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The infinite-model categoricity test for completeness
Statement
In ZFC, let be a consistent sentence theory in an explicitly countable language. Assume every model of is infinite and any two models of of a fixed infinite cardinality are isomorphic. Then is syntactically complete.
Facts & Assumptions
Given: Consistency, the no-finite-model hypothesis, categoricity in infinite , and AC.
Infinite models can be enlarged elementarily to any cardinal at least their size and the language size. (Upward Löwenheim–Skolem, including elementary extensions)
Infinite models have elementary substructures of any infinite size between the language bound and their own size. (Downward Löwenheim–Skolem with parameters)
Syntactic completeness means proving one side of every sentence decision. (Consistency and syntactic completeness)
For countable languages, semantic consequence equals provability, and consistent theories have models. (Completeness for explicitly countable set languages)
AC is assumed for the cardinal-size theorems. (The Axiom of Choice)
Proof
If were not complete in the sense of F3, some sentence would satisfy and . By F4 these mean and . Thus there are models of with and , respectively. The hypothesis on makes both infinite.
For each of , if its size is at most , apply F1; if it is at least , apply F2 with empty parameter set. Since the language is countable and infinite, both language bounds hold. Under A1 this produces of size exactly , elementarily equivalent respectively to . At equality take the structure itself. Consequently and , and both satisfy .
Categoricity gives an isomorphism . It preserves values of terms by induction on terms: variables and constants are preserved, and each function commutes with . Hence it preserves and reflects equality and relation atoms. Negation and conjunction retain this equivalence; existential witnesses transfer forward by and backward by its inverse. Formula induction therefore makes isomorphic structures agree on every sentence, contradicting their opposite decisions of . No such undecided sentence exists, so is complete.
Depends on
Used by
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Dependency tree · two levels
17 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- Weiss–D’Mello, Theorem 6 and full proof, printed pp20–21; syntactic completeness obtained through local countable completeness. (standard reference, not scraped)