4
$\begingroup$

Assuming the GCH, each time you take the powerset of an aleph number, the subscript increases by one. So there is $\aleph_0, \aleph_1, \aleph_2\ldots \aleph_\omega\ldots \aleph_{\epsilon_0}\ldots$ and so on.

This may be a dumb question, but how large of an ordinal has to be present in the subscript below the aleph in order for the resulting aleph number to make it onto the large cardinal hierarchy? For example, how does aleph_church-kleene compare to the smallest inaccessible cardinal?

$\endgroup$
1
  • 5
    $\begingroup$ The comment-sized answer: virtually all large cardinals are fixed points of the aleph function; that is, they're 'numbers' $\kappa$ for which $\kappa = \aleph_\kappa$. $\endgroup$
    – Steven Stadnicki
    Jul 15, 2014 at 18:37

1 Answer 1

Reset to default
7
$\begingroup$

The term "large cardinal" doesn't have an agreed up, concrete definition. When we say that $\kappa$ is a large cardinal we might mean that $\kappa$ is inaccessible and has additional properties; or we might mean that it has properties which imply the consistency of large cardinals.

Some people would refer to the former as large cardinals, and the latter as "large cardinal properties".

If we agree with that convention, then every large cardinal is a weakly inaccessible cardinal at least, which means that it is an $\aleph$-fixed point. Namely $\kappa=\aleph_\kappa$ (and in fact, it is the limit of fixed points, and limits of limits of fixed points and so on).

As to your second question, $\aleph_{\omega_1^{CK}}$ has only countably many $\aleph$'s below it. That's just a tiny fraction of the cardinals you will find below the first $\aleph$ fixed point. If $\kappa$ is inaccessible then it is the $\kappa$-th fixed point, meaning it is unfathomably larger.

Let me add a side remark, $\sf GCH$ has little to do with this. A strongly inaccessible cardinal is also a $\beth$ fixed point (with similar properties as discussed above), where $\beth$ numbers are defined using the power set operation, rather than successor cardinals.

Added:

It may be a bit misleading to think that being a large cardinal is all about having a large index as an $\aleph$ number. Being a large cardinal means having some combinatorial properties which allows you to prove more. Successor of large cardinals are not "large" themselves, because in $\sf ZFC$ successor cardinals are do not posses the basic requirement of a large cardinal, being a regular limit cardinal.

$\endgroup$
28
  • 2
    $\begingroup$ @Steven, while that is true, if $\kappa$ is inaccessible then it is the $\kappa$-th fixed point, so in particular there are plenty of fixed points below the first inaccessible, none of which are regular. $\endgroup$
    – Asaf Karagila ♦
    Jul 15, 2014 at 18:43
  • 1
    $\begingroup$ Regarding "That's just a tiny fraction of the cardinals you will find below the first $\aleph$ fixed point", it's also (this for the OP, of course) just a tiny fraction of the cardinals you will find below ${\aleph}_{{\omega}_{1}}.$ $\endgroup$
    – Dave L. Renfro
    Jul 15, 2014 at 18:54
  • 1
    $\begingroup$ @PanMrož: No. An inaccessible cardinal is a cardinal, which is by definition is a set. The inaccessibility simply means that we cannot access it from "below" in any non-predicative sense (so there are no shorter sequences which are cofinal below it, and power sets operations are not helpful either). As I wrote above, explicitly, we know what is the $\aleph$-index of an inaccessible cardinal $\kappa$: it's $\kappa$ itself. $\endgroup$
    – Asaf Karagila ♦
    Aug 9, 2019 at 14:42
  • 1
    $\begingroup$ @PanMrož Yes. Every cardinal has a successor. $\endgroup$
    – Asaf Karagila ♦
    Aug 9, 2019 at 14:47
  • 1
    $\begingroup$ @PanMrož: Yes. Just like $\omega$ cannot be reached within finite means from finite sets. $\endgroup$
    – Asaf Karagila ♦
    Aug 9, 2019 at 14:55

You must log in to answer this question.

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