Is the conceptual possibility of amorphous infinite sets "evidence against" countabilism?

Question

Countabilism is, roughly, a family of standpoints inclusive of:

1. There is one infinite proper set, of size ℵ₀, and one infinite proper class, ℵ₀^ℵ₀. (See about e.g. "pocket-sized" and Vopenkan set theory in the SEP. For those constructivists/intuitionists for whom even ℕ is not a "completed totality," it trivially follows that the powerset of ℕ is not a "completed totality" either, but so see also Storer[10] as well as Barton and Friedman[21] for similar or sometimes even equivalent propositions.)
2. There is one infinity, as a "completed totality," only, but it is not a set, it is a proper class (or if it is a set, it is a universal one), and this infinity is that of countability already. (Note that the powerset axiom scheme (for infinity) is incompatible with this system.) Suggested by the thematism of Skolem's paradox.

We might parse (1) as weak countabilism, (2) as strong.

At any rate, "everyone knows that a set is infinite if and only if it can be put into a one-to-one correspondence with a proper subset of itself," except nowadays we know that this is not so (actually, this has been known by some for a while longer...), the if-and-only-if parameter is too much, for there is the apparent logical possibility of amorphous sets, which are Dedekind-finite but otherwise infinite. More generally, in set theories open to the possibility of choiceless sets, there are many simple-enough-structure infinite sets that are eventually available that, while not commensurate with the alephs, yet directly (or quantitatively) at least echo the qualitative simplicity of the zeroth aleph. (For more on amorphous and other choiceless sets, see Truss[95] as well as Harrison-Trainor and Kulshreshtha[22].) My question is: if the conception of an infinite set is generically ambivalent in this manner, does countabilism lose some of its apparent plausibility, at least to the extent that strong countabilism is seriously (modally) undermined, while weak countabilism needs to be reconceived as admitting, alongside the zeroth aleph, at least one or so amorphs or other choiceless cardinals, with the proper class of the set world then becoming something like ℵ₀ ∪ A or even 2^{ℵ₀ ∪ A} (for A (co)amorphous^c)?

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^cLet W be the class of well-ordered sets and -W be the class of sets that are not well-ordered. Then let C, for coamorphous, be the complement in -W of A (for A amorphous).

Answer 1

A strong countabilist would likely choose to use at most PA or, say, some theory of hereditarily finite sets, so no 'infinite' objects of any kind to worry about

For a classical weak countabilist, possibly at most full second order arithmetic, maybe some 'faux' two-sort first-order subtheory, but one strong enough to prove the nice theorem

For any infinite set X, there exists a function f such that ∀k∀m (k < m -> f(k) < f(m)) and ∀n(n ∈ X <-> ∃m(f(m) = n))

available already at rather weak theories

Finally, a constructive/predicative/etc. weak countabilist may use some adequate type theory, with pertinent restrictions on construction of function types, etc., etc.

So, in any case, they'd be happily living with/in their chosen theories, which disallow the existence of such amorphous/ill-structured/pathological infinite objects by design, and would not care at all that these may exist in some other 'suspicious' theories

For anyone else, the possibility of amorphous infinite sets may well be evidence against the plausibility of Z(F) and related theories