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

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.

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.

• "Cardinality" should be "cofinality" in two places. Apr 16, 2021 at 17:04
• @AlexKruckman Fixed, thank you. Apr 16, 2021 at 17:05
• A smaller example though is $3$. But yes, infinite cardinals... Apr 16, 2021 at 17:12

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.

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 $$\kappa$$.)

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