I am beginning real analysis and got stuck on the first page (Peano Postulates). It reads as follows, at least in my textbook.

Axiom 1.2.1 (Peano Postulates). There exists a set $\Bbb N$ with an element $1 \in \Bbb N$ and a function $s:\Bbb N \to \Bbb N$ that satisfies the following three properties.

a. There is no $n \in \Bbb N$ such that $s(n) = 1$.

b. The function $s$ is injective.

c. Let $G \subseteq \Bbb N$ be a set. Suppose that $1 \in G$, and that $g \in G \Rightarrow s(g) \in G$. Then $G = \Bbb N$.

Definition 1.2.2. The set of natural numbers, denoted $\Bbb N$, is the set the existence of which is given in the Peano Postulates.

My question is: From my understanding of the postulates, we could construct an infinite set which satisfies the three properties. For example, the odd numbers $\{1,3,5,7, \ldots \}$, or the powers of 5 $\{1,5,25,625 \ldots \}$, could be constructed (with a different $s(n)$, of course, since $s(n)$ is not defined in the postulates anyway). How do these properties uniquely identify the set of the natural numbers?

  • 17
    $\begingroup$ You don't have to be a mathematician to realize that many sets obey the Peano axioms. Most people are accustomed to the fact that both the set of strings $\{1, 2, 3, ...\}$ and the set of strings $\{\text{one, two, three}...\}$ satisfy the axioms if you put the right successor functions on them. $\endgroup$
    – Jack M
    Nov 20, 2019 at 22:33
  • 12
    $\begingroup$ What you did in your examples is essentially just renaming the elements of the set. Function $s(n)$ has a symbol $s$ because it is a successor function. When axiomatically defining the set $\mathbb N$ we don't care about the elements' values yet, just the set's structure, implied by 'being a successor of...'. And all your examples have the same structure (are isomorphic). $\endgroup$
    – CiaPan
    Nov 21, 2019 at 12:24
  • $\begingroup$ What book is that, out of curiosity? I've kind of always wanted an analysis book that started that way. $\endgroup$ Nov 21, 2019 at 18:56
  • 2
    $\begingroup$ @DanielR.Collins: A Google search for "Axiom 1.2.1 (Peano Postulates)" indicates that it's Bloch, Ethan D., The Real Numbers and Real Analysis, Springer (2011): link.springer.com/book/10.1007/978-0-387-72177-4 $\endgroup$
    – ruakh
    Nov 22, 2019 at 5:08
  • 4
    $\begingroup$ But both sets are describeable as $\{1,s(1),s(s(1)),s(s(s(1))),\ldots\}$. We then tend to define $2:=s(1)$, $3:=s(2)$, etc. $\endgroup$ Nov 22, 2019 at 7:48

2 Answers 2


Yes, you can find other sets on which a successor function is defined that satisfies all the Peano axioms.

What makes the natural numbers unique is that you can use the Peano postulates to prove that when you have two such sets you can build a bijection between them that maps one successor function to the other. That means the sets are really "the same" - the elements just have different names.

So you might as well use the traditional names $ 1, 2, 3,\ldots$.

  • 5
    $\begingroup$ This isn't true... You can build a model of PA with any infinite cardinality (meaning, no bijection). We might kind of sort of want to say that all the models "agree" where they're all defined, but even that is really tricky to do. $\endgroup$
    – nomen
    Nov 21, 2019 at 18:28
  • 1
    $\begingroup$ @nomen That's a fair critique of the answer. I've only casual knowledge of mathematical logic / model theory at that level. I think the answer is (essentially) the right one for a student starting to study real analysis beginning with the natural numbers but accepting some standard unspecified set theory. $\endgroup$ Nov 21, 2019 at 22:08
  • 10
    $\begingroup$ @nomen: The question isn't asking about PA. The axioms listed uniquely define the natural numbers (at the price of requiring a set quantifier to state). $\endgroup$
    – Henry
    Nov 21, 2019 at 22:24
  • 2
    $\begingroup$ @nomen You're using PA to mean "first order Peano arithmetic". This is pretty standard nowadays among logicians, but actually Peano originally formulated his axioms in second order logic. Given a surrounding set-theory context (either informal, or say ZF), the second-order Peano postulates characterize $\mathbb{N}$ up to isomorphism. $\endgroup$ Nov 26, 2019 at 20:23

This axioms define the natural numbers up to isomorphism; That is, given another Peano system $S$, there always is a bijective function $f: \mathbb{N}\to S$ such that $f(s(n))=s_1(f(n)); f(1)=1_S$. This substantially means that from the Peano's structure point of view, the two are equal. Given $1, 1_S$, we can just identify as equal the two of them and the other numbers will follow, as if we called the natural numbers with different symbols $1_S$ instead that $1$, but the structure behind is the same (this theorem can actually be formally proved by using the principle of recursion)

A little note, though: this isomorphism is sometimes not very evident end extremely helpful. For example, $\mathbb{N}^2$ can be given a Peano structure, and this can be identified with $\mathbb{N}$ by means of an isomorhpism (namely Cantor pairing function), which is highly non trivial.

  • 1
    $\begingroup$ @user10186512 You are right: I didn't include it in the answer though, since it seems to me that the op is only beginning analysis $\endgroup$
    – user515010
    Nov 21, 2019 at 11:59
  • 6
    $\begingroup$ @nomen Your claim is true only if you take the first order PA (and in such a case, it follows from skolem theorem). However, the 3° axiom stated by the O.P., the principle of induction, is not stated (and neither can be reduced) to a formula in first order language: to see why, note that the op formula ranges on all the subsets of $\mathbb{N}$, while a first order formula is only allowed to vary on a numerable set. In fact, it's not only the isomorphism that does not follow in a first order logic, is the principle of recursion itself. $\endgroup$
    – user515010
    Nov 21, 2019 at 17:39
  • 1
    $\begingroup$ @nomen In ZFC no set contains all of the Neumann ordinals (thanks to the Burali-Forti paradox), and thus it does not seem to me that they can be a model of PA. $\endgroup$
    – user515010
    Nov 21, 2019 at 18:01
  • 5
    $\begingroup$ @nomen Why do you say that it is arbitrary to point out that second order Peano Arithmetic is categorical? It is simply a theorem that up to isomorphism there is only one model of second-order PA. For the first order version of the theory, it is indeed easy to come up with nonstandard models. It could be useful if you add an answer to that effect. $\endgroup$ Nov 21, 2019 at 18:51
  • 2
    $\begingroup$ @nomen Regarding the ordinals being a model of PA, doesn’t $\omega$ violate 3? It contains 0 and is closed under successor, but $\omega \neq \text{Ord}$. $\endgroup$
    – user76284
    Nov 21, 2019 at 20:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .