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!
 A: Yes, the answer is $2^c$. Here's a "proof:" the number of uncountable subsets of $P(\mathbb{Z})$ is $P(P(\mathbb{Z}))$ minus the number of countable subsets of $P(\mathbb{Z})$. The number of those is $(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$, which is $<2^c=\vert P(P(\mathbb{Z}))\vert$.
There are a couple steps there that need to be filled in (counting the number of countable subsets, and showing that "big $-$ small $=$ big"), but these are standard results, and if you haven't seen them before they're good exercises.
A: Start with your definition (after renaming the bound variable) and add a second definition:
$$
A=\{X\in P(P(\mathbb Z)):X \text{ is uncountable}\} \\
B=\{X\in P(P(\mathbb Z)):X \text{ is countable}\}
$$
Consider $f(X) = \overline{X}$ (the complement of $X$ relative to $P(\mathbb Z)$) as a function $f: B \rightarrow A$.  This is a function because every countable subset of $P(\mathbb Z)$ must have an uncountable complement.  Given $X \cup \overline{X} = P(\mathbb Z)$, since the RHS is uncountable, the LHS must also be uncountable, so if $X$ is countable (and in $B$), $\overline{X}$ is uncountable (and in $A$).
The complement is injective by elementary set theory.  Since an injection from $B$ to $A$ exists, $|B| \leq |A|$.  But then we have:
$$
A \cup B = P(P(\mathbb Z)) \\
A \cap B = \emptyset \\
|A| + |B| = |P(P(\mathbb Z))| \\
|A| + |A| \geq |P(P(\mathbb Z))| \\
2|A| \geq |P(P(\mathbb Z))| \\
|A| \geq |P(P(\mathbb Z))|
$$
We already know that:
$$
A \subset P(P(\mathbb Z)) \\
|A| \leq |P(P(\mathbb Z))|
$$
So the equality follows by Schröder–Bernstein.
A: 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 ${\cal P}(\Bbb Z)$.
A: 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.
A: Write $P(\Bbb Z)$ as a disjoint union of sets $A$ and $B$ each of cardinality $c$. There are $2^c$ sets $A\cup X$ with $X\subseteq B$, each of which is uncountable.
A: Rather than inject something into A, another approach is to inject the set of countable subsets into A. This proves that the cardinality of A is at least as large as the cardinality of the set of countable subsets. 
You will then have to argue that given infinite sets A and A', if $|A| \ge |A'|$ then $|A| = |A \cup A'|$ , but that follows pretty easily from transfinite math.
