27

According to here, there is the "standard" model of Peano Arithmetic. This is defined as $0,1,2,...$ in the usual sense. What would be an example of a nonstandard model of Peano Arithmetic? What would a nonstandard amount of time be like?

Mikhail Katz
  • 42,112
  • 3
  • 66
  • 131
  • 7
    Unfortunately it's quite hard to point to or describe a nonstandard model, because of Tennenbaum's Theorem: no countable nonstandard model of Peano arithmetic (PA) can be recursive. Even more: the $+$ and $\times$ operations can't be recursive. So... you can see why your question is very challenging. A nonstandard amount of time would be... very long: infinitely long, as it comes after every finite number of time units $0, 1, 2, ... 17, ... 10^{10}, ... $. – BrianO Jun 06 '16 at 08:01
  • 1
    I am not sure that I agree with the duplicate here. It's one thing to ask for a countable elementary extension, and it's another to ask about non-standard models in general. I think they are sufficiently different, as witnessed by the very different answers. – Asaf Karagila Jun 07 '16 at 06:18
  • "What would a nonstandard amount of time be like?" depends on which nonstandard model. If its elementary equivalent to the standard model, you won't even notice. If not, all sorts of crazy things could happen, like you proving PA inconsistent or something. – Christopher King Nov 16 '17 at 21:19

3 Answers3

42

Peano arithmetic is a first-order theory, and therefore if it has an infinite model---and it has---then it has models of every cardinality.

Not only that, because it has a model which is pointwise definable (every element is definable), then there are non-isomorphic countable models. Which means that you can find models which are not the standard model already at the countable cardinality.

What do these models look like? Well, it's kinda hard to explain. They all have an initial segment that looks like the natural numbers. That much is easy to prove. Also not hard to show is that the rest of the model can be decomposed into $\Bbb Z$-chains. Namely, if $c$ is a non-standard number, then it has a predecessor (since Peano proves that every non-zero element is a successor). So we can define $f(k)=S^k(c)$, as an order isomorphism between the "chunk" of the model that $x$ lives in.

Harder to prove, but still not impossible, is that at least for the countable parts, the models all look the same as far as the order go, they all have an initial segment of $\Bbb N$, followed by $\Bbb{Q\times Z}$ ordered lexicographically.

To produce such models you can use three standard methods:

  1. Compactness. Add a constant $c$, require it to be larger than any numeral, by compactness this is a consistent theory so it has a model. This model cannot be the standard model, because it has an element larger than all the numerals.

  2. Ultrapowers. Take a free ultrafilter $\mathcal U$ over $\Bbb N$, and consider the ultrapower $\Bbb{N^N}/\cal U$. Counting arguments will show you that this ultrapower has cardinality $2^{\aleph_0}$, so it is certainly not isomorphic to $\Bbb N$. If you prefer, you can use the fact that $\mathcal U$ is not a countably-complete ultrafilter, and therefore the ultrapower cannot be well-ordered, so without checking for cardinality it cannot be isomorphic to $\Bbb N$.

  3. Incompleteness. We know that Peano is not a complete theory. Therefore there are statements which are true in $\Bbb N$, but Peano does not prove. Therefore the negation of such statement is consistent with the rest of the axioms of Peano, and must have a model. But this model cannot be isomorphic to $\Bbb N$. The benefit of this method is that it allows you to obtain very different theories of your models, whereas ultrapowers and compactness arguments tend to result in elementarily equivalent models.

Asaf Karagila
  • 393,674
  • 2
    Wow. What a fantastic post, thank you. –  Jun 06 '16 at 07:59
  • You're welcome. I just talked about these things with my students a couple of weeks ago. – Asaf Karagila Jun 06 '16 at 07:59
  • For "compactness". What is $c$ the successor of? –  Jun 06 '16 at 08:08
  • Well... there is no good way to describe the predecessor of $c$. It will not be a numeral. It will be a non-standard element. – Asaf Karagila Jun 06 '16 at 08:12
  • So you are not only adding $c$, you are adding all its predecessors and successors, correct? All of which will be greater than all of $\Bbb{N}$, or just $c$ and the successors? –  Jun 06 '16 at 08:14
  • 4
    You need to end up with a model of Peano. Since $c$ is not a numeral, it will have successors and predecessors, and they will have products and differences and factorials and so on and so forth. You end up adding a whole lot. – Asaf Karagila Jun 06 '16 at 08:16
  • Maybe I remember incorrectly, but cannot you use the Ehrenfeucht-Fraisse games to prove that, for example, $\mathbb{N}$ and $\mathbb{N} \triangleright \mathbb{Z}$ are indistinguishable? (Think of the latter as the set of pairs ${(1, n) \mid n \in \mathbb{N}} \cup {(2, z) \mid z \in \mathbb{Z}}$ ordered lexicographically. – Bakuriu Jun 06 '16 at 10:51
  • 3
    I'm not sure what you mean by that. But probably you are thinking about ordered sets and not models of arithmetic. – Asaf Karagila Jun 06 '16 at 10:59
  • @AsafKaragila what I described is basically what you meant with "adding $c$ plus all its predecessors and successors". In my case $c$ is "a second $0$", so that $\mathbb{Z}$ represents the whole lot of things you are adding. I just wanted to point out that instead of using compactness to deduce that such model exists and is valid you can use Game Theory, in particular show that Duplicator has a winning strategy in the EF game between $\mathbb{N}$ and $\mathbb{N} \triangleright \mathbb{Z}$. – Bakuriu Jun 06 '16 at 11:08
  • 3
    @Bakuriu: But this is not a model of arithmetic. Just a linear order. – Asaf Karagila Jun 06 '16 at 11:10
  • 5
    @Bakuriu To see that this can't possibly work for models of arithmetic, consider the sentence "$\forall x\exists y(x=y+y \vee x=y+y+1)$." This sentence (and others) shows that $\mathbb{N}+\mathbb{Z}$ cannot be the underlying order-type of any model of PA (or much less, even)! In fact, we can prove that the underlying order-type of any (countable) nonstandard model of $PA$ is $\mathbb{N}+\mathbb{Z}\cdot\mathbb{Q}$. Even though this order has computable copies, though, doesn't mean that any of its expansions to a model of arithmetic do! – Noah Schweber Jun 06 '16 at 19:20
  • @NoahSchweber Sorry this is a bit late, but how would you construct such a sentence in the language of $PA$? The symbol $+$ does not occur in the language. – John Gowers Aug 04 '16 at 18:26
  • 2
    @Donkey_2009: The language of PA includes $0$ as a constant, $s$ is an unary function symbol, $+$ and $\cdot$ as binary function symbols, and $\leq$ as a binary relation symbol (and the rest of the first-order logic symbols like $=$ and variables and so on). – Asaf Karagila Aug 04 '16 at 18:42
  • 3
    @Donkey_2009 I believe you're mixing up the language of first-order PA (which does indeed have "$+$") with the language of second-order PA (which doesn't, and doesn't need it because it uses second-order logic). See https://en.wikipedia.org/wiki/Peano_axioms#First-order_theory_of_arithmetic (especially second paragraph: "Therefore, the addition and multiplication operations are directly included in the signature of Peano arithmetic"), and note that because of the general badness of second-order logic, when logicians in the modern era say "PA" they almost always mean first order PA. – Noah Schweber Aug 04 '16 at 19:07
  • If we omit $+$, $\times$ and $\le$ from the signature and remove the relevant axioms, is it true that $\mathbb N+\mathbb Z$ gives a model of the resulting theory? – John Gowers Aug 04 '16 at 20:24
  • And thanks - I was getting confused between the 'Peano axioms' (second order arithmetic) and 'PA' and thought that the only difference between them was the modification of the induction rule from something that talks about sets into the first order axiom scheme. – John Gowers Aug 04 '16 at 20:26
  • 1
    @Donkey_2009: You could consider the second-order version in the "full" language of arithmetic. The point is that they are all definable from $\leq$ in second-order logic. – Asaf Karagila Aug 04 '16 at 20:33
  • There is actually a somewhat specific example of a model of PA (see this). It might be helpful to note. – Christopher King Nov 16 '17 at 21:17
  • Mathematical footnote: the 'standard method 3' shows even a little more: not only that Peano admits nonstandard models, but by the incompleteness theorem it follows that that there cannot even be any set of axioms which would determine $\mathbb{N}$ uniquely up to isomorphism. Historical footnote: this (consequence of) 'standard method 3' is known since at least Gödel's review in Zentralblatt (Band 7, Heft 5) of Skolem's 1933 paper, see footnote in answer-form further below in this thread. – Peter Heinig Jan 29 '18 at 19:52
8

It should be mentioned that one of the most concrete nonstandard models of PA was developed by Skolem in the 1930s in ZF (without the axiom of choice, unlike the constructions mentioned in the other answer). This is roughly in terms of definable functions on $\mathbb N$ ordered by their asymptotic growth; for details see this 2013 publication in Foundations of Science, Section 3.2, page 272.

Mikhail Katz
  • 42,112
  • 3
  • 66
  • 131
  • Other than ultrafilters, neither of the other suggestions on my answer requires the axiom of choice. – Asaf Karagila Jan 18 '18 at 11:43
  • Your use of compactness theorem does not specify that you are referring to countable languages (which would of course be sufficient here). The general compactness theorem does require some form of choice. Similar remarks apply to the other possibility which rely on completeness results. – Mikhail Katz Jan 18 '18 at 12:00
  • נו, העיקר להפריד תבן ממוץ. – Asaf Karagila Jan 18 '18 at 12:01
  • 1
    I am not sure that I understand. Do you claim that you are Skolem? – Asaf Karagila Jan 18 '18 at 12:07
  • 2
    Historical footnote: a reference for Skolem's publication from the 1930s is Skolem, Th.: Über die Unmöglichkeit einer vollständigen Charakterisierung der Zahlenreihe mittels eines endlichen Axiomensystems. Norsk Mat. Forenings Skr., II. Ser. No.1/12, 73-82 (1933). Zbl 0007.19305 – Peter Heinig Jan 29 '18 at 19:39
  • 1
    @PeterHeinig, perhaps a more accessible text is Skolem's paper in German from 1934, listed in the bibliography of the article cited in the answer. – Mikhail Katz Jan 30 '18 at 09:39
7

This is a historical footnote to the nice accepted answer above. More precisely, the method in the yellow box in

enter image description here

seems to be already hinted at in Kurt Gödel's review in Zentralblatt, Band 7, Heft 5, of

Über die Unmöglichkeit einer vollständigen Charakterisierung der Zahlenreihe mittels eines endlichen Axiomensystems. Norsk Mat. Forenings Skr., II. Ser. No.1/12, 73-82 (1933).

Strictly speaking, Gödel does not give an argument in the review, only says that Skolem's result follows easily from his own article on the incompleteness theorem; here is Gödel's review of Skolem's paper, with Gödel's hint highlighted in orange (the right-hand side is my translation):

enter image description here

(the above picture larger)

Peter Heinig
  • 1,258