Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
4  
Yes, this follows from König's Theorem - en.wikipedia.org/wiki/K%C3%B6nig%27s_theorem_%28set_theory%29 –  François G. Dorais Aug 1 '12 at 5:35
add comment

1 Answer

up vote 13 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.)

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.