# What is the cardinality of $(\aleph_\omega)^{\aleph_0}$?

I want to show that $$\aleph_0 \cdot \aleph_1 \cdot \aleph_2 \cdots = (\aleph_\omega)^{\aleph_0}$$. I am not sure why this is true. My thinking is that $$\aleph_n \cdot \aleph_{n+1} = \max\{\aleph_n,\aleph_{n+1}\} = \aleph_{n+1}$$ and so $$\aleph_0 \cdot \aleph_1 \cdot \aleph_2 \cdots = \sup\{\aleph_n \mid n \in \mathbb{N}\} = \aleph_\omega$$. Now we want to show that $$\aleph_\omega = (\aleph_\omega)^{\aleph_0}$$. I read in Jech that for cardinals $$\kappa$$ and $$\lambda$$ with $$\lambda \geq cf(\kappa)$$ we have $$\kappa^\lambda > \kappa$$. Since the cofinality of $$\aleph_\omega$$ is $$\aleph_0$$ then we conclude that $$(\aleph_\omega)^{\aleph_0} > \aleph_\omega$$ and so the result I am trying to prove does not hold. Clearly I have made a mistake somewhere but cannot figure out what it is. Any help is much appreciated.

## 1 Answer

Your mistake is assuming that infinite products are continuous, i.e. that the product is the supremum of its initial segments.

By that logic, $$\aleph_0^{\aleph_0}$$ is $$\sup\{\aleph_0^n\mid n<\omega\}=\aleph_0$$.

As to your question, what is the cardinality of $$(\aleph_\omega)^{\aleph_0}$$? Well, we only know that it is at most $$2^{\aleph_0}\cdot\aleph_{\omega_4}$$. But it could be that $$2^{\aleph_0}<\aleph_\omega$$ and $$(\aleph_\omega)^{\aleph_0}$$ is, assuming the consistency of large cardinal axioms, any $$\aleph_{\alpha+1}$$ where $$\alpha$$ is a countably infinite ordinal.

• Thank you. So I need to figure out what $\aleph_0 \cdot \aleph_1 \cdot \aleph_2 \cdots$ actually is before progressing. – user902930 Mar 20 at 9:11
• Well, it is $\aleph_\omega^{\aleph_0}$. It is clear that $\aleph_\omega^{\aleph_0}$ is an upper bound, for the lower bound, break the product into infinitely many infinite products (e.g. $\prod_{n<\omega}\aleph_{2^n}\cdot\prod_{n<\omega}\aleph_{3^n}\dots$), and observe that each of those is at least $\aleph_\omega$, and therefore a lower bound is replacing each by $\aleph_\omega$ and obtaining $\aleph_\omega^{\aleph_0}$. – Asaf Karagila Mar 20 at 9:38
• Yes, if $2^{\aleph_0}\ge \aleph_{\omega}$ then $(\aleph_\omega)^{\aleph_0}= 2^{\aleph_0}$ and if $2^{\aleph_0}<\aleph_\omega$ then $(\aleph_\omega)^{\aleph_0} <\aleph_{\omega_4}.$ My point was just that this is not the same thing as saying $(\aleph_\omega)^{\aleph_0}$ is strictly less than $2^{\aleph_0}\cdot \aleph_{\omega_4}.$ – spaceisdarkgreen Mar 22 at 16:47
• @spaceisdarkgreen: Ah, right. – Asaf Karagila Mar 22 at 18:22