How many uncountable subsets of power set of integers are there? 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!

Here's a simple injective argument with no prerequisites (except the Schroder-Bernstein theorem at the end, since this only shows that the cardinality is at least $2^\frak{c}$, and it's clearly at most $2^\frak{c}$):
Let $N=\{X\subseteq\Bbb N\mid 0\in X\}$. $N$ is uncountable, because every subset of ${\cal P}(\Bbb N^+)$ can be injected into $N$ by adding $0$. Every $A\subseteq {\cal P}(\Bbb Z)$ such that $N\subseteq A$ is thus also uncountable, so it suffices to prove there are at least $2^\frak{c}$ subsets $N\subseteq A\subseteq {\cal P}(\Bbb Z)$.
Let $\Bbb Z^-=\Bbb Z-\Bbb N$. For each subset $B\subseteq{\cal P}(\Bbb Z^-)$ (of which there are $2^\frak{c}$ many), let $B^*=\{N\cup X\mid X\in B\}$. Since $X$ and $N$ are disjoint (each element of $N$ has $0$ and each element of $X$ does not), the function $X\in B\mapsto N\cup X$ is injective, so $\{B^*\mid B\subseteq{\cal P}(\Bbb Z^-)\}$ also has cardinality $2^\frak{c}$. But now we are done, because every $B^*$ is a distinct uncountable subset of ${\cal P}(\Bbb Z)$.
