The mod finite order on ${}^\omega \omega$ is defined as $f \leq^\ast g$ if and only if $f(n) \leq g(n)$ except for finitely many $n \in \omega$. My question is: can we always extract a cofinal well-founded subset $D \subseteq {}^\omega \omega$?

If that is not always a possibility in ZFC, how can we force that situation? Can we obtain it besides the case where $\mathfrak{b} = \mathfrak{d}$, which is equivalent to the existence of a scale (i.e. a $\leq^\ast$-linearly ordered, $\leq^\ast$-cofinal subset)?

I would also like to know about the generalization where we put the same order on ${}^\lambda \kappa$, where $\lambda$ and $\kappa$ are any two cardinals.


1 Answer 1


The answer is yes, because indeed every partial order admits a well founded cofinal (i.e. dominating) subset. There is no need to consider cardinal characteristics.

Theorem. Every partial order admits a well-founded cofinal subset.

Proof. Build it in a recursive process. At each stage $\alpha$, if there is some element in the partial order not yet dominated, then we can add a dominating element to the subset, such as that element itself. More specifically, let us simply well-order the elements of the original order, and then add the $\alpha$th element to the subcollection, if it is not dominated by any earlier element in that enumeration. The resulting subcollection will be dominating, since every individual of the order is either in the subcollection or dominated by some element in the subcollection. Meanwhile, the subcollection is well-founded, because at no stage did we add an element that was below an element already present in the subset. Thus the collection is graded by the ordinal rank at which it was added — for any nonempty subset of the subcollection, the first-added element will be minimal. $\Box$

A similar argument shows that every dominating family, in any partial order, has a well-founded dominating subfamily. We just perform the same construction, but always choose the dominating element from the fixed dominating family.

  • $\begingroup$ I'm sorry, I was being foolish - I misread multiple things! $\endgroup$ Dec 2, 2022 at 0:47

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