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.
$\begingroup$
$\endgroup$
Add a comment
|
$\begingroup$
$\endgroup$
4
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.
answered Mar 20 at 9:07
-
$\begingroup$ 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
-
