Is $Ded(\kappa)

Is $Ded(\kappa)
2011-08-04T17:12:05Z

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 (http://mathoverflow.net/questions/48231) that asks for the consistency of $Ded(\kappa)<2^\kappa$ (answer is positive)

Answer by mmm for Is $Ded(\kappa)
2011-11-21T15:45:32Z

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.

Answer by Ioannis Souldatos for Is $Ded(\kappa)
2012-05-15T12:45:55Z

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.