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If $T$ is a first order set theory having finitely many axioms, suppose the consistency of $T$ is already known and that $T$ proves existence of naturals, now suppose that $S$ is a schema and that $T+S$ is proved consistent by compactness.

What qualifications $T$ must meet in order for $T+S$ to have an omega model (i.e. a model in which all naturals are standard)?

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    $\begingroup$ "Having an $\omega$-model". $\endgroup$
    – Asaf Karagila
    Jul 22, 2016 at 11:52

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There is no property of $T$ alone that will ensure that $T+S$ always has an $\omega$ model in the circumstances you describe. In fact, there is no computably axiomatizable theory $T$ with the property that $T+S$ has an $\omega$ model whenever it is consistent. To see this, suppose we have such a $T$, and simply let $S=\neg\text{Con}(T)$. It follows that $T+S$ is consistent, but it can have no $\omega$-model, since the proof of a contradiction that $S$ asserts must be nonstandard.

What you need is a property about $T+S$, not just a property about $T$.

Regarding your revised question mentioned in the comment, what we seem to need is a characterization of which theories have $\omega$-models.

Theorem. A theory has an $\omega$-model if and only if it is consistent in $\omega$-logic.

See the Wikipedia entry on $\omega$-logic. Note that being consistent in $\omega$-logic is not the same as being $\omega$-consistent. The latter has to do essentially with a single instance of the $\omega$-deduction rule, whereas consistency in $\omega$-logic allows nested applications of the rule.

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  • $\begingroup$ But T+S in your example is not provable by compactness. Anyhow. OK then take the question to be about what property of T+S. I once thought for example that if T is consistent with for all naturals (phi) then if each instance of S states: the standard natural n fulfills phi, then T+S would have an omega model if T had an omega model. It appears that what gives rise to the nonstandards is having an opposition between a theorem of T and what scheme S is asserting. If we have no such opposition then it appears that T+S would have an omega model if T had one. $\endgroup$ Jul 22, 2016 at 12:53

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