Jump to content

Wikipedia:Reference desk/Archives/Mathematics/2016 September 3

From Wikipedia, the free encyclopedia
Mathematics desk
< September 2 << Aug | September | Oct >> September 4 >
Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.


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]