Must strange sequences wear Russellian socks?

Question

This is an attempt to make more precise a vague guess at the end of this answer of mine. We work in $\mathsf{ZF}$ throughout.

Say that a sequence $\mathcal{A}=(A_i)_{i\in\omega}$ of disjoint sets is strange if each $A_i$ has size $n$ for some fixed $n\in\omega$, the product $\prod\mathcal{A}$ is nonempty, and $\prod\mathcal{A}$ is not of size continuum. Strange sequences can consistently exist; however, the only way I know how to build one is by using a sequence with empty product as a "scaffold" (see my above-linked answer). I'm curious whether this is necessary; here's one precise question along these lines:

Suppose $\mathcal{A}$ is strange. Must there be a sequence of sets $\mathcal{B}=(B_i)_{i\in\omega}$ such that $\prod\mathcal{B}=\emptyset$ and the $B_i$s uniformly inject into some quotient of $\bigsqcup_{i\in \omega}A_i$?

("Uniformly inject" means that a set of injections, one for each $B_i$, exists.)

Answer 1

For $n \ge 2$ and $\langle A_i \rangle$ such that $|A_i| = n$ for all $i,$ TFAE:

1. $|\prod A_i| \neq |\mathbb{R}|.$
2. $\bigsqcup A_i$ is uncountable.
3. $\bigsqcup A_i$ is un-orderable.
4. There are $k \in [2, n]$ and $B_i \in [A_i]^k$ such that $\prod B_i = \emptyset.$

The implications are $4 \rightarrow 3 \rightarrow 1 \rightarrow 2 \rightarrow 3$ are straightforward. We will prove $2 \rightarrow 4$ by induction on $n.$ The base case $n=2$ is clear.

Suppose the claim holds for some $n \ge 2.$ Let $\langle A_i \rangle$ be such that $|A_i|=n+1$ and $\bigsqcup A_i$ is uncountable. If $\prod A_i = \emptyset,$ then we can simply set $B_i = A_i.$ Otherwise, fix $s \in \prod A_i.$ Let $A_i' = A_i \setminus \{s_i\}.$ Clearly, $\bigsqcup A_i' = \bigsqcup A_i \setminus \{s_i: i<\omega\}$ is uncountable. Apply the inductive hypothesis to $\langle A_i' \rangle$ to get our desired $k$ and $\langle B_i \rangle. \square$

Here's some info which shows the above result is essentially sharp. If $\mathbb{R}$ is a countable union of countable sets, then there is a sequence $\langle A_i \rangle$ of countably infinite sets for which $|\prod A_i| >|\mathbb{R}|$ but there is no sequence $\langle B_i \rangle$ of nonempty sets with empty product which uniformly inject into any quotient of $\bigsqcup A_i.$ We have that $\omega_1$ is singular, so let $\langle \alpha_i \rangle$ be cofinal in $\omega_1.$ Set $A_i = \alpha_i.$ Then $|\prod A_i | \ge |\mathbb{R}| + \aleph_1 > |\mathbb{R}|,$ and $|\bigsqcup A_i| = \aleph_1.$

Finally, I'll sketch out a model with a strange sequence of finite sets of unbounded size for which there is no such $\langle B_i \rangle.$ Working in $L,$ build the minimal ZFA model from countable set of atoms $A=\{a_n \}.$ Letting $f$ be the map $f(a_n) = \lfloor \sqrt{n} \rfloor,$ the model $L(A, f, \mathcal{P}(A)) \models ZFA$ satisfies that $A_i = f^{-1}(\{i\})$ is as desired. Standard transfer arguments yield a ZF counterexample.