Question

I believe $\aleph_{\omega}$ has more than $\aleph_{\omega}$ countable subsets but I do not see the proof. I fear it is obvious, but not to me today.

Answer 1

Let $\{A_\alpha:\alpha<\omega_\omega\}$ be all countable subsets of $\omega_\omega$. We build one that is not among them, giving the contradiction. Pick $x_0\in\omega_1$ wich is not in $\bigcup\{A_\alpha:\alpha<\omega\}$. Then choose $\omega_1\leq x_1<\omega_2$ which is not in $\bigcup\{A_\alpha:\alpha<\omega_1\}$, etc. Eventually we get a countable set $\{x_0,x_1,\dots\}$ different from each $A_\alpha$. (This is essentially the argument for proving Konig's inequality.)