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Say that a structure $\mathcal{M}$ is amorphic iff for every finite $\overline{a}\in\mathcal{M}$ and bi-infinite $X\subseteq\mathcal{M}$ there is some automorphism $\alpha\in Aut(\mathcal{M})$ fixing $\overline{a}$ pointwise but not respecting $X$ (that is, either $\alpha[X]\not=X$ or $\alpha[\mathcal{M}\setminus X]\not=\mathcal{M}\setminus X$). An amorphic structure can be "given an amorphous copy" in a symmetric extension of the universe; precisely, there is a symmetric extension $V\subset N\subset V[G]$ and a structure $\mathcal{A}\in N$ such that $\mathcal{A}$'s underlying set is amorphous in $N$ but $V[G]\models\mathcal{A}\cong\mathcal{M}$.

Now say that a countable complete theory $T$ is $\kappa$-amorphic iff there is a $\mathcal{M}\models T$ with $\vert\mathcal{M}\vert=\kappa$, and amorphic iff $T$ is $\kappa$-amorphic for some $\kappa$. For example, the empty theory is trivially $\kappa$-amorphic for every $\kappa$. I'm curious about what implications exist between the various amorphicities. To get started, my main question is:

Does amorphic imply $\omega$-amorphic?

Note that $\omega$-amorphicity is a $\Sigma^1_3(L_\omega)$ property, while on the face of things amorphicity is $\Sigma^0_2(V)$, so this would be a great improvement in terms of the complexity of amorphicity. I vaguely recall a negative result due to Shelah here, but I can't find it (and "due to Shelah" doesn't really narrow the search space much).

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    $\begingroup$ Adding to James' answer, you might want to take a gander at Walczak-Typke, Agatha C. A model theoretic approach to set theory without the axiom of choice. Algebra, logic, set theory, 255–289, Stud. Log. (Lond.), 4, Coll. Publ., London, 2007. $\endgroup$
    – Asaf Karagila
    Commented Jul 14, 2021 at 18:04
  • $\begingroup$ @AsafKaragila Ooh, most definitely! $\endgroup$ Commented Jul 14, 2021 at 18:33

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Amorphicity implies strong minimality and $\omega$-categoricity, which together imply $\kappa$-amorphicity for any $\kappa$.

Assume that $T$ is amorphic. To see that $T$ is $\omega$-categorical, we proceed by induction. It is clear that the type space $S_1(T)$ must be finite, otherwise we could form a bi-infinite set $\bigvee$-definable over a finite set of parameters, which would spoil amorphicity of any given model of $T$. Now suppose that we know that $S_n(T)$ is finite. Assume for the sake of contradiction that $S_{n+1}(T)$ is infinite. Then there must be some type $p(\bar{x}) \in S_n(T)$ with infinitely many extensions to an $(n+1)$-type. Since $S_n(T)$ is finite, $p(\bar{x})$ must be realized in any model of $T$. Fix a model $\mathcal{M}$ and let $\bar{a}$ realize $p(\bar{x})$. If there is some formula $\varphi(x,\bar{a})$ such that $\varphi(\mathcal{M},\bar{a})$ is bi-infinite, then we have a failure of amorphicity, so assume that for every formula $\varphi(x,\bar{y})$, $\varphi(\mathcal{M},\bar{a})$ is either finite or co-finite. Since there are infinitely many extensions of $p(\bar{x})$ to an $(n+1)$-type, we must have that $S_1(\bar{a})$ is infinite and scattered (i.e., has ordinal Cantor-Bendixson rank or equivalently contains no perfect subset), but this implies that there are infinitely many isolated points in $S_1(\bar{a})$, so we can again form a bi-infinite set $\bigvee$-definable over a finite set of parameters, which spoils amorphicity. Therefore it must be the case that $S_{n+1}(T)$ is finite, and thus by induction we have that $S_n(T)$ is finite for all $n < \omega$. So by the Engeler–Ryll-Nardzewski–Svenonius theorem, $T$ is $\omega$-categorical.

It is clear that if $\mathcal{M}$ is amorphic, then it can have no bi-infinite definable subsets, since any such subset defined by a formula $\varphi(x,\bar{a})$ would have $\bar{a}$ witnessing the failure of amorphicity. So we have that $\mathcal{M}$ is minimal. Finally, $\omega$-categorical theories eliminate the quantifier $\exists^\infty$ (i.e., for any formula $\varphi(\bar{x},\bar{y})$, there is a formula $\psi(\bar{y})$ such that for any $\bar{a}$, $\varphi(\bar{x},\bar{a})$ has infinitely many solutions if and only if $\psi(\bar{a})$ holds) and minimal sets are always strongly minimal in such theories, so $T$ is strongly minimal.

To see that strong minimality and $\omega$-categoricity imply $\kappa$-amorphicity for any $\kappa$, note that any such theory is totally categorical and therefore has every model saturated. Saturated models are homogenous, so the only automorphism invariant sets over a finite tuple are those actually definable over it.

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  • $\begingroup$ Dang, I need to actually learn some model theory ever. (BTW re: a previous conversation, have you thought anymore about whether there's a "topologically tame" expansion of R by a countable set?) $\endgroup$ Commented Jul 14, 2021 at 4:41
  • $\begingroup$ @NoahSchweber See my new comment on your original post. $\endgroup$ Commented Jul 15, 2021 at 0:28

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