According to page 25 of the book A First Course in Real Analysis, an inductive set is a set of real numbers such that $0$ is in the set and for every real number $x$ in the set, $x + 1$ is also in the set and a natural number is a real number that every inductive set contains. The problem with that definition is that it is circular because the real numbers are constructed from the natural numbers.

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    $\begingroup$ (1) The author of the book would be helpful. (2) The real numbers can be defined axiomatically, without relying on a construction from the naturals. Specifically, the real numbers are the unique (up to isomorphism) complete ordered field. If you define the reals axiomatically, the argument is not circular. $\endgroup$
    – Xander Henderson
    Sep 20, 2017 at 19:13
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    $\begingroup$ Is that actually given as a definition? Could the author just be saying that it is a property of $ \mathbb R $ – while possibly waving his hands about the identification of a subset of $ \mathbb R $ with $ \mathbb N $ and the question of what those number systems “are”? $\endgroup$
    – PJTraill
    Sep 20, 2017 at 21:50
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    $\begingroup$ Sounds as if the authors are Murray H. Protter and Charles B. Morrey Jr. and the book is ISBN: 978-1-4612-6460-6 (Print) 978-1-4419-8744-0 (Online); taken from link.springer.com/book/10.1007/978-1-4419-8744-0 – you might edit that in. $\endgroup$
    – PJTraill
    Sep 20, 2017 at 21:54
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    $\begingroup$ I think the question title needs clarifying. The question has generated answers along two very different lines, one line answering "what is a natural number?" and the other answering "does this definition in the book work?" $\endgroup$ Sep 21, 2017 at 4:19
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    $\begingroup$ If the book is Protter & Morrey, I agree that the exposition is quite weird: 1. assumed existence of the reals (axiom for fields + axiom of inequality); 2. defined an integer as a real that is either zero, a natural or the negative of a natural; 3. defined the naturals. $\endgroup$ Sep 21, 2017 at 9:55

5 Answers 5


We can define the real numbers axiomatically to be the unique (up to isomorphism) complete ordered field. In this setting, $(\mathbb{R}, +, \cdot, \le)$ must satisfy the properties

  1. $(\mathbb{R},+,\cdot)$ is a field (i.e. both of the operations are commutative and associative, multiplication distributes over addition, every element has an additive inverse, and every nonzero element has a multiplicative inverse).

  2. $(\mathbb{R},\le)$ is a totally ordered set (i.e. the relation $\le$ is reflexive, transitive, and antisymmetric, and for any $a,b\in\mathbb{R}$ with $a\ne b$, then exactly one of $a\le b$ or $b\le a$ holds).

  3. The order is compatible with the field structure in the sense that $a \le b$ implies that $a+c \le b+c$ for all $c$, and if $0 \le a,b$, then $0 \le a\cdot b$. (We could actually deduce the properties of an order from these properties plus the comparability axiom above, but I personally find it easier to attack it this way.)

  4. $(\mathbb{R},+,\cdot,\le)$ is complete, in the sense that if a nonempty subset of $\mathbb{R}$ has an upper bound, then it has a least upper bound.

  5. In this context, we probably have to take the existence of the real numbers as an axiom (i.e. the existence of at least one totally ordered field), as well (thank you for pointing that out, DRF). Alternatively, see the addendum.

It is a bit of work to show that if $(X,+',\cdot',\le')$ satisfies these axioms, then $X$ is isomorphic to $\mathbb{R}$, but it can be done. In any case, we can (and often do) define the real numbers axiomatically in this manner.

Note that there is nothing in the above about natural numbers, rational numbers, Dedekind cuts, or Cauchy completeness. The real numbers are given to you on a silver platter. From this, it is then possible to define the naturals in (more or less) the manner described in the question.

Addendum: Alternatively, instead of declaring the existence of the reals by fiat (item 5, above), we could build them up from scratch via the usual process (i.e. build the integers as equivalence classes of naturals, then the rationals as equivalence classes of integers, then the reals as Dedekind cuts). This construction of the reals will satisfy the above axioms.

From this construction, define an inductive set $S$ to be any subset of $\mathbb{R}$ (where an element of $\mathbb{R}$ is a Dedekind cut, which is...) with the properties that

  1. $0 \in S$, and
  2. if $x\in S$, then $x+1\in S$.

Then define $N$ to be the intersection of all inductive sets, i.e. $$ N = \bigcap_{\text{$S$ an inductive set}} S. $$ This set $N$ is a subset of the real numbers, and behaves (in every meaningful way) as the natural numbers—$N$ is isomorphic to $\mathbb{N}$. So perhaps it is reasonable to call $N$ the set of real-natural numbers. But I don't like having two sets of numbers running around called "natural numbers", so I am going to be lazy, and drop the "real-" part, and just say that $N$ is the set of natural numbers. I'll probably also go ahead and typeset it as $\mathbb{N}$, and write things like $\mathbb{N} \subseteq \mathbb{R}$.

That being said, this "inductive set" definition of the natural numbers comes from a book on analysis, not axiomatic set theory. The goal of the book is to do real analysis (i.e. to study limits, differentiation, integration, etc. in the context of real functions of real nubmers). If you had to start from $\mathsf{ZFC}$ and build up your number systems from there, you could never get to the actual results of analysis in a reasonable page count. You have to build a foundation somewhere. As such, starting from the assumption that the reals exist reasonable, and extracting the natural numbers from the reals is fine.

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    $\begingroup$ Until we know definitely how the book defines $ \mathbb N $ & $ \mathbb R $ we do not know if that resolves the problem stated in the “question”. $\endgroup$
    – PJTraill
    Sep 20, 2017 at 21:58
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    $\begingroup$ Well, there's nothing in the above about natural numbers, but the completeness axiom requires subsets. So you are working in a set theory, most likely. If ZFC, then the natural numbers already exist as an axiom. $\endgroup$ Sep 21, 2017 at 4:16
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    $\begingroup$ Why not just use the peano axioms? $\endgroup$
    – PyRulez
    Sep 21, 2017 at 4:17
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    $\begingroup$ How do you get that what you defined actually exists? Uniqueness isn't hard but existence seems to require a construction. $\endgroup$
    – DRF
    Sep 21, 2017 at 13:10
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    $\begingroup$ @PyRulez Because the original poster wanted to understand how building the naturals out of the reals could be non-circular. As I understand it, the goal isn't to build the reals, but to build the naturals from the reals. $\endgroup$
    – Xander Henderson
    Sep 21, 2017 at 13:36

Suppose we did start with some notion of "natural number" which we used to construct a model of the real numbers.

Then even in this setting, the quoted definition is still not circular, because it's defining a new notion of "natural number" that will henceforth be used instead of the previous notion of "natural number".

We could give the new notion a different name, but there isn't really any point; the new version of "natural numbers" has an obvious isomorphism with the old version so it's not really any different from the old one in any essential way.

There are a number of reasons why an exposition of real analysis might construct the natural numbers from the real numbers; the two most prominent are:

  • It is technically convenient to have the natural numbers be a subset of the real numbers
  • It makes the exposition somewhat more agnostic about foundations; it simply needs the real numbers as a starting point

You have to start somewhere.

And in mathematics, you have to start from some axioms.

If you choose to start from the axioms for the real numbers, then using them you define the natural numbers exactly as you quoted, and using that that definition you can prove that the Peano axioms are true.

If instead you choose to start from the Peano axioms for the natural numbers, then using them you can define the real numbers, and using that definition you can prove that the axioms for the real numbers are true.

Either one of those provides an answer to your question. You may not like that answer, but that's how things go. What you cannot do is to avoid choosing your axioms.

Nonetheless, perhaps there is still a way to break this vicious loop?

Well, actually, there is, by making a different choice of axioms. You can, instead, choose the basic axioms for set theory, also known as the Zermelo-Frankel axioms or ZF for short.

If you choose to start from the ZF axioms, then using them you can define the natural numbers using the Von Neumann definition. Using that definition, you can then prove that the Peano axioms are true. And then, as already said, you can use the Peano axioms (which in this scheme are now theorems instead of axioms) to define the real numbers and use that definition to prove that the axioms for the real numbers are true.

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    $\begingroup$ +1 for recommending Von Neumann. But do you have an issue with the C of ZFC? $\endgroup$ Sep 21, 2017 at 12:09
  • $\begingroup$ Well, the construction of the natural numbers works in ZF, that's all. I'm happy to call it a ZFC theorem, just like I would be happy to refer to a theorem about all groups as a theorem about abelian groups, but it gives more information as I stated it. $\endgroup$
    – Lee Mosher
    Sep 23, 2017 at 13:59

Defining the natural numbers (to within isomorphism) as a distinguished subset of a complete ordered field (COF) seems mathematically slick, but is unsatisfactory in two ways. First, definition should go from the simple to the complex and from the elementary to the advanced—not the other way round. Second, in order for the definition of a COF, and hence of the distinguished subset, to make sense, it must be proved that (A) any two COFs are isomorphic and (B) a COF exists. Let us grant A (which is not exactly trivial) and look at B. The simplest way to establish B is to construct $\Bbb R$ from $\Bbb Q$ via Dedekind cuts or equivalence classes of Cauchy sequences in $\Bbb Q$. But then what is $\Bbb Q$? The simplest way to define it is in terms of equivalence classes of pairs of elements of $\Bbb Z$. And $\Bbb Z$? Why, it is just a set of equivalence classes of pairs of elements of $\Bbb N$. Thus we have come full circle.

Now, it is true that all this set-theoretic structure is unwanted baggage when our aim is to get on and do real analysis, in which we would like to have simply $\Bbb N\subset\Bbb Z\subset\Bbb Q\subset\Bbb R\subset\Bbb C$ (the last one for good measure). But this is not really a problem. Having established existence from the bottom up, we can identify an isomorphic copy of each more basic structure within each more encompassing one, and agree henceforth to use these copies (in $\Bbb C$, say) rather than the originals, for convenience.


Edit: Because of some of the comments, I thought of more to add.

We can define the set of all natural numbers to satisfy the axioms of Peano arithmetic and then from that, we can construct the rational numbers and then the real numbers from the Dedekind cuts of rational numbers. After the set of all real numbers has been defined in this way, it can be proven that this definition of the set of all natural numbers describes the same set as the other definition of the set of all natural numbers. Defining the set of all real numbers as a totally ordered field then defining the natural numbers from it is unsatisfactory because it doesn't prove that a totally ordered field exists. In Zermelo-Frankel set theory, it can probably be proven that constructing the set of all real numbers in the other way gives rise to a totally ordered field and that every totally ordered field is isomorphic to that one.

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    $\begingroup$ If you want to prove the existence of a totally ordered field, you can go through the usual song and dance of defining the natural numbers axiomatically, then building up to the reals via Dedekind cuts or Cauchy completeness. Abstract nonsense tells us that there is at most one (up to isomorphism) totally ordered field, and we have a way of getting such a thing. With that in hand, define the real natural numbers as above, then show that the two definitions are isomorphic (as semigroups?). It still isn't circular. $\endgroup$
    – Xander Henderson
    Sep 20, 2017 at 22:51

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