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Consider cardinality $\alpha = \aleph_0 + 2^{\aleph_0} + 2^{(2^{\aleph_0})}+\dots$, the number of terms is countable. Show that $\alpha$ is the minimal cardinality larger than the cardinalities of sets $\mathbb{N}, \mathcal{P}(\mathbb{N}),\mathcal{P}(\mathcal{P}(\mathbb{N})), \dots.$ Show that $\alpha^{\aleph_0} = 2^\alpha>\alpha$.

I don't know how to solve that and would appreciate help. Here are thoughts I had:

I know that if we add two infinite cardinalities together, the result is equal to the greater cardinality of two; for example, $\aleph_0 + 2^{\aleph_0} = 2^{\aleph_0}$. If we apply this to $\alpha$, we get

two to the power of two to the power of two to the... ...to the power of two to the power of continuum.

This $\alpha$ with vertical dots seems to me to be the cardinality of $\mathcal{P}(\mathcal{P}(\mathcal{P}\dots\mathcal{P}(\mathbb{N})\dots))$. However, since the statement must be correct, $\alpha$ should actually be larger than the cardinality of $\mathcal{P}(\mathcal{P}(\mathcal{P}\dots\mathcal{P}(\mathbb{N})\dots))$. Maybe my mistake is that adding two cardinalities together is not the same as adding a countably infinite number of cardinalities together. Or do I misunderstand something fundamentally?

Thank you.

fragileradius
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    The claim of the problem is that $\alpha$ is larger than the cardinality of any $P^n(\mathbb{N})$ for a finite $n$. You have that $\alpha=\sum_{k=0}^{\infty}|P^k(\mathbb{N})|>\sum_{k=0}^{n+1}|P^k(\mathbb{N})|=|P^{n+1}(\mathbb{N})|>|P^{n}(\mathbb{N})|$. –  Apr 22 '18 at 14:43
  • @deyore: that should be an answer. I was just about to say the same – Ross Millikan Apr 22 '18 at 14:44
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    @fragileradius: If you're interested, you're talking about the Beth Numbers, where $\beth_0 = \aleph_0$ and $\beth_{\alpha+1}=2^{\beth_\alpha}$ for successor $\alpha$ and for limit $\alpha$ it is simply the union of all Beth numbers $<\alpha$. – Isky Mathews Apr 22 '18 at 14:53
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    So, the $\alpha$ in your question $=\beth_\omega$, which is a strong limit ordinal. – Isky Mathews Apr 22 '18 at 14:57
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    This can be seen by the fact that the integers are cofinal in $\omega$, so $\beth_n$ is cofinal in $\beth_\omega$, thus for any $\kappa<\beth_\omega$ there is an $n$ such that $\kappa<\beth_n$ and $2^\kappa\leq\beth_{n+1}$. This should answer your question. – Isky Mathews Apr 22 '18 at 15:01

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