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All non standard models of PA have the natural numbers as the initial segment. There can't be an infinite number of elements less than n where n is a natural n
by yequalsx 9y ago
All non standard models of PA have the natural numbers as the initial segment. There can't be an infinite number of elements less than n where n is a natural number in any nonstandard model of PA.
- dooglius 9y agoIs natural number = standard natural number here? I can see why the standard natural numbers must have a finite number of predecessors (one can explicitly prove in PA all statements of the form (x = 0 || x = S0 || ... || x = n || x > n) for each fixed standard n). But for the same reason, each non-standard number is greater than all standard numbers. So while a non-standard model m might be able to prove a finiteness-implying statement like \exists n (P(x) \implies x <= n), the actual n that makes this statement true could be non-standard, and m could contain infinitely many (standard or non-standard) elements with P(x) true so long as they are all less than some non-standard n. The point is, I don't see how PA can have a satisfactory notion of finite-ness as one could use this to define standardness from within PA (x is standard iff there are finitely many y with y < x) which is not supposed to be possible.
- yequalsx 9y agoThe article is about finite subsets of N. N is the initial segment in every non standard model of PA. Hence your in=bjectiinnin your original post is not valid.