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I'm working through the book Discovering Modern Set Theory by Just and Weese, and this question comes right after this theorem:

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Here's what I've worked out so far:

I believe the cofinality of $\aleph_{\omega_1}$ is $\omega_1$, so I'm allowed to use the theorem, but I'm not sure how to then proceed calculating that sum. I can divide the sum up into some cardinals I can calculate: if $|\alpha|$ is countable, then $|\alpha|^{\aleph_0} = 2^{\aleph_0}$; and if $|\alpha|$ is a successor cardinal $\aleph_{a + 1}$ then I can use the Hausdorff formula $\aleph_{a + 1}^{\aleph_0} = \aleph_a^{\aleph_0}\aleph_{a+1}$ and then using GCH $\aleph_a^{\aleph_0}\aleph_{a+1} = \aleph_a^{\aleph_0}2^{\aleph_a}$. I'm stil left with all the limit cardinals not calculated, and I don't know what the sum of all this would end up being anyways.

Thanks! :)

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1 Answer 1

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If GCH holds, then for $|\alpha|>\lambda,$ we have $|\alpha|^\lambda \le 2^{|\alpha|}= |\alpha|^+,$ so if $\kappa$ is a limit cardinal and $\lambda < \kappa$, then $$\sum_{\alpha < \kappa }|\alpha|^\lambda = \kappa \cdot \sup_{\alpha < \kappa}|\alpha|^\lambda = \kappa \cdot \sup_{\alpha < \kappa}|\alpha|^+=\kappa.$$

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  • $\begingroup$ Would you mind detailing that last step? $\endgroup$
    – violeta
    Commented Jun 22 at 22:39
  • $\begingroup$ @violeta Sure, edited $\endgroup$ Commented Jun 22 at 22:43
  • $\begingroup$ Thank you! I'm still not 100%: shouldn't that first equality be a $\leq$? Is it true that whenever we have an increasing sequence of cardinals indexed by $\kappa$ then their sum is $\kappa$ times the limit of the sequence? $\endgroup$
    – violeta
    Commented Jun 22 at 22:54
  • $\begingroup$ @violeta $\le$ is all we need here, but yes it's true that $\sum_{i\in I}\kappa_i = |I|\cdot \sup_i \kappa_i$ when $\kappa_i > 0$ and the RHS is infinite (increasing is not needed). $\endgroup$ Commented Jun 22 at 23:25
  • $\begingroup$ Makes sense, thank you a lot :) $\endgroup$
    – violeta
    Commented Jun 23 at 0:25

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