Is aleph-omega a Mahlo cardinal?

Question

So I've read up a little on Mahlo cardinals and found a definition along the lines of of $\kappa$ is Mahlo iff it is inaccessible and regular cardinals below it form a stationary set.

Using this definition, I can see why $\aleph_0$ and $\aleph_1$ are not Mahlo, in that the first is not inaccessible because it is countable and that the second is a successor cardinal (and therefore not inaccessible).

But then I get to $\aleph_\omega$... Which is inaccessible, because it is uncountable and a weak limit cardinal. The stationary subset part is tripping me up though. The cardinals below it are $\{\aleph_0, \aleph_1, ..., \aleph_n, ... | n < \omega\}$.

So I guess, is $\aleph_\omega$ Mahlo? If yes, why? And if no, why not? Further, what is the "smallest" Mahlo cardinal?

Answer 1

A Mahlo cardinal has to be regular, which $\aleph_\omega$ is not. $\aleph_\omega =\bigcup\aleph_n$, so $\operatorname{cf}(\aleph_\omega)=\aleph_0$. Every strong inaccessible $\kappa$ satisfies $\kappa=\aleph_\kappa$, but even that is not enough as the lowest $\kappa$ satisfying that has $\operatorname{cf}(\kappa)=\aleph_0$. As we can't prove even that strong inaccessibles exist, we can't say where they are in the $\aleph$ heirarchy.

Answer 2

If $k$ is an uncountable cardinal and $cf(k)<k$ then k has a closed unbounded subset of singular members:

If $\omega=cf(k)<k$ let $f:\omega \to k$ \ $\omega $ be strictly increasing with $\cup_{n\in \omega}f(n)=k.$ Let $s=\{f(n)+1: n\in \omega\}.$ (Ordinal addition.) Then $s$ is club in $k$ and does not even contain any cardinals.

If $\omega<cf(k)<k$ let $t\subset k$ \ $cf(k)$ with $|t|=cf(k)$ and $\cup t=k.$ Let $s=\{x\in k: x=\cup (x\cap t)\}.$ I will leave it as an exercise that $s$ is club in $k$ and that no member of $s$ is regular.

It is a matter of convenience of definition that $\omega$ (i.e. $\aleph_0$) is excluded from being inaccessible and from being Mahlo.