the set $\{1,2,3\}$ and $B$ the set $\{4,5,6\}$. Then $|A \cup B| = |A|+|B| = 6$ since A and B are disjoint. And so  $|\mathbb{P}(A \cup B)| = 2^6$. On the other hand the size of $\mathbb{P}(A) \cup \mathbb{P}(B)$ is at most $|\mathbb{P}(A)|+|\mathbb{P}(B)| = 2^3+2^3=2^4$. Since the two sets don't have the same size, they cannot equal.

Is this correct?
What you said is correct but does not prove the theorem. You only proved that the equation is not in general true. The theorem is that the equation is always false. You need a more general cardinality argument.
What you said is correct but does not prove the theorem. You only proved that the equation is not in general true. The theorem is that the equation is always false. You need a more general cardinality argument.