In set theory, if natural numbers are represented by nested sets that include the empty set, how are the rest of the real numbers represented as sets?

Thanks for the answers. Several answers basically said for irrational numbers that A Dedekind cut is a pair of sets of rational numbers $\{L, R\}$. The set of real numbers is defined to be the set of all Dedekind cuts, where a Dedekind cut is a pair of sets of rational numbers $\{L, R\}$ which have no elements in common, and where all the elements of $L$ are less than any element of $R$. Each Dedekind cut is a real number. This is where I have a problem - surely that can’t be correct. The set $L$ is a set of all rationals, and there must be a rational in the set $L$ that is greater than all other rationals in that set, even if we have no method of determining it. And similarly, there must be a rational in the set $R$ that is less than all other rationals in that set, even if we have no method of determining it. If every irrational number has a corresponding set $L$, then each irrational number has some such corresponding largest element of that set $L$, and then each irrational number has some corresponding rational number. And that would mean that the irrational numbers are countable. So, with Dedekind cuts, the only conclusion is that there must be irrational numbers $x$ which are either greater or lesser than some irrational cut $y$ of the rationals, and between $x$ and $y$ there is no rational number. But that is impossible, so that the Dedekind cuts cannot be the correct representation of the real numbers.

Surely the problem with Dedekind cuts is in using sets of rationals that include all rationals up to a certain rational. But there is an alternative method of representing irrationals can be defined in terms of infinite sets of rational numbers. For example, in binary notation, the non-integer part of $\pi$ is $.00100100\ 00111111\ 01101010\ 10001$. You define a set by: if the nth digit is a $1$, then the natural number $n$ is in the set. And then we have that, for the real numbers between $0$ and $1$, that the set of real numbers is simply the set of all subsets of natural numbers. Each subset corresponds to some real number between $0$ and $1$.

And in this way, all real numbers can be considered to be some set based only on nested sets of the empty set.

But I still haven’t got a satisfactory answer for how negative numbers can be represented in terms only of sets containing the empty set. Any ideas?

  • 7
    $\begingroup$ From the point of view of set-theoretic simplicity, a variant of binary notation is simplest. However, that is not simplest when we want to verify the basic properties of the reals (complete ordered field), so the set-theoretically more complex standard approaches (Dedekind cuts, or Cauchy sequences) are more convenient. $\endgroup$ Sep 8, 2011 at 16:23
  • 1
    $\begingroup$ Peter, in my answer I did that; so did Zhen. First we start with using the usual definition of $\mathbb N$ as sets, then we continue. Either by defining an equivalence relation (thus a negative number is an equivalence class - a set - of pairs of natural numbers (a set of sets of sets...)) or by defining $-n = n\cup\{\mathbb N\}$ as a set. $\endgroup$
    – Asaf Karagila
    Sep 11, 2011 at 11:52
  • 1
    $\begingroup$ Also about the Dedekind cuts constructions, two comments: (1) Not every pair of sets of rationals define a cut; (2) As I indicated in my answer, we can only consider $L$, and we only consider such $L$ which is closed under $<$ and bounded in the rationals. We do not require an explicit supremum, just some bound. $\endgroup$
    – Asaf Karagila
    Sep 11, 2011 at 12:44
  • $\begingroup$ @Peter, I have merged your duplicate account into your current account. If you have trouble logging in, or if you accidentally create duplicate accounts, simply flag one of your own questions for moderator attention, and we will help out. $\endgroup$ Sep 11, 2011 at 15:04
  • 7
    $\begingroup$ I thought I should explicitly point out an error in the question. The statement "there must be a rational in the set L that is greater than all other rationals in that set, even if we have no method of determining it" is false, for trivial reasons that have nothing to do with completeness. The set $L$ of numbers that are strictly less than zero contains no number greater than all other numbers in $L$. (and this is true regardless of whether "number" means "rational number" or "real number".) $\endgroup$ Oct 3, 2012 at 18:00

7 Answers 7


There are a few possibilities, but here is the one approach. Even the starting point—the set of natural numbers $\mathbb{N}$—can be defined in several ways, but the standard definition takes $\mathbb{N}$ to be the set of finite von Neumann ordinals. Let us assume that we do have a set $\mathbb{N}$, a constant $0$, a unary operation $s$, and binary operations $+$ and $\cdot$ satisfying the axioms of second-order Peano arithmetic.

First, we need to construct the set of integers $\mathbb{Z}$. This we can do canonically as follows: we define $\mathbb{Z}$ to be the quotient of $\mathbb{N} \times \mathbb{N}$ by the equivalence relation $$\langle a, b \rangle \sim \langle c, d \rangle \text{ if and only if } a + d = b + c$$ The intended interpretation is that the equivalence class of $\langle a, b \rangle$ represents the integer $a - b$. Arithmetic operations can be defined on $\mathbb{Z}$ in the obvious fashion: $$\langle a, b \rangle + \langle c, d \rangle = \langle a + c, b + d \rangle$$ $$\langle a, b \rangle \cdot \langle c, d \rangle = \langle a c + b d, a d + b c \rangle$$ (Check that these respect the equivalence relation.) Again, this is not the only way to construct $\mathbb{Z}$; we can give a second-order axiomatisation of the integers which is categorical (i.e. any two models are isomorphic). For example, we may replace the set $\mathbb{Z}$ by $\mathbb{N}$, since the two sets are in bijection; the only thing we have to be careful about is to distinguish between the arithmetic operations for $\mathbb{Z}$ and for $\mathbb{N}$. (In other words, $\mathbb{Z}$ is more than just the set of its elements; it is also equipped with operations making it into a ring.)

Next, we need to construct the set of rational numbers $\mathbb{Q}$. This we may do using equivalence relations as well: we can define $\mathbb{Q}$ to be the quotient of $\mathbb{Z} \times (\mathbb{Z} \setminus \{ 0 \})$ by the equivalence relation $$\langle a, b \rangle \sim \langle c, d \rangle \text{ if and only if } a d = b c$$ The intended interpretation is that the equivalence class of $\langle a, b \rangle$ represents the fraction $a / b$. Arithmetic operations are defined by $$\langle a, b \rangle + \langle c, d \rangle = \langle a d + b c, b d \rangle$$ $$\langle a, b \rangle \cdot \langle c, d \rangle = \langle a c, b d \rangle$$ And as before, we can give an axiomatisation of the rational numbers which is categorical.

Now we can construct the set of real numbers $\mathbb{R}$. I describe the construction of Dedekind cuts, which is probably the simplest. A Dedekind cut is a pair of sets of rational numbers $\langle L, R \rangle$, satisfying the following axioms:

  1. If $x < y$, and $y \in L$, then $x \in L$. ($L$ is a lower set.)
  2. If $x < y$, and $x \in R$, then $y \in R$. ($R$ is an upper set.)
  3. If $x \in L$, then there is a $y$ in $L$ greater than $x$. ($L$ is open above.)
  4. If $y \in R$, then there is an $x$ in $R$ less than $y$. ($R$ is open below.)
  5. If $x < y$, then either $x \in L$ or $y \in R$. (The pair $\langle L, R \rangle$ is located.)
  6. For all $x$, we do not have both $x \in L$ and $x \in R$. ($L$ and $R$ are disjoint.)
  7. Neither $L$ nor $R$ are empty. (So $L$ is bounded above by everything in $R$ and $R$ is bounded below by everything in $L$.)

The intended interpretation is that $\langle L, R \rangle$ is the real number $z$ such that $L = \{ x \in \mathbb{Q} : x < z \}$ and $R = \{ y \in \mathbb{Q} : z < y \}$. The set of real numbers is defined to be the set of all Dedekind cuts. (No quotients by equivalence relations!) Arithmetic operations are defined as follows:

  • If $\langle L, R \rangle$ and $\langle L', R' \rangle$ are Dedekind cuts, their sum is defined to be $\langle L + L', R + R' \rangle$, where $L + L' = \{ x + x' : x \in L, x' \in L' \}$ and similarly for $R + R'$.
  • The negative of $\langle L, R \rangle$ is defined to be $\langle -R, -L \rangle$, where $-L = \{ -x : x \in L \}$ and similarly for $-R$.
  • If $\langle L, R \rangle$ and $\langle L', R' \rangle$ are Dedekind cuts, and $0 \notin R$ and $0 \notin R'$ (i.e. they both represent positive numbers), then their product is $\langle L \cdot L' , R \cdot R' \rangle$, where $L \cdot L' = \{ x \cdot x' : x \in L, x' \in L', x \ge 0, x' \ge 0 \} \cup \{ x \in \mathbb{Q} : x < 0 \}$ and $R \cdot R' = \{ y \cdot y' : y \in R, y \in R' \}$. We extend this to negative numbers by the usual laws: $(-z) \cdot z' = -(z \cdot z') = z \cdot -z'$ and $z \cdot z' = (-z) \cdot -z'$.

John Conway gives an alternative approach generalising the Dedekind cuts described above in his book On Numbers and Games. This eventually yields Conway's surreal numbers.

  • $\begingroup$ You have to exclude at least <0,0> in the construction of Q. It would be related to all other pairs so transitivity would fail. $\endgroup$
    – starblue
    Sep 9, 2011 at 6:37

There are two standard constructions, resulting in either equivalence classes of cauchy sequences of rational numbers, or bounded subsets of rational numbers satisfying some additional properties.

You should actually rather think of "the" real numbers as an ordered field satisfying some additional properties (axiomatic approach), than a specific set.

  • 22
    $\begingroup$ The comment that Alexander makes about the axiomatic approach is a subtle point, but entirely worth taking seriously. The importance of the tools we use in mathematics are not their identities, but their properties. While it is very important that one can build models of the real numbers from sets, choosing a single model as canonical is beside the point, and thinking of what the elements of sqrt(2) are is not usually a pertinent question. $\endgroup$ Sep 8, 2011 at 15:27
  • $\begingroup$ @NieldeBeaudrap Yes, but one could be interested about their identities too. $\endgroup$
    – Red Banana
    Jul 24, 2016 at 9:57
  • 1
    $\begingroup$ @Voyska: to give a (somewhat unfair) caricature of your point, one could also be interested in the colour of numbers, or their textures. Indeed, synthaethetic people would attest that these sense data can be pertinent of numbers, and allows them in some case to obtain a more reliable memory of those numbers (e.g. as in being able to recall long portions of the decimal expansion of $\pi$). So are these properties in some robust and meaningful sense a part of the identities of those numbers? Or does this say more about human neurology in individual cases, than numbers as such? $\endgroup$ Jul 25, 2016 at 14:05

First we start with the natural numbers.

  • $0=\varnothing$,
  • $n+1=n\cup\{n\}$.

Now we have the natural numbers. From here we can continue creating the integers (the examples here are just out of the blue to show how this can be done, not something purely canonical.)

  • For $n\in\mathbb N$, define $-n=n\cup\{\mathbb N\}$.

This is unique since $\mathbb N\notin n$ for any $n$. The next step is to define the rational numbers:

  • $\frac{p}{q} = \big\{\langle m,n\rangle\mid mq=np\big\}$

This is, of course, an equivalence relation over the set $\mathbb Z\times\mathbb Z$.

And finally, we reach the definition of the real numbers:

A real number is a set of rational numbers $r$ such that:

  1. If $q\in r$ and $p<q$ then $p\in r$;
  2. There exists some $q\in\mathbb Q$ such that every $p\in r$ is less than $q$.

This can be formulated as subsets of $\mathbb N$ as followed:

Fix some enumeration of $\mathbb Q$, that is $\{q_n\mid n\in\mathbb N\}$. Now consider the real number $r$ as the set of natural numbers which corresponds to the rationals in the above definition.

This is a usual way of looking at the real numbers as subsets of $\mathbb N$ in modern set theory. It is true that not every set of natural numbers is used, but since "enough" of them is being used we can just map the sets of natural numbers so we use them all.

  • $\begingroup$ So, \begin{align}\varnothing = 0 &= s(-1) \\ &= -1 + 1\\ &= -1 \cup \{-1\} = (1\cup \{\mathbb{N}\}) \cup (\{1\cup \{\mathbb{N}\}\})\\ &= \{\{\varnothing, \mathbb{N}\},\{\{\varnothing, \mathbb{N}\}\}\}\end{align} ??? $\endgroup$ May 9, 2016 at 23:14
  • 1
    $\begingroup$ I don't follow your logic. At all. Also, note that the natural numbers embed into the rational numbers, but they are not a subset of them, under this interpretation. $\endgroup$
    – Asaf Karagila
    May 10, 2016 at 3:18
  • $\begingroup$ @AsafKaragila in the rational number axioms did you re-use $p, q, r$, but with different definitions? Also, can you expand on how $r$ as the set of natural numbers which corresponds to the rationals? Do you mean the enumeration for the definition? $\endgroup$
    – Nick
    Jan 21, 2021 at 3:49
  • $\begingroup$ @Nick: Yes, I repurposed the letters. As for the second question, $r$ is a real number. We fix a bijection between the rational numbers and the natural numbers, and then we simply translate the Dedekind cuts (initial segments of $\Bbb Q$) to sets of natural numbers. To be honest, this is not a well written question, I wrote this I was still forming an understanding of the set theoretic world around me, and there's a lot I'd improve upon nowadays. $\endgroup$
    – Asaf Karagila
    Jan 21, 2021 at 11:00

There are a few ways to do this.

  • Dedekind cuts are the representation of real numbers which are the most obviously set-like; it is a representation in which each real number x ∈ ℝ is represented by a pair (ST) of disjoint non-empty open sets S,T ⊂ ℚ, such that

    a. If a ∈ S, then every number b < a is also in S;
    b. If a ∈ T, then every number b > a is also in T.

    (There are alternative, equivalent ways of expressing the idea of a Dedekind cut, e.g. this other example.)

    Every rational number q ∈ ℚ can be represented by S = {a∈ℚ | a<q} and T = {a∈ℚ | a>q}; so that q is the supremum of S and the infimum of T. We identify real numbers with sets in the same way: a real number such as $\sqrt 2$ is identified with the sets S = {a∈ℚ | a2<2 or a<0}, and T = {a∈ℚ | a2>2 and a>0}, so that $\sqrt 2$ is the supremum of S and the infimum of T.

    To use this construction as a "construction in terms of sets", you must represent rational numbers in terms of sets; this is usually done by representing them as equivalence classes of ordered pairs (pq), where q ≠ 0, and in which (pq) ∼ (p', q') if and only if pq' = qp'.

  • Equivalence classes of Cauchy sequences of rational numbers is another way to construct the real numbers. Cauchy sequences (over the thesmevles are represented as sets by identifying a sequence σ as a function σ : ℕ → ℚ; the function is then a set of ordered pairs. The Cauchy sequences form an algebra over the rational numbers, where addition, negation, and multiplication of Cauchy sequences is done point-wise.

    Two Cauchy sequences σ, σ' are considered equivalent if (σ' − σ) converges to 0; real numbers x are then represented by the rational Cauchy sequences converging on x. For instance, $\sqrt 2$ can be represented by the set of Cauchy sequences equivalent to (1, 1.4, 1.41, 1.414, 1.4142, ...) giving the decimal expansion of $\sqrt 2$.

    There is another related representation of real numbers in terms of sequences nested closed intervals; this can be reduced to Cauchy sequences by considering the sequence either of the lower limits or the upper limits of those intervals.

For every construction of the real numbers, there will be a way to describe how to construct a real number from the ground up in terms of sets; it's a good introductory exercise to see how you would do them.


A long time ago I wrote an exposition on the uniqueness (up to unique order-preserving isomorphism) of the real numbers on this very site. Note: It's filled with exercises and it assumes knowledge of abstract algebra and topology.

It also contains a (very short) series of exercises that lead the reader to proving that the "equivalence classes of Cauchy sequences of rational numbers" are a complete ordered field.

In some sense it is a bit of a failure of mathematics that first year math majors can use the least upper bound property of the real numbers and prove many nice theorems about sequences of real numbers and continuous functions, but actually showing that the real numbers exist and are unique (up to...) pretty much requires half of an undergraduate education. It's not that it's a failure of teaching mathematics, it's just that damn hard.


While all of the constructions given here describe the construction of the ordered field of real numbers $\mathbb{R}$, you should be aware that more often than not when a set theorist talks about 'the reals' - particularly in the context of cardinality and large-cardinal theory - they don't generally mean the real numbers, but instead the reals as an abstract entity; in that case a 'real' can refer to either a set of natural numbers (that is, a member of $2^{\mathbb {N}} = \mathcal{P}(\mathbb{N})$) or a function from natural numbers to natural numbers (i.e., a member of $\mathbb{N}^{\mathbb{N}}$), and often the two are used interchangeably, depending on context.


I have written an article where I find the simplest possible representation of real numbers as sets. Namely, real numbers are infinite sets of natural numbers. The order and structure is given explicitly. You can find the details here:



You must log in to answer this question.

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