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I'll state the question about modules, but I'm open to examples in other contexts. I am not an algebraist, so please forgive any non-conventional terminology.

DEFINITION: Let $\mathcal{C}$ be a collection of (countable) modules (for some fixed ring). Given a module $M$ of size $\omega_1$, let's say that $M$ is "strongly $\mathcal{C}$-filtered" iff there is a $\subseteq$-increasing and continuous sequence $\langle M_i \ : \ i < \omega_1 \rangle$ of countable submodules, with union M, such that $M_i \in \mathcal{C}$ and $M_j / M_i \in \mathcal{C}$ whenever $i < j < \omega_1$.

QUESTION: What are some nontrivial examples of classes $\mathcal{C}$ that have the following ``inheritance" property: whenever $|M|=\omega_1$ and $M$ is strongly $\mathcal{C}$-filtered, then every submodule of $M$ of size $\omega_1$ is also strongly $\mathcal{C}$-filtered?

The only example I'm aware of is where $\mathcal{C}$ is the collection of (countable) free abelian groups (i.e. countable free $\mathbb{Z}$-modules). In this situation, for an abelian group $M$ of size $\omega_1$, it's easy to see that $M$ is strongly $\mathcal{C}$-filtered if and only if $M$ is free. And, since subgroups of free abelian groups are always free, $\mathcal{C}$ has the inheritance property requested in the question.

(More generally, $\mathcal{C}$ = "the set of all countable free $R$-modules", where $R$ is a PID, works for the same reason, since submodules of free modules are free if the ring is a PID. Of course in these examples, one could just as well let $\mathcal{C} = \{ F_\omega \}$ where $F_\omega$ is the free object on a countably infinite set of generators. But in the question I don't necessarily require that $\mathcal{C}$ be a singleton, or even a countable, collection).

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  • $\begingroup$ Whitehead's Problem comes to mind... $\endgroup$ – Asaf Karagila Aug 8 at 16:43
  • $\begingroup$ I would be interested in knowing if Whitehead groups of size $\omega_1$ can be characterized by strong $\mathcal{C}$-filtering, for some $\mathcal{C}$ that satisfies the requirements of the question (assuming we're in a universe where there is a non-free Whitehead group, of course). $\endgroup$ – Sean Cox Aug 8 at 19:09
  • $\begingroup$ Well, in the proof that $\lozenge$ implies that every Whitehead group is free there is a characterization of what and where things fail. To my understanding that was the origin of club guessing, too. $\endgroup$ – Asaf Karagila Aug 8 at 20:30

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