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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.

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Yes, this follows from König's Theorem - – François G. Dorais Aug 1 '12 at 5:35
up vote 14 down vote accepted

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.)

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Thanks for both answers. Since you have shown it to me, Konig's inequality will be obvious to me from now on. – Colin McLarty Aug 1 '12 at 13:16

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