Minimal non-standard number in non-standard models of PA
Excuse me, if the question sounds too naive.
From Gödel's incompleteness theorem we know that there would be non-standard models where the Gödel sentence would be false. These models will have an initial segment isomorphic to standard natural numbers. Will there be a minimal non-standard number in such models such that every number smaller than it is a standard natural number and every number bigger than it would be non-standard ?
Since non-standard model would be a model of arithmetic then I think there should be a minimal non-standard number, but then maybe my concept is unclear about it. Any help?
2 answers
You are accessing this answer with a direct link, so it's being shown above all other answers regardless of its score. You can return to the normal view.
Quite the opposite; in no non-standard model of Peano arithmetic is there a minimal non-standard number.
Consider the formula \(\phi(x) = \left(x = 0\right) \vee \exists y \left(x = S(y)\right)\). The first-order induction axiom for \(\phi\) is
\[ \phi(0) \wedge \forall x \bigl(\phi(x) \Rightarrow \phi(S(x))\bigr) \Rightarrow \forall x \phi(x) \]
\(\phi(0)\) is trivially true in any model. If we have \(\phi(x)\), and there exists a \(y\) such that \(x = S(y)\), then \(S(x) = S(S(y))\), giving us \(\phi(S(x))\). From the induction axiom, \(\phi\) holds for all numbers—every number is 0 or has a predecessor, in any model with first-order Peano induction.
The non-existence of a minimal non-standard number follows by contradiction: any number with a standard predecessor is itself standard, any number with a non-standard predecessor is not minimal, and any number with no predecessor is zero.
The nonstandard models would be elementarily equivalent to the standard models, i.e. they would satisfy the same first-order formulas. In particular, for every first-order formula with one free variable, the set of numbers satisfying it would have a least element.
However, for every number $x>0,$ the number $x-1$ exists. Thus for every nonstandard number $x$, the number $x-1$ exists. It is a nonstandard number. Hence there can be no minimal nonstandard number.
Hence we must conclude that the set of all nonstandard numbers cannot be defined by any first-order formula.
0 comment threads
2 comment threads