What is the cardinality of the set of all subsets of $P(\mathbb N)$?

Question

Let $A$ be the set of all subsets of $P(\mathbb N)$. For example the set $\{\emptyset, \{1,2,3\}, \mathbb N -\{1,2,4\}, \mathbb N\}$ is an element of $A$. What is the cardinality of $A$?

I'm inclined to think it's $c$ but I'm not sure. Isn't $A$ the same as $P(\mathbb N)$? We know that $|P(\mathbb N)|= c$.

Answer 1

I think when you wrote $\Bbb C$ you meant $c$. Because $c$ is a cardinal, while $\Bbb C$ is not.

Anyway, the cardinality of $P(P(\Bbb N))$ is $2^c=2^{2^{\aleph_0}}$. Not that that really says much, the definition of that notation is essentially just "the cardinality of $P(P(\Bbb N))$". But no, $2^c\ne c$; in fact $2^c>c$.

Answer 2

By your definition, $A$ is just $P(P(\mathbb{N}))$. Recall Cantor's Theorem, which states that for any set $S$, $|P(S)| > S$. Thus the cardinality of $A$ cannot be $c$, since $|P(\mathbb{N})| = c$.