is $ded^{*}(\kappa)< ded(\kappa)$ consistent?

Question

Hello,

I wonder if anyone knows this.

Definition:

1. $ded\left(\lambda\right)$ is the supremum of all sizes of linear orders with a dense subset of size $\lambda$.

2. $ded^{*}\left(\lambda\right)$ is the supremum of all sizes of branches of the same hight of set theoretic trees with $\lambda$ many nodes.

(i.e. a partial order such that for every $x$, the set of elements smaller than $x$ is well-ordered)

It is known that $\lambda< ded^{*}\left(\lambda\right)\leq ded\left(\lambda\right)\leq2^{\lambda}$.

Can it be that $ded^{*}\left(\lambda\right)< ded\left(\lambda\right)$?

Answer 1

EDIT: I added some more comments at the end.

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I am not sure if this exactly what you want, but here is a theorem that might help:

Definition Assume $\mu\le\kappa\le\lambda$ and $\mu$ is a regular cardinal. (1) Let $D(\kappa,\lambda)$ iff there is a linear order of size $\lambda$ and a dense set of size $\kappa$. (2) Let $D(\kappa,\lambda,\mu)$ iff there is a linear order of size $\lambda$, character $\mu$ that has a dense subset of size $\kappa$.

Then

Theorem (1) $D(\kappa,\lambda,\mu)$ iff there is a tree of height $\mu$, cardinality $\le\kappa$ and at least $\lambda$ branches of length $\mu$. (2) $D(\kappa,\lambda)$ iff there is a tree of height $\le\kappa$, cardinality $\le\kappa$ and at least $\lambda$ branches.

This is from J.E. Baumgartner, Almost-disjoint sets the dense set problem and the partition calculus, Annals of Mathematical Logic, Volume 9, Issue 4, May 1976, Pages 401-439

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Let me combine and expand on the definitions: $$ded(λ)=sup_κD(λ,κ)$$ $$ded(λ,μ)=sup_κD(λ,κ,μ)$$ $$ded^*(\lambda,\mu)=\sup_\kappa \mbox{ There is a tree T with height $\mu$ and $\lambda$ many nodes}$$

$$\mbox{and the set of $\mu$−branches through T has size $\kappa$}$$

$$ded^∗(λ)=\sup_μded^∗(λ,μ).$$ I am not sure if this is your intended definition or not. Assume so.

Then by Baumgartner's theorem, part (1): $$ded^*(\lambda,\mu)=\sup_\kappa D(\lambda, \kappa,\mu)=ded(\lambda,\mu)$$ and $$ded^*(λ)=\sup_μ ded(λ,μ).$$

So the question becomes whether or not $$ded^*(\lambda)=\sup_μ ded(λ,μ)\le ded(\lambda).$$

I hope I am not misinterpreting your definition. I have to give it some thought for this last inequality, but it is not obvious to me why it should hold. On the other hand, I don't have a counterexample either.