For something like P(A) u P(B) ⊆ P(A u B) and P(A u B) ⊄ P(A) u P(B)
proving the first part was fine, I just have to use the definition of power sets to show that both P(A) and P(B) correspond to P(A u B).
But I have problem showing the second part. I wanted to prove it by reaching some sort of contradiction, by assuming that P(A u B) is a subset of P(A) u P(B).
so by definition of power sets: P(A u B) = {Y|Y ⊆ (A u B)} so Y ⊆ (A u B), and ∀y ∈ Y, y ∈ (A u B).
then y ∈ A or y ∈ B
⇒
Y ⊆ A or Y ⊆ B
⇒ Y ∈ P(A) or Y ∈ P(B)
and then I end up with Y ∈ P(A) u P(B)?
..... but since the statement was suppose to be false, I should have arrived at a contradiction instead? I think I am misunderstanding a concept or something, but I don't know what went wrong.
Sorry if I made some dumb mistake or something is written in an improper format, I'm still self learning proofs.
The problem with your attempt is that the fact that $y \in A$ or $y \in B$ for all $y \in Y$ does not imply that $Y \subseteq A$ or $Y \subseteq B$. It is easy to see why.
It is enough to show that an element of $\mathcal{P} (A \cup B)$ is not contained in $\mathcal{P} (A) \cup \mathcal{P} (B)$. Consider the case where $A \ne B$ and the set $A \cup B$.