Why $A \subseteq \mathscr P(A)$?

Question

I'm trying to prove the following:

Suppose $A ⊆ \mathscr P (A).$ Prove that $\mathscr P(A) ⊆ \mathscr P ( \mathscr P(A)).$

How can you suppose that $A \subseteq \mathscr P(A)$?

If my understanding is correct, $\mathscr P(A)$ denotes power set of $A$, or in other words, set that contains all the subsets of $A$

Let $A = \{1,2\}$, then $\mathscr P(A) = \{\emptyset, \{1\},\{2\}, \{1,2\}\}$

$\{1,2\}$ is not a subset of $\mathscr P(A)$, but $\{\{1,2\}\}$ is. We can only say that $A \in \mathscr P(A)$

So again, given that $\mathscr P(A)$ denotes power set of $A$, how is it possible to assume that $A \subseteq \mathscr P(A)$?

Answer 1

If it makes matters less confusing, an equivalent way of stating the problem is:

Let $A$ be a set. Prove that if $A \subseteq \mathcal{P}(A)$, then $\mathcal{P}(A) \subseteq \mathcal{P}(\mathcal{P}(A))$.

This doesn't mean that $A \subseteq \mathcal{P}(A)$ is true for all sets $A$; it just means that for the set $A$ being discussed, the assumption is made that $A \subseteq \mathcal{P}(A)$, in order to derive the conclusion that $\mathcal{P}(A) \subseteq \mathcal{P}(\mathcal{P}(A))$.

And indeed $A \subseteq \mathcal{P}(A)$ is a reasonable assumption to make. For example, it's true when $A = \varnothing$ or $A = \{ \varnothing, \{ \varnothing \} \}$ (or any other von Neumann ordinal) or $A = V_{\omega}$, or more generally any (pure) transitive set.

The fact that you found a counterexample to the hypothesis just means that the conclusion of the theorem doesn't apply to that set.

Answer 2

First consider $A_0 = \mathbb{N}$ and $A_n = \mathscr{P}(A_{n-1})$. Then consider $B = \bigcup_{n \in \mathbb{N}}{A_n} \dots$

For your question, suppose $x \in \mathscr{P}(A)$. Then $x \subset A$ . But then $x \subset A \subset \mathscr{P}(A) $ and indeed $x \in \mathscr{P}(\mathscr{P}(A)) $.