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I know perfectly well the Peano axioms, but if they were sufficient for defining $\mathbb N$, there would be no controversy whether $0$ is a member of $\mathbb N$ or not because $\mathbb N$ is isomorphic to a host of sub- and supersets of $\mathbb Z$.

So it seems that mathematicians, when specifying $\mathbb N$, rarely mean the set of natural numbers according to its definition but rather some particular embedding into sets defined by different axiomatic systems.

Doesn't this sort of misstate then the role of the Peano axioms as the defining characteristics of $\mathbb N$? It seems like most of the time the Peano axioms are not more than a retrofit to a subset of $\mathbb Z$ or other sets defined by a different axiomatic system with one or several axioms incompatible with the Peano axioms.

Asaf Karagila
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user266061
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    The decision to include or exclude zero as a "natural number" is not a problem for the foundations of mathematics. Some authors prefer to include it, some exclude it, and some (like myself) are inconsistent in finding it convenient for some applications to have it, and other times not. As long as the author makes the decision clear for purposes of a discussion/publication, either approach is okay. – hardmath Sep 01 '15 at 13:39
  • I think you're going to run into a problem assuming there's a standard definition that anyone uses. Most mathematicians just run with either ${1,2,...}$ or ${0,1,2,...}$ based on what's most useful at the moment and don't think much more about it. – walkar Sep 01 '15 at 13:41
  • If $0$ occurs in your Peano axioms the $0\in\mathbb N$. If you replace $1$ for $0$ in them, then $0\notin \mathbb N$. – Hagen von Eitzen Sep 01 '15 at 13:41
  • Perhaps I could say "a number is a natural number if it's the sum of four squares." (It would be odd to take it as a definition, but it could be taken as a definition if you want it to.) That particular definition is interesting in that it only uses the concepts of $+$ and $\times$, rather than something like $>$. (Of course, by this definition, $0$ is natural. You could easily change it if you don't want to count $0$ as a natural.) – Akiva Weinberger Sep 01 '15 at 18:16
  • @columbus8myhw: So every complex number is a natural number? :-) – Asaf Karagila Sep 01 '15 at 19:16
  • @AsafKaragila Meh, I wrote that I was working in $\Bbb Z$, but then I wanted to edit some stuff before pressing 'enter' and I accidentally got rid of that comment. Pfft. (In any case, $\Bbb N$ isn't first-order definable in $(\Bbb C,+,\times)$, is it?) – Akiva Weinberger Sep 01 '15 at 19:21
  • @columbus8myhw: No, the theory of the complex numbers is complete, decidable and consistent. So it cannot possibly interpret enough of the arithmetic of the natural numbers. In any case, context is important of course. I'm just being a smartypants. :-) – Asaf Karagila Sep 01 '15 at 19:23
  • @AsafKaragila Really? It's complete and decidable? I need to go learn some model theory. – Akiva Weinberger Sep 01 '15 at 19:25
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    @columbus8myhw: $\mathsf{ACF}_p$ is complete for every choice of characteristics (prime or $0$, of course). And the theory really just says that you have a field (of a certain characteristics), and that every polynomial has a root. This can be easily done with finitely many axioms and two schemata (one for $p>0$). So it's easily complete and recursively enumerable. But those two things together imply decidability! – Asaf Karagila Sep 01 '15 at 19:35

6 Answers6

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How is a chair defined? Surely something as simple and fundamental as a chair should have a reasonable definition.

In the dining room, the table is certainly not a chair. Because you don't eat from a chair. But in the university lobby, sometimes I would sit on the table when talking to someone who's standing up. And maybe I decided to sit on the floor and eat off the chair (with plates, mind you), because I needed to hide from someone.

So what is a chair? Does the number of legs define it? Does its location in the house define it? Does anything else?

Well. The honest truth is that a chair is an abstract idea, and it is not universally and globally well-defined. At some times a chair is something like this, and at other times it could be a completely different thing. The important thing is that within the context of the conversation, if I say "chair" it will be clear what I meant by that term.

And now we finally approach the point I am trying to make. $\Bbb N$ is an abstract object. When I talk to a fellow set theorist, it is clear that when I say $\Bbb N$, I mean the set of finite von Neumann ordinals (often called $\omega$). When I talk to someone who's doing a lot of category theory it is clear, at least to him I guess, that $\Bbb N$ is the initial object in the category of free monoids. When I talk to someone whose main interest is analysis, she might have a clear and intuitive as what are the real numbers, and from that she will derive the meaning of $\Bbb N$ as the smallest inductive set.

Sometimes it will be beneficial and useful to include $0$ as a natural number. For example if a set is finite if and only if its cardinality is a natural number, then surely $0=|\varnothing|$ is a natural number.

Sometimes it will be cumbersome to include $0$, when I was to talk often about sequences that look like $\frac1n$ having to add "for $n>0$" everywhere is a terrible crime against yourself, the reader and the listener. So just excluding $0$ works better.

Since you brought up the Peano arithmetic axioms, it is perfectly reasonable to develop those in a language which include $0$, and in a language which does not include $0$ (in which case $1$ is the minimal element, and you need to modify a few axioms accordingly).

So what is $\Bbb N$? It is an abstract idea, which you can implement, or realize in many various ways. Some are more concrete, some are less concrete. But it is an abstract object, that you know satisfy some properties, and perhaps the most important one is that it has a linear ordering that (1) has no maximal element; and (2) every proper initial segment is finite.

Why am I choosing the order here rather than the arithmetic structure? Because from this order we can define (in the broad sense of the word) the arithmetic structure, and everything else. With or without $0$. And this order gives us a uniqueness property. Every two linear orders with these two properties are isomorphic (and the isomorphism is unique, too).

So it really doesn't matter what is the set $\Bbb N$, because there are way too many options for interpreting $\Bbb N$ as a concrete set. It matters that we can use induction on that set, and that we know there is at least one set like that.

Asaf Karagila
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  • "When I talk to someone who's doing a lot of category theory it is clear, at least to him I guess, that $\mathbb{N}$ is the initial object in the category of free monoids." I enjoyed this statement immensely. A small correction though: "$\mathbb{N}$ is the initial object in the category of pointed monoids" or, better yet, "$\mathbb{N}$ is the free monoid on a point." Unfortunately, the language of universal properties can often hide the simplicity of the underlying idea: in this case, the idea is that $$\mathbb{N} = {0,1,1+1,(1+1)+1,1+(1+1),(1+1)+(1+1),((1+1)+1)+1,\ldots}$$ with an... – goblin GONE May 19 '16 at 05:53
  • ... appropriately chosen notion of equality; for example, we want $(1+1)+1$ to equal $1+(1+1)$, etc. For some reason, although students of category theory usually think about things as sets of formal expressions with appropriately chosen equality relations, we're in the nasty habit of hiding this intuition under the language of universal properties. I think we should just be honest, and admit that we're thinking about structures as sets of formal expressions. – goblin GONE May 19 '16 at 05:54
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    I hate universal properties... :-P – Asaf Karagila May 19 '16 at 05:57
  • I'm sympathetic :) – goblin GONE May 19 '16 at 05:58
  • To me, or to universal properties? – Asaf Karagila May 19 '16 at 05:59
  • I'm sympathetic to you. I dislike UP's too, except when I can rethink what's going on via sets of expressions with appropriately chosen equality relations. Otherwise, I just get really confused and mostly can't prove anything non-trivial from the UP. – goblin GONE May 19 '16 at 06:08
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Well, usually they do mean "the set," just different mathematicians mean different sets. It's like a language difference. There are lots of cases in math of this sort. For example, "the standard normal distribution" means something different to different mathematicians. Basically, good luck trying to sort this out.

Historically, the natural numbers were essentially the ordinals. So, if you could say "I came in $n$th place in a race" then $n$ was a natural number.

The distinction between ordinal number and cardinal (counting) number is essentially the logical error that led to problems with the number zero. When you count a pile of beans like a child would, you are assigning an ordinal to each bean. "This is the first bean, this is the second bean, etc." The last ordinal used is then treated like a cardinal number.

(Aside: It is actually not 100% intuitive that this counting technique works - that no matter how we assign the ordinals to a finite set, we get the same last ordinal. We need to be taught that this works. There is a Sesame Street video that counts kids on the screen, then has the kids run around into a different configuration, and counts them again - essentially assigning different ordinals to each kid each time, demonstrating that the final ordinal is always the same, no matter how you assign them. This counting technique breaks when trying to count infinite sets.)

Since there was no ordinal $0$, the difficulty with "counting" empty sets was that there was no last bean in the process. So they'd simply say, "there are no beans," rather than giving a count of zero.

Set theory starts ordinals at $0$, instead, for a variety of reasons. This means that, in a race, the person who places $n$th was beaten by $n$ people. So $0$th place would be beaten by zero people, $1$st place would be beaten by one person, etc. Nobody but mathematicians treats zero as an ordinal, but it does have its advantages. In set theory, the $n$th ordinal is a set of the previous ordinals - you can see that as "my race result is the set of results of the people who beat me."

So in set theory: $\emptyset$ is an ordinal, and if $x$ is an ordinal, then $x\cup\{x\}$ is the "next ordinal" - the person in the race with the result directly after me was beaten by me and everybody who beat me. Set theory assumes there exists a set closed under this operation $x\mapsto x\cup\{x\}$, and then shows there is a unique minimal such set, which is called the natural numbers - minimality essentially being equivalent to induction.

(Then we get into infinite cardinals and ordinals, where the relationship becomes way more complicated. Race results, for example, is really only a good metaphor in the finite case.)

Thomas Andrews
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  • I think of an ordinal more like of a queue itself rather that of particular places in the queue, even though the particular places are standardly encoded by all strictly smaller ordinals. I.e. there is nothing like “0th”, but zero as ordinal represents an empty queue. – user87690 Sep 01 '15 at 14:22
  • That works, but I think it is much more natural to compare two people in a queue than compare two queues with potentially different people. For example, a "successor queue" can certainly be defined the queue resulting from adding an element to the end, but finding of one queue's ordinal is the successor of another's is a bit obscure. It's certainly not a bad alternative, but it has trade-offs. @user87690 – Thomas Andrews Sep 01 '15 at 14:28
  • Typically an ordinal means both – a queue type and a place in a queue. But $0$ means empty queue and also first place in any nonempty queue, but not zeroth place. – user87690 Sep 01 '15 at 14:41
  • @user87690: You can combine these together by asking "what is the type of all those preceding a place in the queue?", then $0$ is exactly the type of the set of predecessors of the zeroth place. – Asaf Karagila Sep 01 '15 at 14:44
  • @AsafKaragila: I would say that $0$ is exactly the type of the set of prececessors of the first place. – user87690 Sep 01 '15 at 14:48
  • @user87690: If you look at the line to the bathroom in a party, then you need to ask yourself. Is $0$ the guy first in line, or the guy that in the bathroom? (in either case, I wouldn't want to be $\omega$.) But this is just a silly debate about whose convention is better. And unfortunately(?) I have other things to do, so maybe next time you're in town we can hammer this over a beer. – Asaf Karagila Sep 01 '15 at 14:51
  • I'm personally agnostic on the matter. I don't think there is a "better." As a pedagogical question, it depends on the student, and having multiple metaphors can be better than having one. @AsafKaragila – Thomas Andrews Sep 01 '15 at 14:57
  • @AsafKaragila: Yes, I know it's usually a silly debate about conventions. I just thought that the usual conflict is whether the first element of a sequence has index zero or index one, but I (maybe naively) thought that there is no or little disagreement in that the first element of a sequence is first and not zeroth. – user87690 Sep 01 '15 at 15:01
  • @user87690: $0$ is the person using the bathroom. Ask the first guy in line "How many people are before you in the queue?" and he will answer $0$. Ask him "How many people need to finish using the bathroom before you can enter?" and he will answer $1$. I hope this makes it clearer, and that's all the time we have for now... :-) – Asaf Karagila Sep 01 '15 at 15:02
  • About "the distinction between ordinal number and cardinal (counting) number [and] the logical error that led to problems with the number zero", and the "silly debate about conventions" ... There is a point of view according to which $1$ is the number of satellites of Earth, while $0$ is the number of satellites of Mercury: clearly, the two are not the same and seem not (to me) "conventional" at all. – Mauro ALLEGRANZA Sep 02 '15 at 09:48
  • @MauroALLEGRANZA You are confusing things The "silly debate" in comments was about queues, and what it means to be the zeroth member of a queue, I believe. The quoted text from my answer is about why the ancients had a problem with treating $0$ as a cardinal number - namely, a confusing of cardinals with ordinals due to the way we use ordinals to count cardinals. – Thomas Andrews Sep 02 '15 at 10:39
  • @MauroALLEGRANZA Nobody is saying that $0$ is not a number. The question of whether and why we have two notions of natural numbers goes back to the reason there is a distinction between cardinal and ordinal numbers. – Thomas Andrews Sep 02 '15 at 10:52
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The original Peano's axioms (1889, page 1) started from $1$ :

Axiomata

$1 \in N$ [...].

This "tradition" is followed by many mathematicians, like :

Axiom 1 : $1$ is a natural number.

In the mathematical logic tradition, following G.Frege (see e.g. The Foundations of Arithmetic (1884), transl J.L.Austin, page 90) and B.Russell (see Principles of Mathematics (1903), §120. Peano’s indefinables and primitive propositions) is common to "start from" $0$.

See : Kurt Gödel (1931). "Über formal unentscheidbare Sätze der Principia Mathematica und verwandter Systeme, I". Monatshefte für Mathematik 38, page 177 :

The primitive signs of the system $P$ are the following:

I. Constants: "$\sim$" (not), "$\lor$" (or), "$\Pi$" (for all), "$0$" (zero), "$f$" (the successor of), "(", ")" (parentheses); [...]

The following formulas (I-V) are called axioms [...] :

I.1. $\sim (fx_1 = 0 )$,

[...].

You can see also :

Mauro ALLEGRANZA
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I don't think that Peano axioms are sufficient for defining $\mathbb{N}$. For one thing, Peano axioms form a theory, not a particular structure. Additionally, the theory of Peano arithmetics is not complete, so it even doesn't define a structure up to elementary equivalence.

On the other hand, there is a standard model of Peano arithmetics in the standard set theory (ZFC), namely the smallest infinite ordinal – as described in other answers.

Also, you can easily construct a structure that is different but isomorphic to $\mathbb{N}$. But that is trivial and matematitians often think of structures “up to isomorphism”. So they often do not make difference for example between zero as a natural number and zero as an integer, even though they are strictly speaking different objects (different sets) according to the standard set-theoretic formal construction.

user87690
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In the set theory course I took, natural numbers were not defined by Peano's axioms. One can define an inductive set in this manner: $I$ is inductive set if

$$\emptyset\in I\ \wedge\ \forall x\in I\ ((x\cup \{x\})\in I) $$

We say that $n$ is natural number if $n\in I$ for any inductive set $I$. Some examples of natural numbers are:

\begin{align} 0 &= \emptyset\\ 1 &= \{\emptyset\}\\ 2 &= \{\emptyset,\{\emptyset\}\}\\ &\ \ \vdots\\ n &= \{0,1,\ldots ,n-1\} \end{align}

But, we still don't know if the set of all natural numbers exists. This is guaranteed by Axiom of infinity:

There exists an inductive set.

and with this we can prove existence of set $\mathbb N = \{n\ |\ n\ \text{is natural}\}$. Also, we can see that $\mathbb N$ is the smallest inductive set in the sense of set inclusion.

Ennar
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You have to start somewhere.

Some treatments of mathematics start with Peano's axioms for the natural numbers, and use that to define other number systems which extend the natural numbers --- first integers, then rational numbers, real numbers, complex numbers,...

But that's not the only choice. You could equally well decide to start with the axioms for the real numbers, and use that to define other number systems: first natural numbers, integers, and rationals within the real numbers, then the extension to complex numbers …

The above two methods can be thought of as dependent on "naive set theory". Because of that, perhaps a more theoretically satisfying place to start is with axioms for set theory such as ZF (the Zermelo Frankel axioms) or ZFC (the Zermelo Frankel axioms with the axiom of Choice tacked on). In these treatments, the natural numbers are usually defined as the set of finite ordinals (the existence of which is one of the ZF axioms, and the Peano axioms must then be proved), in which case the number zero, which is defined as the empty set, would be included. Alternatively, the natural numbers could be defined as the finite nonempty ordinals, in which case zero would not be included.

As you can see, there's different choices of where to start, and different people may make different choices.

Lee Mosher
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  • But you actually cannot start with real numbers and then define naturals, since the set of natural numbers is not definable in the reals, is it? – user87690 Sep 01 '15 at 14:30
  • @user87690: But you can. The natural number $1$ is the multiplicative identity of $\mathbb{R}$, and the set of natural numbers $\mathbb{N}$ is defined to be the intersection of all subsets of $\mathbb{R}$ that contain $1$ and are closed under addition. – Lee Mosher Sep 01 '15 at 14:32
  • But then you are quantifying over sets. By definable I meant first-order defineable. I.e. I meant that you cannot start just with reals without an underlying set theory. On the other hand, you can start just with Peano arithmetics and define everything including set theory in it (if I'm not wrong). – user87690 Sep 01 '15 at 14:37