Understanding nonstandard Peano arithmetic

Question

I've had the idea of nonstandard Peano arithmetic introduced to me in the comments of this question. The concept that we could write down the axioms which produce the natural numbers and also produce something else from them is totally crazy to me! I did not expect this.

I've been doing some reading here, to try and understand what these non-standard models of the Peano axioms are. It's difficult for me to follow the details of this paper, but I've understood that the most simple non-standard model of the Peano axioms is $$ M = \{\mathbb{N}, +, ×, ≤, 0, 1, c\} $$

I understand this to be the natural numbers along with an additional constant $c$, is that correct?

So from this understanding, it makes sense to me to ask the questions,

• What is $n+c$ for $n$ a natural number?
• What is $n\cdot c$ for $n$ a natural number?
• What is $c+c$?
• What is $c\cdot c$?
• Is $c≤n$ for $n$ a natural number?

Are these sensible questions? If not, is there some similar concrete way to understand this model?

Answer 1

Delving into the proof of completeness theorem, as Noah suggests, is one good way to gain intuition about nonstandard models of PA. Here's another approach.

First, we should see why there are nonstandard models of PA. It's not hard to find proofs that "Peano's Axioms" have a unique model, up to isomorphism. (Peano's predecessor Dedekind may have given the first such proof, in his paper "Was sind und was sollen die Zahlen" ("The nature and meaning of numbers")). This doesn't contradict the existence of nonstandard models of PA, because "PA" nowadays doesn't mean Peano's Axioms, but the first-order theory of arithmetic. Peano's induction axiom says that an arbitrary subset of $\mathbb{N}$ is $\mathbb{N}$ if it contains 0 and is closed under succession. This is a second-order axiom, meaning it talks about arbitrary sets of natural numbers.

The induction axioms of PA talk only about subsets of the system that can be defined by a first-order formula: a formula built up with +, $\times$, etc., and the basic constructs of logic ($\forall$, $\neg$, $\wedge$, etc.) A nonstandard model, call it $N$, will have subset that is isomorphic to $\mathbb{N}$; we might as well call it $\mathbb{N}$. This subset violates the second-order version of induction in $N$, since it's a subset closed under succession and containing 0 but not all of $N$. But it's "invisible" to first-order logic: it can't be defined by a first-order formula.

There is a weaker set of first-order axioms known as PA$^-$. (See here for a list of its axioms.) Basically this includes all the "algebraic" and "ordering" properties of PA, but omits all the induction axioms. So for example we have $x+y=y+x$ and $x<y\rightarrow x+z<y+z$ as axioms. Here it's much easier to construct a nonstandard model. Start with $\mathbb{N}$, and add a constant $c$; you may think of $c$ as a kind of "infinite number". Because you have $c$, you also must have $c\pm n$ for all $n\in\mathbb{N}$, $c^2$, $2c^2$, etc. In fact, it's not hard to see that the model must contain elements corresponding to all polynomials of the form $$a_nc^n+\cdots+a_0,\qquad a_n,\cdots,a_0\in\mathbb{Z},\;a_n>0$$ And that's enough! The set of such polynomials, together with 0, with the natural definitions for $+,\times,<$, is a model of PA$^-$ (call it $M$).

The induction axioms do not hold in $M$. For example, you can prove by induction that $$\forall x\exists y(y+y=x\vee y+y+1=x)$$ in other words, every number is even or odd. But there is no such $y$ in $M$ when $x=c$.

What to do, if we want a nonstandard model of PA? One idea: extend $M$ to a model $N$ of PA. Say we add a new constant $d$ and stipulate that $d+d+1=c$. (Or $d+d=c$, but not both!) Just as adjoining $c$ forced us to add lots of other elements, adjoining $d$ will force us to add a lot more. And that's just for one instance of one induction axiom! Still, the basic idea is to keep adjoining new elements whenever PA proves that something should exist, and it doesn't yet.

It is kind of remarkable that this approach can be made to work. With the model $M$, each polynomial defined a unique element, and different polynomials defined different elements. Also it was not hard to decide on the meaning of $+,\times,<$ for the elements of $M$. This becomes much more problematic with the wholesale introduction of constants designed to satisfy the induction axioms.

That it does work is the beauty of Henkin's construction, that Noah sketched. But we have to sacrifice something. The model $M$ of PA$^-$ is recursive, i.e., you could write a computer program to implement it. Tennenbaum's theorem says that no recursive nonstandard model of PA exists. In fact, you can't even make a model where the + operation is computable, not to say $\times$. (And vice versa.)

Plug: I've been blogging about nonstandard models of PA with a couple of other people (John Baez and Bruce Smith).

Answer 2

I've understood that the most simple non-standard model of the Peano axioms is $$ M = \{\mathbb{N}, +, ×, ≤, 0, 1, c\} $$ I understand this to be the natural numbers along with an additional constant $c$, is that correct?

It sounds like your $M$ consists of only the usual naturals and the single extra element $c$. But we know this cannot be the case: $\mathsf{PA}$ proves for example that every number has a distinct successor, so $c+1$ needs to be something other than $c$ (and clearly can't be a standard natural). The usual ways of showing that a nonstandard model exists do involve introducing a new constant $c$ and building a model around it, but that doesn't mean that that model consists only of $c$ and the standard numbers.

Indeed, there is no simplest nonstandard model of $\mathsf{PA}$ in any sense I can think of, and Tennenbaum's theorem in fact rules out nicely-describable models altogether in a fairly strong way.

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I suspect the confusion arises from considering the "snappy" construction of a nonstandard model of $\mathsf{PA}$:

Let $T$ be a complete consistent theory in the language of arithmetic + a new constant symbol $c$, containing $\mathsf{PA}\cup\{c>\underline{n}: n\in\mathbb{N}\}$. By the completeness theorem, $T$ has a model $M$; but $M$, or rather the reduct to the language of arithmetic of $M$, is clearly a nonstandard model of $\mathsf{PA}$.

The issue is that this hides a hugely complicated step: the actual model consruction invoked when we apply the completeness theorem. To see what's going on here, and why it doesn't result in a simple picture at all, we have to look at how it's proved. I think the Henkinization argument is best here (Godel's original argument was different):

Henkin argued as follows. For any theory $T$ whatsoever, we can look at the set $Term_T$ of closed terms in the language of $T$ modulo $T$-provable equality, and in a natural way turn that set into a structure $\mathfrak{Term}_T$ in the language of $T$. This is the term structure of $T$.

However, in general we do not have $\mathfrak{Term}_T\models T$ - this is a fun exercise. To get around this, Henkin now turns to the task of proving the following:

Every consistent theory $T$ has some extension $S$ - possibly in a larger language - such that $\mathfrak{Term}_S\models S$.

A fortiori (the reduct to the language of $T$ of) the structure $\mathfrak{Term}_S$ will satisfy $T$, so if we can show this we'll be done. Note by the exercise above this reduct is in general not the same as $\mathfrak{Term}_T$: $S$ is playing an essential role here.

The passage from a theory $T$ to the structure $\mathfrak{Term}_T$ is totally explicit; however, the passage from a theory $T$ to an appropriate extension $S$ if necessary is not. It's this second step which is elided in the snappy construction above, and the elision of which may lead one to underestimate the complexity of the resulting situation.

(For a bit more on this, see here.)

Answer 3

The appearance of nonstandard models of Peano arithmetic can be understood intuitively as an entropic effect. You see, one of the Peano axioms is called the induction axiom. It gives us sufficient conditions to determine if a set of natural numbers encompasses all of them - namely:

• Any set of natural numbers $A$, which has the two properties that $0 \in A$, and for every $n \in A$, its successor $S(n)$ is also in $A$, contains all the natural numbers.

In effect, this process is precisely to "weed" such non-standard numbers out by taking each way you can form, in some candidate for a model of the natural numbers, such a set containing 0, some natural numbers and all their successors, and then throwing away whatever numbers aren't included in it, and repeating this over an over (technically, a "supertask"), until all the unwanted numbers are flushed.

Non-standard models appear when we weaken this statement from "any set" to "any describable set" in some description language. (Typically, this description language is first-order logic.) In this case, there are "indescribable" sets, i.e. random sets (hence "entropic", from before) in some regard, that have no better way to describe them than to list out every element, or do some other equivalently infinitely detailed process. But "describability" implies a finite description, hence those sets are not covered. Some nonstandard numbers make it through.

That also answers your question regarding what such a model "looks like". We can't really write one down explicitly, for this very reason: the sets containing the nonstandard numbers, while also meeting the hypotheses of the above principle, have to be sneaky enough in what they do and don't include in order to "evade" every single statement we could pose in our description language that might suss out the nonstandard numbers in them. Worse, we can't just switch to another, because it can be shown that answering this kind of question about languages is one of those "undecidable" problems, in the same sense as the Halting Problem. No physically possible computing machine can perform the necessary computation to generate such a model.

However, all is not entirely hopeless. Consider the case of a countable nonstandard Peano model - it has "as many" (in the sense of infinite cardinality) naturals as the standard one, but it also includes nonstandard numbers. One can prove that the basic order structure of the model is that beyond all the natural numbers, there is a "furry infinityey nebulaline" of nonstandard, "infinitely big" natural numbers looming out there that consists of doubly-unended chains of such that are themselves ordered densely in the same way the rational numbers are. In particular, you might be tempted to say that

1. a standard natural number is of the form $n$, for $n$ a natural number,
2. a nonstandard natural number is of the form $0_q + m$, where $q$ is a rational number and $m$ is an integer.

The "origin" constants $0_q$ are ordered so that if $q_1 < q_2$ as rationals, then $0_{q_1} < 0_{q_2}$, and every $0_q$ also is beyond every natural $n$, i.e. $n < 0_q$ for every standard $n$.

What is indescribable is what the subscripts for sums $0_{q_1} + 0_{q_2}$ and products $0_{q_1}0_{q_2}$ have to be, to ensure that the weakened induction axiom holds.

(This also works for uncountable models, too - only here now the index set of constants becomes some larger, uncountable set, and the "nebula" becomes denser, hazhyer, and worse. Only now, it's even worse to try and "describe" the sums and products.)