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.



Why does $\aleph_{\omega}$ have more than $\aleph_{\omega}$ countable subsets?
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.) 


