@AsafKaragila I see, nice.

Incidentally, it would be interesting to see if just "Infinite = Dedekind-infinite" is sufficient to get "generically no amorphous sets."

If I'm not mistaken, an infinite set $X$ is a universe for any such $T$ iff $X$ is orderable and there is a bijection between $X$ and $X^2.$ The forward direction is by considering $T$ to be (a finite fragment of) PA and the backward direction by taking the E.M. model generated with $X$ as a set of order-indiscernibles. This would mean expansive models are precisely those satisfying choice by Tarski's characterization of choice.

@AsafKaragila The argument in that paper relies heavily on the permutation model structure. It doesn't give an argument for the claim I suggested.

@AsafKaragila Yeah, I'm thinking the same. I wonder if it wouldn't be too hard to prove AC in $\text{ZFA} + \text{AC}_{\text{WO}} + \text{AC}^{\text{WO}} + \text{WF} \models \text{AC}.$ (WF the well-founded kernel).

@ZuhairAl-Johar Probably a lot, though I haven't thought about this sort of question in a while. You might find this paper relevant since it examines similar questions in the ZF setting: math.bu.edu/people/aki/7.pdf

In fact, local choice isn't even needed: just over NBG, global choice fails iff $|Ord| < |Ord \cup \bigcup_{\alpha} \{\text{well-orderings of } V_{\alpha} \}| < |V|.$

@DmytroTaranovsky The first is inconsistent: under NBG + local choice, if $|Ord| < |V|,$ then $Ord \cup \bigcup_{\alpha} \{\text{well-orderings of } V_{\alpha}\}$ is of strictly intermediate cardinality.

See here for a model (without choice) in which a union of a chain of countable sets of reals is always countable: math.stackexchange.com/a/4423643/210610

To be somewhat pedantic, this is provable from countable choice but not from ZF alone. In Cohen's model with an infinite, Dedekind finite set of reals, there is a complete probability space for which the associated measure algebra is not complete. In particular, there is a family of characteristic functions on this space with no supremum.

@AsafKaragila It's actually equivalent to $\text{AC}_{\omega}(\mathbb{R}).$ I've updated my answer.

@Lorenzo From an enumeration of $(2^{-n}, 1) \setminus X$ one can use diagonalization to canonically choose a real from each $X \cap I$ ($I$ a subinterval).

@AsafKaragila It's not clear to me if that's enough. That $\langle r_i \rangle$ is encoded by a real in some $S_n$ depends on their union being all of $\mathbb{R}.$

Turing reducible.