# Existence of regular cardinals larger than an arbitrary cardinal in von Neumann universes without axiom of choice

It is well known that the von Neumann universes $V_{\alpha}$ is a model of ZF(C) when $\alpha$ is an inaccessible cardinal. In the following let $V$ be such a model of ZF(C). It is also well known (see corollary 5.3 of Set Theory, The Third Edition, by Thomas Jech) that assuming axiom of choice every successor aleph cardinal, $\aleph_{\beta+1}$, is a regular cardinal for any ordinal number $\beta$. This means that in ZFC (together with the necessary assumption on existence of the inaccessible cardinal necessary to construct $V$) we have:

For any cardinal number $\alpha \in V$ in the universe there is a regular cardinal $\beta \in V$ in the same universe, such that $\alpha < \beta$.

Does this statement hold in ZF? In other words, is it provable without the axiom of choice that for any given cardinal number in a universe there a strictly larger and regular cardinal in that universe?

Edit: The answer to the question above is no as pointed out in the comments by Mohammad Golshani. The question then is what other axiom, weaker than the axiom of choice allows us to prove the statement above?

• No, by Gitik's work All uncountable cardinals can be singular Aug 22, 2017 at 7:01
• I see. Then, the next question is, is it known that some axiom weaker than the axiom of choice would allow us to prove such a theorem? Aug 22, 2017 at 7:08
• Related post on math.SE: Existence of a regular uncountable $\aleph_{\alpha}$ without $\mathsf{AC}$ Aug 22, 2017 at 7:15
• The axiom WISC ncatlab.org/nlab/show/WISC implies that in ZF there are arbitrarily large regular cardinals (van den Berg, WISC is independent from ZF (2012) staff.fnwi.uva.nl/b.vandenberg3/papers/WISC.pdf, reworked as Theorem 5.1 in arXiv:1207.0959, where WISC is renamed AMC, unwisely IMHO, due to a name clash). +1 modulo the obvious correction, which I suggest you work into the main question rather than leave as a comment. Aug 22, 2017 at 7:51
• It is unclear to me whether or not you even want to postulate the existence of universe. If not, then any model without inaccessible cardinals will automatically satisfy this granted there is a proper class of regular cardinals. It also seems to me that you are really asking about something else, and you "try to formulate it in set theoretic terms". But I think it will be easier for everyone if you just write it out. Aug 23, 2017 at 7:05

Actually, if you assume that every set is inside a universe, then you can get something slightly weaker, but you do get a class of inaccessible cardinals which are regular. So this is already something.

Patterns of singular cardinals can be difficult to obtain sometimes, but we have no reason to believe that they are necessarily inconsistent.

Specially, it seems that you are asking for a model where all regular cardinals are strongly inaccessible as well, and there is a class of regular cardinals. Then every set is in a universe, but you can't find arbitrarily large cardinals in any given universe (assuming there is no inaccessible limit of inaccessible cardinals, of course, but we can always chop off the universe at the least such inaccessible for such result).

This was shown to be consistent by Arthur Apter , although I did not read the paper in detail to say whether or not each of these regular limit cardinals is also inaccessible (in the sense that $V_\kappa$ is a model of $\sf ZF_2$, or a universe as you might want to call it). It does seem to be the case, though.

And of course, it is consistent that all cardinals have countable cofinal, there there are no universes, but no uncountable regular cardinals either (See  and ).

1. Apter, Arthur W., A cardinal pattern inspired by $\text{AD}$, Math. Log. Q. 42, No.2, 211-218 (1996). ZBL0857.03030.

2. Gitik, M., All uncountable cardinals can be singular, Isr. J. Math. 35, 61-88 (1980). ZBL0439.03036.

3. Gitik, Moti, Regular cardinals in models of ZF, Trans. Am. Math. Soc. 290, 41-68 (1985). ZBL0589.03033.

This statement is FORM 159 in the book Consequences of the Axiom of Choice by Paul Howard and Jean E. Rubin. In that book (consult also the website Consequences of the Axiom of Choice Project Homepage), you can find some equivalent forms of this statement or some relationships between it and other weaker forms of the axiom of choice.

For instance, this statement is equivalent to

For every variety $W$ and set $X$, $W$ has a free algebra on $X$.

And it is a consequence of

$\mathsf{AC}_{\mathrm{wo}}$: There is a choice function on every well-orderable set.

It is known that $\mathsf{AC}_{\mathrm{wo}}$ does not imply $\mathsf{DC}_{\omega_1}$. Hence $\mathsf{AC}_{\mathrm{wo}}$ is an example from which we can prove the statement you stated and from which we cannot prove the axiom of choice.

• WISC is a lot weaker statement that also implies "For every variety W and set X, W has a free algebra on X.", see my comment to the OP. Aug 23, 2017 at 2:02
• @Roberts: Does WISC imply the axiom of countable choices or the axiom of dependent choices? Aug 23, 2017 at 5:02
• WISC does not imply anything of that sort. It might be worth mentioning that $\sf AC_{WO}$ implies $\sf DC$. Aug 23, 2017 at 7:02
• As Asaf said, WISC is very nearly the weakest choice principle there is. @AsafKaragila I wonder if the weaker principle about local smallness of the anafunctor bicategory still implies some of these set-theoretic principles. Worth thinking about when I get around to writing that paper up... Aug 23, 2017 at 7:27
• @David: We had this discussion before; yes, it's about time that you sat down to write everything. Aug 23, 2017 at 7:30