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I am trying to calculate $\kappa^\lambda = \aleph_{\omega_1}^{\aleph_0}$.

I know that if $\kappa$ is a limit cardinal and $0 < \lambda < \mathrm{cf}(\kappa)$ then $\kappa^{\lambda} = \displaystyle \sum_{\alpha < \kappa} |\alpha|^{\lambda}$.

Hence $ \aleph_{\omega_1}^{\aleph_0} = \displaystyle \sum_{\alpha < \aleph_{\omega_1}} |\alpha|^{\aleph_0}$.

I also know that if $\kappa, \lambda$ are infinite cardinals then $(\kappa^+)^\lambda = \kappa^\lambda \cdot \kappa^+$ so that $\aleph_\alpha^{\aleph_0} = \aleph_\alpha \cdot \aleph_{\alpha-1}^{\aleph_0} = \aleph_\alpha \cdot \aleph_{\alpha-1} \cdot \aleph_{\alpha-2}^{\aleph_0} = \dots = \aleph_\alpha \cdot \aleph_{\alpha-1} \cdot \dots \cdot \aleph_{0}^{\aleph_0} = \aleph_\alpha$.

Hence $ \aleph_{\omega_1}^{\aleph_0} = \displaystyle \sum_{\alpha < \aleph_{\omega_1}} |\alpha|^{\aleph_0} = \sum_{\alpha < \aleph_{\omega_1}} |\alpha| = \sum_{\alpha < \omega_1} \aleph_{\alpha}$.

Since for infinite cardinals $\lambda \le \kappa$ we have that $\lambda + \kappa = \kappa$, $\displaystyle \sum_{\alpha < \omega_1} \aleph_{\alpha} = \sup_{\alpha < \omega_1} \aleph_{\alpha} = \aleph_{\omega_1}$.

Hence $\aleph_{\omega_1}^{\aleph_0} = \aleph_{\omega_1}$. Is this correct? This is an exercise in Just/Weese and the hint is "Assume GCH". I don't think I have used GCH so I suspect I am missing something. Thanks for your help.


Edit

I have used a different method to compute it and have reached the same result (although I still don't know whether this is correct):

By Tarski's theorem, $\aleph_{\omega_1}^{\aleph_0} = \displaystyle \sum_{\alpha < \aleph_{\omega_1}} |\alpha|^{\aleph_0}$.

Since $|\alpha|^{\aleph_0} \le \aleph_{\omega_1}$ for all $\alpha < \aleph_{\omega_1}$ we get $\displaystyle \sum_{\alpha < \aleph_{\omega_1}} |\alpha|^{\aleph_0} \leq \aleph_{\omega_1}$. Of course, $\aleph_{\omega_1} \leq \aleph_{\omega_1}^{\aleph_0}$.

Hence $\aleph_{\omega_1} = \aleph_{\omega_1}^{\aleph_0}$.

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    $\begingroup$ $\aleph_\alpha^{\aleph_0} = \aleph_\alpha \cdot \aleph_{\alpha-1}^{\aleph_0} = \aleph_\alpha \cdot \aleph_{\alpha-1} \cdot \aleph_{\alpha-2}^{\aleph_0} = \dots = \aleph_\alpha \cdot \aleph_{\alpha-1} \cdot \ldots \cdot \aleph_{0}^{\aleph_0} = \aleph_\alpha$ is possible only for $\alpha=\omega+n$ for some $n\in\omega$. (And I really don’t understand the downvote that I just cancelled.) $\endgroup$ Commented Dec 17, 2012 at 10:08
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    $\begingroup$ I would change the notation to $\alpha+1$ and $\alpha$ rather than $\alpha$ and $\alpha-1$. You can't really write $\omega+\omega-1$. $\endgroup$
    – Asaf Karagila
    Commented Dec 17, 2012 at 10:13
  • $\begingroup$ @BrianM.Scott Thank you! But I could do $$ \aleph_{\omega_1}^{\aleph_0} = \displaystyle \sum_{\alpha < \aleph_{\omega_1}} |\alpha|^{\aleph_0} = \sum_{\alpha < \aleph_0} |\alpha|^{\aleph_0} + \sum_{\alpha < \aleph_{\omega_1}} |\alpha|^{\aleph_0} = \aleph_1 + \sum_{\alpha < \aleph_{\omega_1}} |\alpha|^{\aleph_0} $$ and still arrive at the same result. (In fact, I considered this splitting up of the sum to reach what I posted above.) $\endgroup$ Commented Dec 17, 2012 at 10:14
  • $\begingroup$ @AsafKaragila You are right. $\endgroup$ Commented Dec 17, 2012 at 10:15
  • $\begingroup$ I don’t see how this decomposition helps you: it just tells you that $\omega_1\le(\omega_{\omega_1})^\omega$, which you knew already. $\endgroup$ Commented Dec 17, 2012 at 10:17

2 Answers 2

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The reason you have to use GCH is that it is consistent that $2^{\aleph_0}=\aleph_{\omega_2}$. In this case the proof fails, simply because $\aleph_0^{\aleph_0}$ is very big compared to $\aleph_{\omega_1}$.

I suppose that the authors chose to use GCH in order to ensure that no cardinal misbehaves in this aspect. It might be the case, after all, that $\aleph_\omega^{\aleph_0}$ is very big (if things go bad enough).

Furthermore what you have written about the exponentiation is true in the case where $\kappa$ is a strong limit. It might not be the case for $\aleph_{\omega_1}$, but it is always the case under GCH.


Okay so we assumed GCH, now we have that $\kappa^{\aleph_0}\leq\kappa^+$, and equality holds if and only if $\operatorname{cf}(\kappa)=\aleph_0$.

Given $\aleph_{\alpha}^{\aleph_0}$ if $\alpha$ is a limit then this simply equal to $\aleph_{\alpha+1}$; but if not then there is some $\delta$ which is a limit and $n\in\omega$ such that $\alpha=\delta+n$. Now show by induction, as you did in your post (the part where you subtract from the index) that $\aleph_\alpha^{\aleph_0}=\aleph_\alpha\cdot\ldots\cdot\aleph_\delta^{\aleph_0}=\aleph_\alpha$.

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  • $\begingroup$ So the result is correct? (I found a different way to reach the same result and added it to my question.) $\endgroup$ Commented Dec 17, 2012 at 11:41
  • $\begingroup$ @Matt: Yes. Assuming $\aleph_{\omega_1}$ is a strong limit, which follows from GCH, the result is correct. Note that you only used that in your approach, which is fine. $\endgroup$
    – Asaf Karagila
    Commented Dec 17, 2012 at 11:42
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You want $(\omega_{\omega_1})^\omega$, and you’ve established that $$(\omega_{\omega_1})^\omega=\sum_{\alpha<\omega_{\omega_1}}|\alpha|^\omega\;.\tag{1}$$ Now what if $2^\omega=\omega_{\omega_1+1}$? This is consistent with ZFC, and if it holds, then clearly the righthand side of $(1)$ is at least $\omega_{\omega_1+1}>\omega_{\omega_1}$.

Added: You can extend $(1)$:

$$(\omega_{\omega_1})^\omega=\sum_{\alpha<\omega_{\omega_1}}|\alpha|^\omega\le\sum_{\xi<\omega_1}\omega_\xi^+\cdot(\omega_\xi)^\omega\le\omega_{\omega_1}\cdot\sum_{\xi<\omega_1}(\omega_\xi)^\omega\le\omega_{\omega_1}\cdot\sup_{\xi<\omega_1}(\omega_\xi)^\omega\le(\omega_{\omega_1})^\omega\;,$$

so $$(\omega_{\omega_1})^\omega=\omega_{\omega_1}\cdot\sup_{\xi<\omega_1}(\omega_\xi)^\omega\;.$$

Under GCH you have $(\omega_{\xi+1})^\omega=\omega_{\xi+1}$ and $(\omega_\xi)^\omega=\omega_\xi^+=\omega_{\xi+1}$ when $\xi<\omega_1$ is a limit ordinal, so

$$(\omega_{\omega_1})^\omega=\omega_{\omega_1}\cdot\sup_{\xi<\omega_1}\omega_{\xi+1}=\omega_{\omega_1}\;.$$

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  • $\begingroup$ @Matt: Apparently you were going back and forth with Asaf while I was typing this, but here it is for what it’s worth. $\endgroup$ Commented Dec 17, 2012 at 11:57
  • $\begingroup$ I didn't get pinged! Thanks for replying and editing your answer! $\endgroup$ Commented Dec 17, 2012 at 19:25

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