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A beautiful and surprising (to me at least) result around the axiom of choice is that, while full $\mathsf{AC}$ is preserved by forcing, a model of $\mathsf{ZF}$ + "There are no amorphous sets" may have a (set-)generic extension in which there do exist amorphous sets. This was proved by Monro, On generic extensions without the axiom of choice; see also Asaf Karagila's summary here.

I'm interested in not-too-strong sufficient conditions on a c.t.m. $\mathcal{M}\models\mathsf{ZF}$ to have no generic extensions in which amorphous sets exist. Specifically, I'm curious if the following model-theoretic condition does the job:

Say that a c.t.m. $\mathcal{M}\models\mathsf{ZF}$ is expansive iff for every first-order theory $T\in \mathcal{M}$ in a finite language with infinite models, every infinite set in $\mathcal{M}$ is the underlying set of a model of $T$ in $\mathcal{M}$.

(Note that such an $\mathcal{M}$ does correctly compute whether such a $T$ has infinite models.) For instance, expansiveness prevents the existence of infinite Dedekind-finite sets, since we can take $T$ to be the theory of a discrete linear order. My question, then, is how expansiveness, amorphousness in particular, and forcing interact:

  • Is every set-generic extension of an expansive c.t.m. also expansive?

  • If not, is there a set-generic extension of an expansive c.t.m. which has amorphous sets?

(Incidentally, it's not immediately clear to me that expansiveness is any stronger than "Infinite = Dedekind-infinite"!)

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    $\begingroup$ 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. $\endgroup$ Oct 24, 2022 at 2:05
  • $\begingroup$ Incidentally, it would be interesting to see if just "Infinite = Dedekind-infinite" is sufficient to get "generically no amorphous sets." $\endgroup$ Oct 24, 2022 at 3:26
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    $\begingroup$ Related? mathoverflow.net/a/412400/7206 $\endgroup$
    – Asaf Karagila
    Oct 24, 2022 at 5:56
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    $\begingroup$ Why ask about CTM instead of just asking about the theory ZF + "it is not forceable that there is an amorphous set"? The CTM aspect seems irrelevant. $\endgroup$ Oct 26, 2022 at 12:29
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    $\begingroup$ @ElliotGlazer: No, it does not. Indeed, not even DC is enough. Feferman's model, which is $L(\Bbb R)$ of the Cohen model satisfies DC, and by abstract nonsense results, the Cohen model is a generic extension of it (we add a set of reals, each real is in the model, and the set is in a generic extension). So even if DC holds, there can be a generic extension with an amorphous set. Alternatively, my paper with Jonathan Schilhan shows even higher DC won't do in a direct proof. $\endgroup$
    – Asaf Karagila
    Oct 28, 2022 at 1:00

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Turning my comment into an answer, an $X$ which is the universe of any finitely axiomatized theory with an infinite model must be orderable, and there must be a bijection between between $X$ and $X^2.$ Both of these follow from $X$ satisfying a large finite fragment of PA. In fact this is a characterization of such $X$ by the E.M. model construction, treating $X$ equipped with an arbitrary order as the generating set of indiscernibles. The resulting model is equinumerous with $X$ since $|X|=|X|^2$ implies $|X|=|X|^{<\omega}.$

Just from $|X|=|X|^2,$ we can deduce that a model is expansive precisely if it models choice, by Tarski's equivalence of AC with every infinite set being in bijection with its square. Trivially, such models satisfy generic non-existence of amorphous sets.

There are still a lot of open questions regarding generic non-existence of amorphous sets. Asaf's posts summarize the state of knowledge regarding this principle: it's strictly stronger than nonexistence of amorphous sets, since it fails in the Cohen model, and it's implied by "all sets are almost well-orderable." I don't even know how to show the latter is strictly stronger.

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