Prove that $\mathcal{P}(A)\cup \mathcal{P}(B) \subseteq \mathcal{P}(A\cup B)$.

Solution 1:

Here's an easier approach (anyway, a different one). We'll use these two general facts: $$\text{If } X \subseteq Y \text{ then } \mathscr{P}(X) \subseteq \mathscr{P}(Y) \tag{1}.$$

$$\text{If } X \subseteq Z \text{ and } Y \subseteq Z \text{ then } X \cup Y \subseteq Z \tag{2}.$$

These are (very) easy to prove if you haven't already proved them in your course. ("Proving" them amounts to unpacking the definitions of $\subseteq, \cup$ and $\mathscr{P}(.)$, which reveals that they're trivial. In both cases, the converse is true too.) Using these truisms, the result follows simply:

Necessarily $A \subseteq A \cup B$, so by (1): $$\mathscr{P}(A) \subseteq \mathscr{P}(A \cup B) \text{;} \tag{a} $$ similarly, $B \subseteq A \cup B$, so again by (1): $$\mathscr{P}(B) \subseteq \mathscr{P}(A \cup B) \text{.} \tag{b} $$ From (a) and (b), using (2), we conclude: $$ \mathscr{P}(A) \cup \mathscr{P}(B) \subseteq \mathscr{P}(A \cup B) \text{.} $$