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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 $0 = a < b = 1$]

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  • $\begingroup$ The $c$ is going to depend on the particular ultrafilter. Either the all $0$'s sequence or the all $1$'s sequence will work. $\endgroup$
    – James E Hanson
    Commented Dec 27, 2023 at 8:23
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    $\begingroup$ "take a quotent that gives the ultrapower: identify sequences when they agree except on finitely many components" : this is an incorrect description of the ultrapower. One has to identify sequences if they agree on a dominant set there dominant does not have to be cofinite. You are confusing ultrafilters and Frechet filters. $\endgroup$
    – Mikhail Katz
    Commented Dec 27, 2023 at 9:28
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    $\begingroup$ @JamesHanson Actually, $c$ has to be (equivalent to) the all $1$s sequence, since either $a=1$ and $b=0$ or the reverse (depending on the ultrafilter). $\endgroup$
    – Alex Kruckman
    Commented Dec 27, 2023 at 16:16

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Amazing. Thank you to commenter @Mikhail-katz. Looks like my confusion is due to a Wikipedia error from mid-2019 that I will go correct. Thank you!

This is super helpful as now I see that it's quite subtle to prove that the resulting construction satisfies all the induction axioms.

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    $\begingroup$ Note that modding out by the equivalence relation induced by the Frechet filter (or any other filter) does make sense, it just doesn't preserve all sentences. In general, it's quite interesting to look for what properties a given filter does(n't) preserve! $\endgroup$
    – Noah Schweber
    Commented Dec 27, 2023 at 21:49
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    $\begingroup$ You write that it's quite subtle to prove that the resulting construction satisfies the induction axioms. This is true, it's in no way immediate from the construction, but it does follow immediately from a very general theorem, about ultraproducts: Łoś theorem. Łoś's theorem itself is straightforward to prove (it's a simple induction on the complexity of formulas). The ultraproduct construction is cooked up exactly to make Łoś's theorem true. $\endgroup$
    – Alex Kruckman
    Commented Dec 27, 2023 at 23:15
  • $\begingroup$ Dave, you may want to correct the formulation of your question. $\endgroup$
    – Mikhail Katz
    Commented Dec 28, 2023 at 10:25
  • $\begingroup$ Thanks Mikhail, done! $\endgroup$
    – Dave Pritchard
    Commented Dec 31, 2023 at 2:11

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