How can induction work on non-standard natural numbers?

Question

When we consider the Peano axioms minus the induction scheme, we can have strange, but still quite understandable models in which there are "parallel strands" of numbers, as I imagine in the picture below:

$\quad\quad\quad$

This mental image makes it at least plausible that induction might not work in all models of this set of axioms: "knocking over the domino at zero, there is no reason that any domino in the parallel strand will ever fall over".

But if we add the induction scheme, we still have non-standard models besides $\Bbb N$. I cannot wrap my head around how any of these might work. Yes, I know, you append $\Bbb Q$-many copies of $\Bbb Z$ and so on. But the real problem for me is, how can induction work (that is, prove statements about all non-standard numbers) if "no domino in a parallel strand is ever tipped over by the chain of dominos starting from $0$"? In such a model, is it just "coincidence" that all statements satisfied by $0$ and its successors also hold for the additional numbers?

Answer 1

If $M$ is a nonstandard model of the Peano axioms then the induction scheme applies to "proofs by induction" that can be proved from PA. So it is not quite true that "all statements satisfied by $0$ and its successors also hold for the additional numbers", depending on what you mean by "statement" and "holds". (However, since you distinguish between the "successors of $0$" and "other numbers", I assume that in this case by "successors of $0$" you mean those numbers in the same successor chain of $0$.) Here are some examples.

Example 1. Let $a$ be a nonstandard element of $M$ and consider the statement $P(x)$ defined by $x<a$. Then $P(x)$ holds in the model $M$ for $0$ and its successors, but not for every element in $M$.

In that example, the statement does not preserve the successor function since $P(a-1)$ holds but $P(a)$ fails. (Thanks to AlexKruckman for clarifying this.)

Example 2. Let $P(x)$ be the statement "$x=s^n(0)$ for some (standard) $n\geq 0$", where $s$ denotes the successor function. Then, $P(x)$ is true for $0$ and all of its successors, but not true of every element in $M$.

Note that this example precisely illustrates how something can hold for all dominoes knocked over from $0$, but no dominoes in a parallel strand. But in this case I've cheated because $P(x)$ is not a first-order statement.

Example 3. Let $Q$ be a first-order sentence that is true in $\mathbb{N}$ but not provable from PA, and consider the statement $P(x)$ defined by $(x=x)\wedge Q$. Then, in the standard model $\mathbb{N}$, $P(x)$ is true of $0$ and all successors. But $P(x)$ may not be true of any element in $M$.

In the last example, we know that $\mathbb{N}$ satisfies $\forall xP(x)$. This would pass to $M$ if $M$ were a model of the complete theory of $\mathbb{N}$. But since $Q$ is not provable from PA, there is a model $M$ of PA such where $Q$ fails, and in such a model we would in fact have $\forall x\neg P(x)$.

The accurate statement is as follows.

Suppose $P(x)$ is a first-order statement (without parameters) and PA proves that $P(x)$ holds for $0$ and is preserved under successors, i.e., $PA\vdash P(0)\wedge \forall x(P(x)\rightarrow P(x+1))$. Then $P(a)$ holds for every $a$ in $M$.

The previous statement is true, but it is no "coincidence". Indeed, PA contains the axiom $$ (P(0)\wedge \forall x(P(x)\rightarrow P(x+1)))\rightarrow\forall xP(x) $$ and so the combined assumptions imply that PA proves $\forall x P(x)$. So this holds in $M$ since $M$ is a model of PA. In other words, there is no point in trying to extend the "domino" metaphor in this case, because the reason $\forall x P(x)$ holds in $M$ is simply because $M$ is assumed to be a model of PA, and hence satisfies every theorem that PA can prove.

Answer 2

I think the big problem here is posing the question faithfully.

In one sense of course (and I think the other answers are addressing this) it's no coincidence that nonstandard models of $\mathsf{PA}$ satisfy the first-order induction scheme: that's literally part of the definition of "model of $\mathsf{PA}$." But despite this not all models feel the same. Per the domino picture, the (second-order) principle of well-foundedness provides a "satisfying justification" for the first-order induction scheme in $\mathbb{N}$. Nonstandard models of $\mathsf{PA}$ - which of course includes the first-order induction scheme - are obviously not well-founded. Put another way:

No nonstandard model of $\mathsf{PA}$ can satisfy the first-order induction scheme for "the right reason."

And this is exactly the cost of trying to "first-orderize" a categorical description of an infinite structure: you'll wind up (per compactness) opening the door to structures which satisfy your approximate version for wrong reasons.

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So as soon as we see that well-foundedness characterizes $\mathbb{N}$ up to isomorphism, we know that any of its "first-orderizations" will have "unintended models." Should we use the term "coincidence" at this point? I'm not sure: unintended phenomena are not always coincidences!

I think one reasonable question which arises at this point is whether some first-order approximation to true well-foundedness could itself be a higher-order principle of interest on its own:

Are there any "reasonably simple" second-order sentences which (over the discrete ordered semiring axioms, say) imply the first-order induction scheme but which are strictly weaker than well-foundedness?

(The "reasonably simple"-bit is important: "Every first-order $\mathsf{PA}$ axiom is true" can be expressed by a single second-order sentence, but is pretty silly.) Tentatively I'll say no, that the only "natural" justification for the first-order induction scheme is the idea of genuine well-foundedness. At the same time I can't quite bring myself to use the word "coincidence" here: the passage from second-order to first-order induction is definitely not arbitrary, and so being a nonstandard model of $\mathsf{PA}$ still feels very meaningful to me (moreso than, say, being a model of $I\Sigma_{17}$ - why $17$ and not $18$?). But here we move into the realm of mathematical aesthetics, so I'll stop there.

Answer 3

See Boolos etc., Computability and Logic (5th ed, 2007) Ch.25 Nonstandard Models, page 304:

the elements of the domain of any nonstandard model of arithmetic are going to be linearly ordered by LESS THAN. This ordering will have an initial segment that is isomorphic to the usual ordering of natural numbers, followed by a sequence of blocks, each of which is isomorphic to the usual ordering of the integers (negative, zero, and positive). There is neither an earliest nor a latest block, and between any two blocks there lies a third. Thus the ordering of the blocks is what was called a dense linear ordering without endpoints, and so, as shown there, it is isomorphic to the usual ordering of the rational numbers.

And see page 303:

The standard NUMBERS are precisely those that can be obtained from ZERO by applying the SUCCESSOR operation a finite number of times.

Thus, in a nutshell, induction works because it is an axiom. "Domino picture" (every number can be "computed" starting from $0$ after a finite number of steps) is not an axiom.

Also worth resading is: Martin Goldstern & Haim Judah, The Incompleteness Phenomenon: A New Course in Mathematical Logic, Ch.2.3 Nonstandard Models of Arithmetic.