Assuming the axiom of choice we have that successor cardinals are regular. However as one of the first examples of uses of forcing show, it is consistent relative to $\sf ZF$ that $\omega_1$ is singular. On the other hand, Schindler proved that if there are two consecutive singular cardinals, then there is an inner model with a Woodin cardinal.

There has been quite some recent research into all sort of patterns of singular successor cardinals from large cardinal assumptions.

But how much can we do without any large cardinal assumption? E.g. can we have two singular successors, perhaps with infinitely many cardinals between them, without large cardinals? Can we have a proper class of singular successors?

I don't expect the exact line of "there be large cardinals" to be known, but do we know anything except the Feferman-Levy model (or the Truss model, where he mimics the classic Solovay model construction, and shows that if we start with a singular then $\omega_1$ is singular)?


2 Answers 2


The thesis by Dimitriou "Symmetric Models, Singular Cardinal Patterns, and Indiscernibles" contains, among many other interesting results, a proof of the following theorem:

Theorem. If $V$ is a model of $ZFC$, $\kappa_0$ is a regular cardinal in $V$ and $\rho$ is an ordinal in $V$ , then there is a model of $ZF$ with a sequence of successive alternating singular and regular cardinals that starts at $\kappa_0$ and that contains $\rho-$many singular cardinals.

(you can find her thesis by searching in google)

  • $\begingroup$ Thank you! I am familiar with her thesis, but I never read all the theorems. It seems that I missed that one, or thought there were large cardinal assumptions involved. Thank you! $\endgroup$
    – Asaf Karagila
    Aug 8, 2013 at 11:38
  • 1
    $\begingroup$ That's theorem 2.10 in that thesis and it doesn't have any large cardinal assumptions. Also, this can be done for any definable class sized pattern of singular/regular cardinals, again without large cardinal assumptions as long as there are no successive singulars. $\endgroup$
    – Ioanna
    Sep 4, 2013 at 4:28

I think the following is an easy way to produce a model with two singular successors. Start with a model of $V=L$, and let $\kappa$ be a regular cardinal. First force with $Col(\kappa^{+ \omega+1},< \! \kappa^{+\omega + \omega})$ and consider the submodel $W = V(\mathcal{P}(\kappa^{+\omega+1})^*)$, where $\mathcal{P}(\kappa^{+\omega+1})^* = \bigcup \lbrace \mathcal{P}(\kappa^{+\omega+1})^{V[G_\alpha]} : \alpha < \kappa^{+\omega+\omega} \rbrace$. Then do the same with $Col(\kappa,< \! \kappa^{+\omega})$ over $W$, producing $W(\mathcal{P}(\kappa)^*)$. In this model we have $\kappa$ is regular, $\kappa^+$ is singular, $\kappa^{+2}$ is regular, and $\kappa^{+3}$ is singular. To argue that this works you use automorphisms of the Levy collapse.


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