4
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Thank you. So I need to figure out what $\aleph_0 \cdot \aleph_1 \cdot \aleph_2 \cdots$ actually is before progressing. $\endgroup$ – user902930 Mar 20 at 9:11
  • 2
    $\begingroup$ 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 $\aleph_\omega^{\aleph_0}$. $\endgroup$ – Asaf Karagila Mar 20 at 9:38
  • $\begingroup$ 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 $2^{\aleph_0}\cdot \aleph_{\omega_4}.$ $\endgroup$ – spaceisdarkgreen Mar 22 at 16:47
  • $\begingroup$ @spaceisdarkgreen: Ah, right. $\endgroup$ – Asaf Karagila Mar 22 at 18:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.