Given a decreasing family of sets and partitions with a refinement condition, is there a monotonous choice function from the partitions?

Question

I have been trying to show the statement below using the $AC$ but I am starting to think that it is not strong enough to do it.

Context: Let $\Gamma$ be an uncountable linearly ordered set with a smallest element (not necessarily well-ordered).

For each $\alpha\in\Gamma$, let $C_\alpha$ be a non-empty set such that $C_\alpha\supsetneqq C_\beta$ whenever $\alpha<\beta$. For each $\alpha\in\Gamma$, let $P(C_\alpha)$ be a partition of $C_\alpha$ such that: whenever $\alpha<\beta$, for all $B\in P(C_\beta)$, there exists $A\in P(C_\alpha)$ such that $B\subsetneqq A$.

Statement: There exists $\{A_\alpha\}_{\alpha\in\Gamma}$ such that $A_\alpha\in P(C_\alpha)$ and $A_\alpha\supsetneqq A_\beta$ whenever $\alpha<\beta$.

By using the AC we can see that there exists $\{A_\alpha\}_{\alpha\in\Gamma}$ such that $A_\alpha\in P(C_\alpha)$, but (I think) there is nothing to ensure the monotonicity condition: $A_\alpha\supsetneqq A_\beta$ whenever $\alpha<\beta$.

Maybe the statement is a well-known result or conjecture that I am not aware of, I would appreciate some answer or reference.

Answer 1

Let me first ignore your requirements that $\Gamma$ is uncountable and that the containments are strict, since these requirements are wholly artificial and have nothing to do with what's going on, and just make counterexamples a bit messier. Here, then, is the prototypical counterexample. Let $\Gamma=\mathbb{N}$, let $C_n=\{m\in\mathbb{N}:m\geq n\}$, and let $P(C_n)$ be the partition of $C_n$ into singleton sets. There is then no sequence of the sort you ask for since each singleton set $\{k\}$ in any of the partitions stops existing once you reach $C_{k+1}$.

OK, now if you insist, we can bulk this example up to meet all your requirements. Let's take $\Gamma=[0,\infty)\subset\mathbb{R}$, and let $C_x=\{(y,z)\in\mathbb{R}^2:y,z\geq x\}$, and let $P(C_x)$ be the partition into sets of the form $\{y\}\times[x,\infty)$. Now $\Gamma$ is uncountable and all the inclusions are strict, but the sequence you ask for fails to exist for the same reason as in the first example.

Answer 2

The answer is negative. This is not provable, even when assuming the axiom of choice. Even under the assumption that $\Gamma$ is a well-ordered set.

Let $T$ be a tree of height $\omega_1$ without a branch (either an Aronszajn tree, assuming choice; or any counterexample to $\sf DC_{\omega_1}$ otherwise).

Let $C_\alpha$ be $T\setminus T\restriction\alpha$, namely all the nodes of height at least $\alpha$. The partition is easy, $P(C_\alpha)$ is simply the set of subtrees above each node in the $\alpha$th level of $T$.

Easily, the sets $C_\alpha$ are descending, and the partitions refine each other. But now, if $A_\alpha$ is as descending choice sequence, by the fact that each $A_\alpha$ has a unique root, this would give us a branch. But the assumption on $T$ is that it has no uncountable branches, which is a contradiction.