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Q&A

Minimal non-standard number in non-standard models of PA

+4
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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?

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x-post https://www.reddit.com/r/math/comments/1ap6clo/minimal_nonstandard_number_in_nonstandard_model... (2 comments)
X-post https://www.reddit.com/r/modeltheory/comments/1ap68ix/minimal_nonstandard_number_in_nonstandar... (1 comment)

2 answers

+4
−0

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.

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but PA proves that every set definable in its language must have a minimal element, so will atleast set of non-standard numbers definable in PA have a minimal element ? (7 comments)
+0
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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.

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