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September 3[edit]

Minimal Model, with regard to Peano system:[edit]

The set N of finite naturals, is known to be the minimal model of Peano system, in the sense that the model N has no proper sub-model of Peano system (BTW, N must be a sub-model of every model of Peano system, this being a stronger aspect of the minimality of N).

Let U be a consistent union of Peano system along with another set of axioms. Does U necessarily have a minimal model (i.e. a model having no proper sub-model of U)? For example, let S be an infinite set of axioms such that - for every finite natural n - the n-th axiom states that ω is greater than n. Does the union of Peano system along with S have a minimal model M (i.e. a model having no proper sub-model of that union)? If it does, then: does that hold also if one replaces S by another set of axioms whose union with Peano system is consistent? HOTmag (talk) 19:27, 3 September 2016 (UTC)[reply]

Hmm, interesting. Of course if the axioms you add are all true, then the answer is yes, because the naturals are still the minimal model.

But if you add a false axiom, still consistent with PA, then I don't know. For example, suppose you add the false axiom ~Con(PA). You still have a consistent theory (because otherwise PA would prove Con(PA), violating second incompleteness). Any model of PA + ~Con(PA) is necessarily illfounded; it has a copy of the natural numbers, and then above it it has infinitely many order-copies of the integers, the copies being ordered by a dense linear order.

So there's even a question what "minimal" means here. Do you mean minimal in the sense that no other such model embeds homomorphically into it? My guess is that there's no such model, but I don't see a proof off the top of my head. --Trovatore (talk) 20:26, 3 September 2016 (UTC)[reply]

By a "minimal model of U ", I mean a model of U - having no proper sub-model of U. Thanks to your question, a clarification has just been added to my original post. HOTmag (talk) 21:03, 3 September 2016 (UTC)[reply]