Set of all infinite subsets of natural numbers is equipotent with the power set of natural numbers I am trying to prove the following statement:
Set of all infinite subsets of natural numbers is equipotent with the power set of natutal numbers.
My thought is
Let the set of all infinite subsets of natural numbers be $S$.
We need to show that there is a bijection from $S$ to the power set of natural numbers. But I have no clue on how to start. Any help would be appreciated.
 A: A different approach is to say that $S$ injects into $P(\Bbb N)$ because it is a subset.  To inject $P(\Bbb N)$ into $S$, take any subset of $\Bbb N$ and double its members to get a set of even numbers.  Take the union of that set with all the odd numbers.  This is an infinite set of naturals, so is a member of $S$.  We have injections both ways, so can use the Schroeder-Bernstein theorem to show there is a bijection.
A: Clearly it suffices to give an injection from the subsets of $\mathbb N$ to the infinite subsets of $\mathbb Z$.
To do this just send $A$ to $A\cup \{-1,-2,\dots\}$
A: Form the disjoint union:
$$A:=\bigcup_{n=1}^{\infty} \mathbb N^n$$
Show that $B:=\{T\subset \mathbb N: T \text{ is finite}\}$ injects into $A$. 
Note that $A$ is the countable union of countable sets, so it is countable.
We have:
$$2^{\mathbb N} = S \cup B$$
Hence, since $B$ is countable, we get $|S| = |2^{\mathbb N}|$.
A: For every finite $A$ that is a subset of $N$, notice that $N/A$ is infinite. However, NOT for all infinite $B$ that is a subset of $N$, $N/B$ is finite. Therefore, the set infinite subsets of $N$ is denser than the set of finite subsets of $N$. Then, the set of finite subsets of $N$ must be countably infinite and the set of infinite subsets of $N$ must be uncountably infinite but cannot be any denser than their union (or, namely, the power set of $N$). So, the set of infinite subsets of $N$ is equipotent with the power set of $N$.
