Are there countably or uncountably many infinite subsets of the positive even integers? Let $S$ be the set of all infinite subsets of $\mathbb N$ such that $S$ consists only of even numbers.
Is $S$ countable or uncountable?
I know that set $F$ of all finite subsets of $\mathbb N$ is countable but from that I am not able to deduce that $S$ is uncountable since it looks hard to find a bijection between $S$ and $P(\mathbb N)\setminus F$. Also I am not finding the way at the moment to find any bijection between $S$ and $[0,1]$ to show that $S$ is uncountable nor I can find any bijection between $S$ and $\mathbb N$ or $S$ and $\mathbb Q$ to show that it is countable. So I am thinking is there some clever way to show what is the cardinality of $S$ by avoiding bijectivity arguments?
So can you help me?
 A: Notice that by dividing by two, you get all infinite subsets of $\mathbb{N}$. Now to make a bijection from $]0,1]$ to this set, write real numbers in base two, and for each real, get the set of positions of $1$ in de binary expansion.
You have to write numbers of the form $\frac{n}{2^p}$ with infinitely many $1$ digits (they have two binary expansions, one finite, one infinite). Otherwise, the image of such a real by this application would not fall into the set of infinite sequences of integers (it would have only finitely many $1$).
A: The set of even numbers is countably infinite, and so equinumerous to $\Bbb N$ (explicit bijection given by $n\mapsto\frac12n$). With $F$ the set of finite subsets of $\Bbb N$, as you say, we can use the map $f(n)=\frac12n$ to get a bijection $g:S\to P(\Bbb N)\setminus F$, given by $$g(A)=\{f(n):n\in A\}.$$ I leave it to you to prove that it's bijective.
A: Use the Cantor-Bernstein theorem, that is (in the simplified version)
$$|A| \leq |B| \text{ and } |B| \leq |A| \text{ implies } |A| = |B|.$$
We have that $S \subset P(\mathbb{N})$, so $|S| \leq |P(\mathbb{N})|$. Moreover, we have an injection $f : P(\mathbb{N}) \to S$ given by 
$$ f(A) = \{4k \mid k \in A\} \cup \{4k+2 \mid k \in \mathbb{N}\}. $$
Hence, we have $|P(\mathbb{N})| \leq |S|$, so $|S| = |P(\mathbb{N})|$.
I hope this helps ;-)
A: Hint: If $\{x_n\mid n\in\Bbb N\}$ is a set of even integers then $\{\frac{x_n}2\mid n\in\Bbb N\}$ is a sequence of integers. Show that this operation is in fact a bijection between $S$ and the set of all infinite subsets of integers. The latter is more familiar.
