Cardinal arithmetic gone wrong?

Question

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.

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

Answer 1

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.

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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 $\aleph_\alpha^{\aleph_0}=\aleph_\alpha\cdot\ldots\cdot\aleph_\delta^{\aleph_0}=\aleph_\alpha$.

Answer 2

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

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