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Hello,

I want to ask if anyone can tells us what is known (consistently) about $Ded(\kappa)$, $\kappa$ an infinite cardinal.

Definition If there is a dense linear order w/o endpoints of size $\lambda$ with a dense subset of size $\kappa$ then we write $D(\kappa,\lambda)$. Then $Ded(\kappa)=\sup_\lambda \{D(\kappa,\lambda)\}$.

Known theorems: (1) $Ded(\kappa)\le 2^\kappa$ and under GCH $Ded(\kappa)$ is always equal to $2^\kappa$. (2) If $\mu$ is the least cardinal such that $\kappa^\mu>\kappa$, then $D(\kappa,\kappa^\mu)$ holds, which implies in particular that $Ded(\kappa)\ge \kappa^\mu$.

Questions (1) Can we prove that $Ded(\kappa)< Ded(\kappa)^\omega$ is consistent? (2) If $\mu$ a cardinal between $\omega$ and $\kappa$, can we prove that $Ded(\kappa)=\kappa^\mu$ is consistent?

Note 1: Following Keisler $Ded(\kappa)$, $Ded(\kappa)^\omega$ are two of the six possible "stability functions", the other four being $\kappa$, $\kappa+2^\omega$, $\kappa^\omega$ and $2^\kappa$. Stability functions give us the number of types of a theory $T$ over models of power $\kappa$. For more on this consult The Stability Function of a Theory by H. Jerome Keisler, The Journal of Symbolic Logic, Vol. 43, No. 3 (Sep., 1978), pp. 481-486

Note 2: There is a similar question posted on MathOverflow (Given a cardinal k, what's the biggest dense linear order with a dense subset of size k?) that asks for the consistency of $Ded(\kappa)<2^\kappa$ (answer is positive)

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  • $\begingroup$ My answer to (1) was fatally flawed, so please un-accept my answer, which I think should be deleted. Andy Voellmer's comment about (2) is still correct though: you can force $2^\omega=2^\kappa=\kappa^+$ (with GCH in the ground model) to show that (2) is consistently true. $\endgroup$
    – David Milovich
    Commented Aug 5, 2011 at 6:25
  • $\begingroup$ Perhaps you could add to the list of known things that if $\kappa=\kappa^{\omega}$, then $Ded(\kappa)=Ded(\kappa)^{\omega}$. This is an observation of Kunen mentioned in the paper of Keisler. $\endgroup$
    – Artem Chernikov
    Commented Aug 6, 2011 at 1:23
  • $\begingroup$ Good point. $Ded(\kappa)<Ded(\kappa)^\omega$ can be consistent only if $\kappa<\kappa^\omega<Ded(\kappa)<2^\kappa$. This is remarked by Keisler in SIX CLASSES OF THEORIES, J. Austral. Math. Soc. 21 (Series A) (1976), 257-266. He attributes the proof to Kunen, but I didn't find a reference. (To obtain the article follow the link: journals.cambridge.org/… ) This is where the question $\endgroup$
    – Ioannis Souldatos
    Commented Aug 8, 2011 at 14:13
  • $\begingroup$ ...The paper SIX CLASSES OF THEORIES is where the question "Is $Ded(\kappa)<Ded(\kappa)^\omega$ consistent?" appears. $\endgroup$
    – Ioannis Souldatos
    Commented Aug 8, 2011 at 14:14

2 Answers 2

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I just found the following paper on arXiv: "On non-forking spectra" by Artem Chernikov, Itay Kaplan and Saharon Shelah ( http://arxiv.org/abs/1205.3101 ). They claim that it is consistent that $Ded(\kappa)< Ded(\kappa)^\omega$, therefore answering this question positively.

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I think a recent result of Itay Kaplan claims that this 6th class does indeed exist... perhaps you should ask him directly, i am not sure i have a link.

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  • $\begingroup$ I am not sure what you mean. Do you mean that $Ded(\kappa)^\omega< 2^\kappa$ is consistent? $\endgroup$
    – Ioannis Souldatos
    Commented Nov 22, 2011 at 14:41
  • $\begingroup$ sorry, i should have said 'all 6 classes are different'. So it is Ded(k)<Ded(k)^w is consistent. though i may be mistaken, you should ask him. $\endgroup$
    – mmm
    Commented Nov 22, 2011 at 15:03

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