# Given a cardinal k, what's the biggest dense linear order with a dense subset of size k?

It's not hard to show that for any cardinal $\kappa$, there is no dense linear order without endpoints (DLO) of size greater than $2^{\kappa}$ that has a dense subset of size $\kappa$. But one can show that if $\mu$ is the least cardinal such that $\kappa^{\mu} > \kappa$, then there's a DLO of size $\kappa^{\mu}$ with a dense subset of size $\kappa$, so in particular there's a dense linear order of size greater than $\kappa$ with a dense subset of size $\kappa$. Thus in models of GCH, the answer to the question in the title is "as big as possible." Also from ZFC alone, the case $\kappa = \omega$ also has the answer "as big as possible": just consider $\mathbb{Q}$ and $\mathbb{R}$.

So the question is: Is it consistent that for some $\kappa$, there is no DLO of size $2^{\kappa}$ with a dense subset of size $\kappa$?

Apparently the answer's supposed to be "yes," and the relevant notion is that of a "Dedekind number," but searches for "Dedekind number" haven't yielded anything relevant, so if someone could point me to a relevent reference, that'd be great too.

Unless I made a mistake, I've shown that a DLO $(L,\leq)$ is a DLO with a dense subset of size $\kappa$ iff it is (up to isomorphism) some set $X \subset \mathcal{P}(\kappa)$ ordered by inclusion such that $(X,\subseteq)$ satisfies the following property:

$(*)\ \ \forall \alpha < \kappa$, the set $\{ x \in X : \alpha \notin x \}$ has a least upper bound, let's call it $x _{\alpha}$, in $(X,\subseteq)$

Furthermore, I've shown that given $X \subset \mathcal{P}(\kappa)$ which is a DLO under inclusion (and $\cup X = \kappa$, and $|X| > \kappa$), it can be modified to give $X' \subset \mathcal{P}(\kappa)$ which is also a DLO under inclusion, has the same size as $X$, and has a dense subset of size $\kappa$: First, obtain $X_0$ from $X$ by removing, for each $\alpha < \kappa$, the greatest lower bound in X (if it exists) of the collection $\{x \in X : \alpha \in x\}$; Next obtain $X'$ from $X_0$ by adding, for each $\alpha < \kappa$, the set $x _{\alpha} = \cup \{x \in X_0 : \alpha \notin x\}$.

So an equivalent reformulation of the question is: Is it consistent that for some $\kappa$, there is no $X \subset \mathcal{P}(\kappa)$ of size $2^{\kappa}$ which forms a DLO under inclusion?

• Out of curiosity, what's the source of the belief that "Apparently the answer's supposed to be yes,' and the relevant notion is that of a Dedekind number,'" ? – Ed Dean Dec 4 '10 at 3:26
• Thomas Scanlon recalled reading about it in someone's PhD thesis here at Berkeley. It should have occurred to me sooner, the second formulation of the question is just another way of saying that a DLO can be associated with its set of Dedekind cuts, that's why it makes sense to call them 'Dedekind numbers.' – Amit Kumar Gupta Dec 4 '10 at 10:03

The supremum of the cardinalities of linear orders with a dense subset of size $\kappa$ is called $\rm{ded}(\kappa$) in Jerry Keisler's paper "Six classes of theories" (J. Australian Math. Soc. 21 (1976) 256-266), where it (along with its $\omega$th power) is part of the answer to a fundamental question in stability theory. I don't have access to the paper at the moment, but the review on MathSciNet mentions that (Keisler mentions that) Bill Mitchell showed the consistency of $(\rm{ded}(\omega_1))^\omega<2^{\omega_1}$; the reference given there is Annals of Math. Logic 5 (1972) 21-46.
This is very late, but in 2012, Artem Chernikov, Itay Kaplan and Saharon Shelah posted on arXiv a paper titled "On non-forking spectra" ( http://arxiv.org/abs/1205.3101 ). They claim that it is consistent that $Ded(\kappa)< Ded(\kappa)^\omega$. In particular, it is consistent that $Ded(\kappa)<2^\kappa$. It is still open whether both inequalities can be made strict: $Ded(\kappa)<Ded(\kappa)^\omega<2^\kappa$.