Not a duplicate of How to prove $P(A) \cup P(B) \subseteq P(A \cup B) $.
This is the exercise $3.5.7$ from the book How to Prove it, by Velleman ($2^{\text{nd}}$ edition).
Prove that, for any sets $A$ and $B$, $\mathscr P(A)\cup\mathscr P(B)\subseteq \mathscr P(A\cup B)$.
Here is my proof:
Let $X$ be an arbitrary element of $\mathscr P(A)\cup\mathscr P(B)$. This means $X\in\mathscr P(A)$ or $X\in \mathscr P(B)$. Let $x$ be an arbitrary element of $X$. Now we consider two different cases.
Case $1.$ Suppose $X\in\mathscr P(A)$. So $X\subseteq A$ and since $x\in X$, $x\in A$. Thus $x\in A\cup B$.
Case $2.$ Suppose $X\in\mathscr P(B)$. So $X\subseteq B$ and since $x\in X$, $x\in B$. Thus $x\in A\cup B$.
Since the above cases are exhaustive, $x\in A\cup B$. Thus if $x\in X$ then $x\in A\cup B$. since $x$ is arbitrary, $\forall x(x\in X\rightarrow x\in A\cup B)$ and so $X\subseteq A\cup B$. Ergo $X\in\mathscr P(A\cup B)$. Therefore if $X\in \mathscr P(A)\cup\mathscr P(B)$ then $X\in \mathscr P(A\cup B)$. Since $X$ is arbitrary, $\forall X\Bigr(X\in\mathscr P(A)\cup\mathscr P(B)\rightarrow X\in \mathscr P(A\cup B)\Bigr)$ and so $\mathscr P(A)\cup\mathscr P(B)\subseteq \mathscr P(A\cup B)$. $Q.E.D.$
Is my proof valid$?$
Thanks for your attention.
Best Answer
Your proof is nice and rigorous. Although a more concise proof would be:
Let $A$ and $B$ be any sets. Let $X \in \mathscr{P}(A) \cup \mathscr{P}(B)$. Then, $X \in \mathscr{P}(A)$ or $X \in \mathscr{P}(B)$. First, suppose that $X \in \mathscr{P}(A)$. Hence, $X \subseteq A$. Then, we have that $X \subseteq A \cup B$, so $X \in \mathscr{P}(A \cup B)$. Next, suppose that $X \in \mathscr{P}(B)$. Hence, $X \subseteq B$. Then, we have that $X \subseteq A \cup B$, so $X \in \mathscr{P}(A \cup B)$. In both cases, we have that, if $X \in \mathscr{P}(A) \cup \mathscr{P}(B)$, then $X \in \mathscr{P}(A \cup B)$. Therefore $\mathscr{P}(A) \cup \mathscr{P}(B) \subseteq \mathscr{P}(A \cup B)$. $\square$