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Talk:Amorphous set

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Moving towards a model of an amorphous set in V

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An amorphous set can be made from urelements, if they are available. But can one be made in the Von Neumann universe V which consists of the well-founded pure sets? Let us try to make one of minimal rank. It cannot be of finite rank because then it would have to be a finite set since each Vn for finite n is finite. Nor could it be of rank ω because then its elements could be partitioned by their rank into parts none of which is infinite. So let us try to put A (the amorphous set) at rank ω+1. So all its elements will be of rank ω. For simplicity, we will make them sparse subsets of the natural numbers ω. We will define a sparse sequence of natural numbers { sn | n ∈ ω } and the elements of A will be the co-finite subsets of that sequence. To eliminate the non-trivial partitions of those elements, we will use the a variant of the downward Lowenheim-Skolem theorem to remove them.

Suppose we are working in a universe which satisfies ZFC and has a strongly inaccessible cardinal Κ. Let R be a well-ordering on VΚ. We may tweak it to put the sets of lower rank before the sets of higher rank. Let sn+1 be the smallest natural number greater than sn which cannot be defined in VΚ by a formula in the language of =, ∈, R having fewer than 2n symbols.

We will select a subset of VΚ at each stage n∈ω of the construction. After which, we will apply the Mostowski collapse lemma to the resulting countable model to get a countable transitive model of ZF. At stage 0, we will include the elements of Vω. At stage n+1, we add A, An (the subset of the sparse sequence excluding those in positions coded by the binary representation of n), and the results of applying the axioms of ZF (no choice) to the sets in stage n using R to pick the exact value when there would otherwise be an ambiguity. In particular, for extentionality we add the R minimal element of x-y (if any) and for regularity we add the R minimal element of the given non-empty set. OK? JRSpriggs (talk) 23:02, 24 September 2022 (UTC)[reply]

The remaining issue is whether the sequence sn will remain undefinable in the resulting model. JRSpriggs (talk) 23:33, 24 September 2022 (UTC)[reply]

Now, I see that it fails. The union of A would recover the sparse sequence. And applying the axiom of separation to A and a member of the sparse sequence would partition the elements of A into two infinite subsets: co-finite subsets of the sequence including the chosen element, and co-finite subsets excluding the chosen element.
From this, we can infer the following lessons: we need to use a rank higher than ω+1 for A; instead of finite or co-finite subsets of something, we should use entirely disjoint sets as elements of A; otherwise the properties of the elements should be as similar as possible (their transitive closures should have a similar structure as graphs); perhaps, we should eliminate the axiom of choice first before trying to construct A. JRSpriggs (talk) 16:01, 25 September 2022 (UTC)[reply]