8
$\begingroup$

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)

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$ 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. $\endgroup$ 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 $\endgroup$ 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. $\endgroup$ Aug 8, 2011 at 14:14

2 Answers 2

5
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ I am not sure what you mean. Do you mean that $Ded(\kappa)^\omega< 2^\kappa$ is consistent? $\endgroup$ 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
    Nov 22, 2011 at 15:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.