5
$\begingroup$

So, I know that the existence of a scale (that is, a linear cofinal set in $({}^\omega \omega , \leq^\ast )$, where $\leq^\ast $ is eventual domination, is equivalent to $\mathfrak{b} = \mathfrak{d}$, and that we can obtain the latter through Martin's Axiom (see Hechler forcing).

My question is, what about cofinal trees of a certain height and width? A scale is a tree of width $1$ and height $\mathfrak{d}$. If $\mathfrak{b} < \mathfrak{d}$, can I at least force the existence of a cofinal tree of width $< \mathfrak{d}$?

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ Nice question! One trivial observation: cofinal trees of width $<\frak{b}$ are equivalent to a scale, since you can replace a level by a function dominating it. $\endgroup$
    – Joel David Hamkins
    Feb 1 at 13:13
  • $\begingroup$ Do you really want trees, or do you want to go into a more PCF-y direction and consider directed sets instead? $\endgroup$
    – Asaf Karagila ♦
    Feb 1 at 13:26
  • 2
    $\begingroup$ I'm less sure of that comment above now, since perhaps that replacing process gets in trouble at limit stages. $\endgroup$
    – Joel David Hamkins
    Feb 1 at 13:30

1 Answer 1

Reset to default
5
$\begingroup$

Hechler proved that you can force the existence of a cofinal subset of $\omega^\omega$ having any shape you like (subject to one or two necessary restrictions):

Theorem: (Hechler) Suppose that $(P,\leq)$ is a partially ordered set with the property that every countable subset of $P$ has a strict upper bound in $P$. Then there is a ccc forcing notion which adds a cofinal subset of $(\omega^\omega,\leq^*)$ that is order isomorphic to $(P,\leq)$.

The result can be found in:

Hechler, Stephen H., On the existence of certain cofinal subsets of ${}^\omega\omega$, Axiom. Set Theor., Proc. Symp. Los Angeles 1967, 155-173 (1974). ZBL0326.02047.

EDIT:

In my original answer to this question, I said that Hechler's result implies that, yes, any reasonable-looking tree can be the order type of a cofinal subset of $\omega^\omega$. K. P. Hart pointed out to me in an email that this is not in fact the case.

The poset $(\omega^\omega,\leq^*)$ is (countably) directed, which means that it cannot contain two incompatible elements. The same is true for any cofinal subset of this order. Thus the only trees that are allowed as the order type of cofinal subsets of $(\omega^\omega,\leq^*)$ are the unbranching trees with uncountable cofinality: i.e., the type of $\mathfrak{d}$ when $\mathfrak{b} = \mathfrak{d}$. (Or at least the type of some ordinal $\kappa$ with cofinality $\mathfrak{d}$, which can be attained by packing extra functions into a scale in certain models of $\mathfrak{c} > \mathfrak{d} = \mathfrak{b}$.)

But that being said, if $(T,\leq)$ is any tree-like poset such that every branch has uncountable cofinality, then $T$ can be used to generated a countably directed poset $(P,\leq)$, namely the poset of downward-closed subsets of $T$ that do not contain a branch, ordered by inclusion. (I'm envisioning an upward-growing tree when I say "downward closed" here -- feel free to reverse it according to your own preferences.) And this poset can, in a Hechler-style forcing extension, be isomorphic to the order type of a cofinal subset of $(\omega^\omega,\leq^*)$.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Would $\mathfrak{b} < \mathfrak{d}$ possibly hold in such models? Is there any relation between the values of the cardinal characteristics and the cofinal posets that you can have? $\endgroup$
    – Matteo Casarosa
    Feb 1 at 13:45
  • $\begingroup$ @MatteoCasarosa: Yes, this is certainly possible in such models. In fact, the values of $\mathfrak{b}$ and $\mathfrak{d}$ are determined by the poset $(P,\leq)$: $\mathfrak{d}$ is the cofinality of the poset (the smallest size of a cofinal family), and $\mathfrak{b}$ is the additivity of the poset (the smallest size a subset with no upper bound). $\endgroup$
    – Will Brian
    Feb 1 at 13:48
  • $\begingroup$ Does $\mathfrak{c}$ depend in some way on the poset P that I've used? In particular, can it be $\mathfrak{b} < \mathfrak{d} = \mathfrak{c}$? For which posets? $\endgroup$
    – Matteo Casarosa
    Feb 1 at 18:26
  • $\begingroup$ In general I don't think $\mathfrak{c}$ can be determined from $P$. For example, you can have a cofinal set of order type $\omega_1$ whenever $\mathfrak{b} = \mathfrak{d} = \aleph_1$, regardless of the value of $\mathfrak{c}$. But that being said, I think Hechler's paper addresses this too, and allows you to get $\mathfrak{c}$ equal to any value that is compatible with a given poset $P$. $\endgroup$
    – Will Brian
    Feb 1 at 18:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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