2
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ $\aleph_\omega$ is not even inaccessible. Note that it is not a regular cardinal since it has cofinality $\aleph_0$. $\endgroup$ Commented Jul 14, 2017 at 14:43
  • 2
    $\begingroup$ Why did this get downvoted? It's a basic issue (cofinality), but a basic issue that trips practically everyone up when first learning this stuff. Seems like a good question to me. $\endgroup$ Commented Jul 14, 2017 at 17:28
  • $\begingroup$ An inaccessible, whether weak or strong, must also be regular, which $\aleph_{\omega}$ certainly is not. $\endgroup$
    – BrianO
    Commented Jul 14, 2017 at 19:16

2 Answers 2

5
$\begingroup$

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.

$\endgroup$
9
  • 1
    $\begingroup$ Your last sentence is not quite accurate. We can easily show inaccessible cardinals are $\aleph$ fixed points. $\endgroup$
    – Asaf Karagila
    Commented Jul 15, 2017 at 7:38
  • $\begingroup$ @AsafKaragila assuming that strong inaccessibles exist, let's say we take the first aleph fixed point (where $\kappa = \aleph_\kappa$). It is inaccessible for sure, but also regular right? Why wouldn't this be Mahlo? Does $cf(\kappa)$ not equal $\kappa$ in this case? $\endgroup$ Commented Jul 17, 2017 at 3:27
  • 2
    $\begingroup$ @user3684314: No, the first fixed point is not inaccessible. The first fixed point has cofinality $\omega$. And inaccessible cardinal is not just any fixed point, if $\kappa$ is inaccessible, then there is a club of order type $\kappa$ of fixed points before it. $\endgroup$
    – Asaf Karagila
    Commented Jul 17, 2017 at 4:57
  • $\begingroup$ It isn't even inaccessible. It is the limit of a countable sequence. $\endgroup$ Commented Jul 17, 2017 at 4:59
  • 3
    $\begingroup$ Yes, every inaccessible is a fixed point but not every fixed point is an inaccessible. An inaccessible has to be regular, with $cf(\kappa)=\kappa$ but the first fixed point has cofinality $\aleph_0$ $\endgroup$ Commented Jul 17, 2017 at 17:46
2
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .