Set theory (Jech) $\text{p.}\;27:$

It is an open problem whether one can prove without the axiom of choice that there exists a regular uncountable $\aleph_{\alpha}\;($the informed guess is that one cannot$)$.

My question is on the remark in brackets, in particular the use of the word informed: could someone briefly expand on the reasons for believing that this is likely to be impossible?

  • 1
    $\begingroup$ Which edition of Set Theory is this from? The first one, from the 1970s? $\endgroup$ – Asaf Karagila Apr 3 '14 at 17:45

This is not an open problem anymore.

In 1980 Moti Gitik published a solution. Starting with a proper class of strongly compact cardinals, one can construct a model of $\sf ZF$ in which every limit ordinal has countable cofinality. Therefore there are no uncountable regular $\aleph$'s.

M. Gitik, All uncountable cardinals can be singular, Israel J. Math. 35 (1980), no. 1-2, 61--88.

The informed guess comes from understanding the Feferman-Levy model in which $\omega_1$ is a countable union of countable ordinals (i.e. has a countable cofinality).

While one can show without using the axiom of choice that for an infinite ordinal $\alpha$ it holds that $|\alpha|=|\alpha\times\alpha|$, to use that to show that $\omega_1$ is regular, one has to take a countable list of countable ordinals and prove their union is countable. We do that by choosing enumerations for these ordinals, and then the union is an enumerated union, which is indeed countable.

But as the Feferman-Levy model showed, one has to use the axiom of choice to choose these enumerations. The result translates to larger cardinals as well.

And so Gitik's work shows that indeed assuming large cardinals this is impossible to boot. One might point out that large cardinals are necessary. It was [recently] shown that if $\kappa$ and $\kappa^+$ are singular then there is an inner model with a Woodin cardinal. So to have all uncountable $\aleph$ singular, one would have to use at least a proper class of Woodin cardinals, and we expect this to be even more than just that.

Ralf-Dieter Schindler, Successive weakly compact or singular cardinals, J. Symbolic Logic 64 (1999), no. 1, 139--146.

  • $\begingroup$ "So to have all uncountable $\aleph$ singular, one would have to use at least a proper class of Woodin cardinals". This is very much expected, but it is not clear (or even known). The best known bound is in The strength of choiceless patterns of singular and weakly compact cardinals by Schindler and Busche (2009), it proves that we need infinitely many Woodin cardinals, but it is still significantly short of your claim. $\endgroup$ – Andrés E. Caicedo Apr 3 '14 at 19:41
  • $\begingroup$ Andres, do we know how to get two successive singulars from one Woodin cardinal? $\endgroup$ – Asaf Karagila Apr 3 '14 at 19:49
  • $\begingroup$ Hmm... Starting from $\mathsf{AD}$ we can force all uncountable cardinals below $\Theta$ to be singular. This is a result of Apter. I do not know whether anything more precise is known when only two successive cardinals are targeted. $\endgroup$ – Andrés E. Caicedo Apr 3 '14 at 20:04
  • $\begingroup$ Yeah, I know about Apter's result on the topic. I have a feeling that every time this topic comes up, we have the same comment discussion. I ought to know it by now. ;-) $\endgroup$ – Asaf Karagila Apr 3 '14 at 20:10
  • $\begingroup$ @Andres: In which of Apter's papers can the result you mention be found? $\endgroup$ – Thomas Benjamin May 6 '14 at 2:59

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