Why are there $2^{\aleph_0}$ injections from $\omega$ to $\omega_1?$

Question

I have to prove that there are $2^{\aleph_0}$ injections from $\omega$ to $\omega_1.$ I can see that there is a bijection between this set and the set of pairs: (permutation of $\omega$, infinitely countable subset of $\omega_1$), which means that the standard "finite" formula works, and there are $$\aleph_0!\cdot{\aleph_1 \choose\aleph_0}$$

of such injections. I know that $\aleph_0!=2^{\aleph_0},$ but I can't see how the other factor is $2^{\aleph_0}$ too.

Answer 1

Since every countable subset of $\aleph_1$ is bounded, you can get a handle on it as follows:

$${\aleph_1 \choose\aleph_0} = \bigcup_{\alpha < \omega_1}{\alpha \choose\aleph_0}$$