Suppose $A \subseteq \mathscr{P}(A)$. Prove that $\mathscr{P}(A) \subseteq \mathscr{P}(\mathscr{P}(A))$

Question

This problem is from Velleman's "How To Prove It". The text mentions that we are able to condense certain parts of proofs, but I am unsure of which parts to condense while maintaining clarity. Here is my attempt at the proof:

Suppose $A \subseteq \mathscr{P}(A)$. Prove that $\mathscr{P}(A) \subseteq \mathscr{P}(\mathscr{P}(A))$

Let $y$ be arbitrary and suppose $y \in \mathscr{P}(A)$. This means that $y \subseteq A$. Let $w$ be arbitrary and suppose $w \in y$. Since $y \subseteq A$ and $w \in y$, we have that $w \in A$. But then since $A \subseteq \mathscr{P}(A)$ and $w\in A$, $w \in \mathscr{P}(A)$. Therefore, if $w \in y$, then $w \in \mathscr{P}(A)$. Since $w$ was arbitrary, we have proven that $\forall w(w\in y \implies w \in \mathscr{P}(A))$. Thus, $y \subseteq \mathscr{P}(A)$ and $y \in \mathscr{P}(\mathscr{P}(A))$. Since y was arbitrary, we have proven that $\forall y (y \in \mathscr{P}(A) \implies y \in \mathscr{P}(\mathscr{P}(A)))$. This means that $\mathscr{P}(A) \subseteq \mathscr{P}(\mathscr{P}(A))$ as required. $\square$

The next questions asks

The hypothesis of the theorem proven in exercise 4 is $A \subseteq \mathscr{P}(A)$. Can you think of a set $A$ for which this hypothesis is true? Can you think of another?

If we let $S = \{\emptyset \}$, then $\mathscr{P}(S) = \{ \emptyset, \{\emptyset\}\}$. So $S \subseteq \mathscr{P}(S)$. Also, if we let $S = \emptyset$, then $\mathscr{P}(S) = \{\emptyset\}$. Again, $S \subseteq \mathscr{P}(S)$. Are these the only correct possibilities?

Answer 1

The proof is good. I think it's worth noting that this is a special (and perhaps more confusing) case of the more general rule: $$A \subseteq B \implies \mathscr{P}(A) \subseteq \mathscr{P}(B).$$ I suggest trying to prove the above implication, and compare/contrast with your proof. This may be the "condensing" that Velleman is suggesting you do.

As for your second question, I can name another: $$\{\emptyset, \{\emptyset\},\{\emptyset,\{\emptyset\}\}\}.$$ Indeed, we can form an infinite sequence of such sets, by observing that, if $A$ has this property, then $$A \cup \{A\}$$ also has this property.

Answer 2

A set $A$ for which $A \subseteq \mathcal{P}(A)$ is called a transitive set. There are many transitive sets besides the empty set and the singleton containing only the empty set; indeed, the statement being proved is exactly that the power set of a transitive set is again transitive.