Prove that for any set A, A = $\cup$$\mathscr{P}$(A)
I am having trouble with determining what my goal should be. On the one hand, I could prove $$A \subseteq \cup \mathscr{P}(A) \land \cup \mathscr{P}(A) \subseteq A$$ Or, $$ \forall A \forall x (x \in A \iff x \in \cup \mathscr{P}(A))$$
My issue is that proving the first goal suggests that I only must assume an arbitrary element of A (and then an arbitrary element of $\cup$$\mathscr{P}(A)$) and not also an arbitrary set A. Now if the question is asking to show this relation for any set A, must I not make it clear that I assuming some arbitrary set A? Thus, the second goal seems more appropriate. However, if it may be assumed that A is arbitrary, I suppose the first goal will be fine. My issue is that I cannot tell if that assumption is appropriate.
I'm going to interpret $\cup \mathcal{P}(A)$ as the union of all the sets in $\mathcal{P}(A)$.
$\mathcal{P}(A)$ is the set of all subsets of $A$. Thus for all $x \in A$, $x \in \{x\} \in \mathcal{P}(A)$. Hence $x \in \cup \mathcal{P}(A)$.
Similarly, for all $x \in \cup \mathcal{P}(A)$, $x$ belongs to some subset $S$ of $A$. Since $x \in S \subseteq A$, then $x \in A$.