Consider the set $ \{ X \subseteq \omega_{1} \ | \text{ such that } |X| = \aleph_{0} \} $
I know $\omega$ is in this set. But then I thought about it and realized that {2,3,4,... } was also in this set, and it has cardinality $ \omega $ . The answer turns out to be $2^{\aleph_{0}}$. Is there a bijection from the set in question to $P(\omega) $ or is there a straight forward counting argument.
edit: Changed $2^{\omega} $ to $ 2^{\aleph_{0}} $.
The quickest way to prove this is the computation $$ 2^{\aleph_0}\leq |X|\leq{\aleph_1}^{\aleph_0}\leq(2^{\aleph_0})^{\aleph_0} =2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0}. $$ An alternative approach is to first observe that a countable subset of $\omega_1$ cannot be cofinal in $\omega_1$, so your set $X$ is the union, over all infiniteordinals $\alpha<\omega_1$, of the sets $X_\alpha$ of infinite subsets of $\alpha$. Since each such $\alpha$ is countably infinite, each $X_\alpha$ has the cardinality of the continuum. So $X$ is the union of $\aleph_1$ sets of size $2^{\aleph_0}$, and therefore it also has size $2^{\aleph_0}$.