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This question is motivated by Vaananen's paper Generalized quantifiers in models of set theory. Say that a (set-sized, regular) logic $\mathcal{L}$ is

  • unbounded if there are $\mathcal{L}$-sentences $\varphi,\psi$ each with arbitrarily large models but with disjoint classes of cardinalities of models, and

  • strongly unbounded if there is an $\mathcal{L}$-sentence $\theta$ in a language containing a unary relation symbol $A$ and a binary relation symbol $<$ such that $(i)$ whenever $\mathcal{M}\models\theta$ the subreduct $(A,<)^\mathcal{M}$ is a well-ordering and $(ii)$ there is no bound on the lengths of well-orderings of the form $(A,<)^\mathcal{M}$ for $\mathcal{M}\models\varphi$.

It turns out that strongly unbounded logics are indeed unbounded. The proof is a simple consequence of the nonexistence of a descending sequence of cardinalities. A linear order $I$ is a well-order iff there is a family of sets $X_i$ ($i\in I$) of sets such that $i<_Ij \implies \vert X_i\vert<\vert X_j\vert$, and so if $\mathcal{L}$ is strongly unbounded via $\varphi,\psi$ we can take as our $\theta$ an appropriately-constructed sentence describing a linear order $(A,<)$ with a collection of pairs of sets $X_a,Y_a$ for each $a\in A$ such that $X_a$ is (the domain of) a model of $\varphi$ and $Y_a$ is (the domain of) a model of $\psi$, and if $a\le_Ab$ then there are injections from $X_a$ to $Y_a$ and from $Y_a$ to $X_b$.

I'm curious whether this trick is avoidable. Or, put another way:

Question 1: Over $\mathsf{ZF+DC}$, does "Every strongly unbounded logic is unbounded" imply "There is no descending sequence of cardinalities"?

(See Asaf Karagila's comment below for why I've included Dependent Choice here. Meanwhile, note that it is currently open whether "There is no descending sequence of cardinalities" is equivalent to $\mathsf{AC}$, see Howard/Tachtsis.)

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  • $\begingroup$ The problem with "no descending sequence of cardinals" is that it's a tricky statement. In ZF+DC and ZF+DF it's the same thing as saying "no decreasing sequence of sets of smaller cardinalities". But there's a weird gap between the two things, where it's not clear if you can always arrange from a sequence of cardinals a sequence of sets. Or at the very least, it is not clear to me. The point is that in many places in the literature, "no descending sequence of cardinals" is actually formulated with concrete sets, rather than cardinals. $\endgroup$
    – Asaf Karagila ♦
    Jan 29, 2023 at 1:49
  • $\begingroup$ @AsafKaragila That's a good point. To make this more answerable, I've tamed the base theory a bit. $\endgroup$
    – Noah Schweber
    Jan 29, 2023 at 1:50
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    $\begingroup$ (BTW, DF in my comment meant "There is an infinite Dedekind-finite set".) $\endgroup$
    – Asaf Karagila ♦
    Jan 29, 2023 at 1:50

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