The question is to determine how many uncountable subsets of ${P(\mathbb Z)}$ are there.
I think that the answer is $2^c$.
Let $A=\{B\in P(P(\mathbb Z)):B \text{ is uncountable}\}$
$P(P(\mathbb Z))$ has $2^c$ elements, so cardinality of $A$ is at most $2^c$.
Of course, I'm having trouble with the lower bound and I'm trying to find an injective function from some set of cardinality $2^c$ into $A$.
If anybody has any idea, I'd be very grateful!
Best Answer
Try to first argue this about $\Bbb R$. How many uncountable subsets does $\Bbb R$ have?
Then note that because $\Bbb R$ and $\mathcal P(\Bbb Z)$ have the same cardinality, any bijection witnessing this would map uncountable subsets of $\Bbb R$ to uncountable subsets of $\mathcal P(\Bbb Z)$. So the answer would be the same.