Prove that P (A ∪ B) = P (A) ∪ P(B) is true iff B ⊆ A or A ⊆ B.

Question

Prove that P (A ∪ B) = P (A) ∪ P(B) is true iff B ⊆ A or A ⊆ B.

I know that in general, P (A ∪ B) = P (A) ∪ P(B) is not true (thus the iff is needed here).

I'm having trouble proving it formally.

Thanks in advance!

Answer 1

Let $a \in A \setminus B$ and $b \in B \setminus A$. Then $\{a,b\}\in P(A\cup B)$ but $\{a,b\} \not \in P(A)$ and $\{a,b\} \not \in P(B)$.

And if $A \subset B$ then $A\cup B = B$ and $P(A) \cup P(B) = P(B)$.

====

I suppose proving that $X \subset Y \implies P(X) \subset P(Y)$ requires a bit of thought. But not too much.

Answer 2

If $\wp(A\cup B)=\wp(A)\cup\wp(B)$ then from $A\cup B\in\wp(A\cup B)$ it follows that: $$A\cup B\in\wp(A)\cup\wp(B)$$ so that: $$A\cup B\in\wp(A)\text{ or }A\cup B\in\wp(B)$$ or equivalently: $$A\cup B\subseteq A\text{ or }A\cup B\subseteq B$$ or equivalently: $$B\subseteq A\text{ or }A\subseteq B$$