Are all cardinals of the form $\aleph_0$ or $2^\alpha$?

Question

Assuming the Generalized Continuum Hypothesis (GCH), this is, for any cardinal $\lambda$ there is no cardinal $\kappa$ such that $\lambda<\kappa<2^\lambda$.

Is it true that any cardinal is of the form of $\aleph_0$ or $2^\lambda$ for some cardinal $\lambda$?

This is what I have been able to show:

An equivalent statement of the GCH, is that for any ordinal $\alpha$, $\aleph_{\alpha+1}=2^{\aleph_\alpha}$.

Another way to state this is that $\aleph_\alpha=\beth_\alpha$ for any ordinal $\alpha$. Where $\beth_\alpha$ the the beth numbers and are defined as $\beth_0:=\aleph_0$, $\beth_{\alpha+1}:=2^{\beth_{\alpha}}$.

Now, it is stated that by Zermelo’s theorem, every cardinal is an aleph. So by the GCH, every cardinal is a beth number and thus any cardinal is either $\beth_0=\aleph_0$ or $\beth_\alpha$.

Here we have two options: If $\alpha$ is a successor then clearly $\beth_\alpha=2^{\beth_{\alpha-1}}$.

But what if $\alpha$ is a limit ordinal? Is then $\beth_\alpha=2^{\lambda}$ for some cardinal $\lambda$. I don't think so. But I don't know how to prove this.

Answer 1

The answer is no, regardless of the truth of the generalized continuum hypothesis. Indeed, one can show that $\aleph_\omega$, or any other infinite cardinal of countable cofinality, cannot be equal to $2^\lambda$ for any $\lambda$. This is because König's theorem implies that $2^\lambda$ must have cofinality strictly greater than $\lambda$, so in particular cannot be countable.

Answer 2

No, if the GCH holds then no limit cardinal is of the form $2^\kappa.$ We know what $2^\kappa$ is for every $\kappa.$ It is $\kappa^+,$ which is a successor cardinal.

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Wojowu points out in their answer that cardinals with countable cofinality always have this property, even when we don't assume GCH. Another example is any strong limit cardinal, i.e. a cardinal of the form $\beth_{\alpha}$ for a limit ordinal $\alpha.$ An infinite cardinal $\kappa$ is a strong limit iff for all $\lambda < \kappa,$ $2^\lambda < \kappa.$ So it can't be of the form $2^\lambda$ since $2^\lambda < \kappa$ for any $\lambda < \kappa$ and $2^\lambda > \kappa$ for any $\lambda\ge \kappa.$

Notice this is just a ZFC-only version of the result quoted for GCH since under GCH, every limit cardinal is a strong limit. It also shows that there are examples of all cofinalities (since if $\kappa$ is some infinite cardinal $\beth_\kappa$ is a strong limit cardinal with the same cofinality as $\kappa$.)

Answer 3

Under GCH, $2^\kappa$ is never a limit cardinal, since it is always the successor of $\kappa$.