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While thinking of natural numbers as anything that satisfies the Peano axioms, I was left wondering, what if I take the successor function $S(x)$ to be anything other than $x\to x+1$?

Some examples were
$0,1,2, \ldots $
Zero, one, two, $\ldots$
$\{\}$, $\{\{\}\}$, $\ldots$

I started seeing the natural numbers as a category of objects linearly connected by the successor function as the arrow. And all of the different relabelings of the objects as essentially equivalent.

So I started thinking of more examples of the same linear structure, for example, initial element $1$ with $S(x) = 2x$. This can form a sequence $1,2,4,\dots$, yet this does not correspond to what we generally consider to be the natural numbers.

I went further. What stops me from considering bijections from the naturals to other sets, such as the integers or the rational numbers? Yet we say that the rational numbers are not the same as the natural numbers.

What is going on?

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    $\begingroup$ Hi Povilas -- this site is for research level mathematics, and as such this post is off-topic here. That being said you may have better luck over at math.stackexchange; good luck! $\endgroup$
    – Alec Rhea
    Commented Jul 12, 2023 at 1:25
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    $\begingroup$ If Joel is fine with it I am too. (+1 and apologies for the misunderstanding, although I was never a downvoter :) $\endgroup$
    – Alec Rhea
    Commented Jul 12, 2023 at 2:31
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    $\begingroup$ While it was probably appropriate for this question to be closed, I hope the discussion shows you that it was a good question, just probably not suitable for MO. I hope you'll keep following the impulses that lead you to ask these questions; they're good questions, and soon enough will lead you into new and interesting questions that are appropriate for MO. Best of luck! $\endgroup$
    – LSpice
    Commented Jul 12, 2023 at 14:51
  • $\begingroup$ @AlecRhea, while this question may not strictly speaking meet the relevant guidelines, since it did receive an excellent answer, would it be worth trying to reopen it? $\endgroup$ Commented Jul 13, 2023 at 8:34
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    $\begingroup$ Whether or not the question is re-opened, @‍Povilas, I think you can still accept the answer, and it might be nice to do so (since it's hard to imagine it being improved upon!). $\endgroup$
    – LSpice
    Commented Jul 13, 2023 at 14:03

1 Answer 1

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The main lesson is that it doesn't matter at all which particular objects you take as the numbers and what function you use as the successor function, as long as your system fulfills the right structural features.

It was Dedekind who first identified this structural theory in 1888 and proved his famous categoricity theorem, which showed that there is up to isomorphism only one model of his theory. His axioms were as follows:

Dedekind arithmetic. For a set $N$ of objects that we will call the natural numbers, with a distinguished element $0$ in $N$ called zero, and a unary function $S:N\to N$ called the successor function, the following three axioms:

  • zero is not a successor, that is, $0\neq Sx$ for all $x\in N$
  • the successor function is one-to-one, that is, $Sx=Sy\to x=y$, and
  • the successor function generates every number from zero, in the sense that if $X\subseteq N$ is a set for which $0\in X$ and $x\in X$ implies $Sx\in X$, then every number is in $X$, that is, $X=N$.

Notice that the axioms do not refer to any essential properties of the elements of $N$ or the function $S$---they can be anything at all, as long as the system fulfills the stated structural properties.

But furthermore, as emphasized in the comments by KP Hart, one doesn't really have the natural numbers and the operation $x+1$ and so forth until after introducing Dedekind's theory and then developing number theory on the basis of it, defining $x+y$ and $x\cdot y$ and $x^y$ by recursion, as he did and as Peano did so successfully after him, using his theory.

Theorem. (Dedekind 1888) Any two models of Dedekind arithmetic are isomorphic.

If you have your natural number structure $\langle\mathbb{N},0,S\rangle$ and I have my natural number structure $\langle \bar{\mathbb{N}},\bar 0,\bar S\rangle$, then there will be an isomorphism $\pi:\mathbb{N}\to\bar{\mathbb{N}}$, a bijection of the sets such that $\pi(0)=\bar 0$ and $\pi(Sx)=\bar S(\pi(x))$ for every $x\in\mathbb{N}$.

Indeed, the requirement of the isomorphism tells you exactly how to define $\pi$, namely, by recursion, setting $\pi(0)=\bar 0$ and $\pi(Sx)=\bar S(\pi(x))$. So, in order to prove his theorem, Dedekind first proved that definition by recursion is legitimate in any model of his theory. And then the categoricity result drops out of it.

In my view, Dedekind's theorem is the beginning of the philosophy of structuralism in mathematics, the view that we in mathematics we should treat all our mathematical ideas and structure as invariant under isomorphism. We should not seek to find out what numbers "really" are, to find their essence, but rather to find the structural features that characterize the numbers as we intend them. This is just what Dedekind did.

We now have categorical accounts for all of our familiar structures---the reals are up to isomorphism the unique complete ordered field (Huntington 1904); the complex numbers are up to isomorphism the unique algebraic closure of the real field; the various ranks of the set-theoretic hierarchy also have categorical characterizations (Zermelo 1930). And so on. These categorical characterizations show us how to do mathematics in an isomorphism invariant manner.

There is nothing mathematical at stake in having a different copy of the natural number structure.

(I have noticed that some people are downvoting your question, because perhaps it isn't asked in the best way. But I find the topic both philosophically interesting and deep, and so I have posted this answer. Perhaps the question will get migrated to math.stackexchange.)

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    $\begingroup$ Hi LSpice , the different categorical descriptions of the complex numbers refer to different structures. The one you mention is a characterisation of the complex field as a pure field. Joel's description characterises it as a topological field (because it is a degree 2 extension of the reals, and so uniquely inherits the product topology). $\endgroup$ Commented Jul 12, 2023 at 8:00
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    $\begingroup$ Re, there are many subtleties that I get wrong— @JonathanKirby's comment was particularly helpful in understanding your emphasis here—but I'm pretty sure the complex numbers don't have characteristic two. 😄 $\endgroup$
    – LSpice
    Commented Jul 12, 2023 at 14:49
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    $\begingroup$ D'oh, Oops! Of course I meant characteristic zero. But actually I disagree with Jonathan, since the categoricity result itself shows that the topological aspect is inherently part of the structure. There is only one such field, and it supports that topological structure. The topological nature is not unique, of course, just as there are many copies of the real field to be found within the complex field. $\endgroup$ Commented Jul 12, 2023 at 15:18
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    $\begingroup$ @JoelDavidHamkins: Is there a structuralist view on whether we should consider something like $\mathbb N\subseteq \mathbb Z$ to be true? Of course, this is not literally true under the most common set-theoretic definitions of these structures, but I wonder if a structuralist would say something like: "since the subset relation is not invariant under isomorphism, it is not interesting/relevant/meaningful whether $\mathbb N\subseteq\mathbb Z$: rather, it is relevant that any model of the group $\mathbb Z$ contains a model of Dedekind arithmetic." $\endgroup$
    – Joe Lamond
    Commented Jul 12, 2023 at 21:10
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    $\begingroup$ @Joe In my view, the set-theoretic constructions are not definitions, but existence proofs that a copy of the desired structure can be found. It is clear in set theory that without loss one can have $\mathbb{N}\subset\mathbb{Z}$, if desired for some set-theoretic purpose. And regardless of one's foundations, whether set theory, type theory, category theory, etc., this sort of change in context involves some type-changing aparatus, inherently, when we want to view a given structure as a part of another. It is basically the same move regardless of the foundational system. $\endgroup$ Commented Jul 12, 2023 at 22:40

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