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Corrections to misleading stuff in question, as proposed by responder
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Dave Pritchard
Dave Pritchard
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I've read that one nonstandard model of arithmetic is:

  • take $\mathbb{N}^\mathbb{N}$, the set of countably infinite sequences of natural numbers
  • take a quotent that gives the ultrapower: identify sequences when they agree except on finitely many components [correction: this should say, for some fixed $\mathbb{N}$-ultrafilter, identify sequences when they agree on a member of the ultrafilter]
  • then define arithmetic componentwise

But I think my mental model is wrong because I arrive at sequences that seem to violate the axioms of first-order Peano axioms. Take $a:=(0, 1, 0, 1, ...)$ and $b:=(1, 0, 1, 0, ...)$. They should satisfy the trichotomy property and so either $a=b+c$ or $b=a+c$ for some $c$, but clearly there is no such $c.$ Where am I going wrong? [correction: it was the wrong definition of ultrapower. With the correct definition, we know that for every subset $S$ of $\mathbf{N}$, an ultrafilter either contains $S$ or its complement; if the ultrafilter contains the set of even whole numbers then $1 = a > b = 0$, while otherwise it contains the set of odd whole numbers and $0 = a < b = 1$]

I've read that one nonstandard model of arithmetic is:

  • take $\mathbb{N}^\mathbb{N}$, the set of countably infinite sequences of natural numbers
  • take a quotent that gives the ultrapower: identify sequences when they agree except on finitely many components
  • then define arithmetic componentwise

But I think my mental model is wrong because I arrive at sequences that seem to violate the axioms of first-order Peano axioms. Take $a:=(0, 1, 0, 1, ...)$ and $b:=(1, 0, 1, 0, ...)$. They should satisfy the trichotomy property and so either $a=b+c$ or $b=a+c$ for some $c$, but clearly there is no such $c.$ Where am I going wrong?

I've read that one nonstandard model of arithmetic is:

  • take $\mathbb{N}^\mathbb{N}$, the set of countably infinite sequences of natural numbers
  • take a quotent that gives the ultrapower: identify sequences when they agree except on finitely many components [correction: this should say, for some fixed $\mathbb{N}$-ultrafilter, identify sequences when they agree on a member of the ultrafilter]
  • then define arithmetic componentwise

But I think my mental model is wrong because I arrive at sequences that seem to violate the axioms of first-order Peano axioms. Take $a:=(0, 1, 0, 1, ...)$ and $b:=(1, 0, 1, 0, ...)$. They should satisfy the trichotomy property and so either $a=b+c$ or $b=a+c$ for some $c$, but clearly there is no such $c.$ Where am I going wrong? [correction: it was the wrong definition of ultrapower. With the correct definition, we know that for every subset $S$ of $\mathbf{N}$, an ultrafilter either contains $S$ or its complement; if the ultrafilter contains the set of even whole numbers then $1 = a > b = 0$, while otherwise it contains the set of odd whole numbers and $0 = a < b = 1$]

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Dave Pritchard
Dave Pritchard
  • 1.3k
  • 10
  • 21

Seeking clarification of ultrapower nonstandard model of arithmetic

I've read that one nonstandard model of arithmetic is:

  • take $\mathbb{N}^\mathbb{N}$, the set of countably infinite sequences of natural numbers
  • take a quotent that gives the ultrapower: identify sequences when they agree except on finitely many components
  • then define arithmetic componentwise

But I think my mental model is wrong because I arrive at sequences that seem to violate the axioms of first-order Peano axioms. Take $a:=(0, 1, 0, 1, ...)$ and $b:=(1, 0, 1, 0, ...)$. They should satisfy the trichotomy property and so either $a=b+c$ or $b=a+c$ for some $c$, but clearly there is no such $c.$ Where am I going wrong?