The completeness of \(\mathbb R\) is a fundamental property of the real numbers. It expresses the fact that the real line has no “gaps”: every nonempty set of real numbers that is bounded above possesses a real supremum.
This property sets \(\mathbb R\) sharply apart from the set of rational numbers \(\mathbb Q\). Within the rationals there are nonempty, bounded-above sets that have no rational supremum. In \(\mathbb R\), by contrast, this can never happen.
In what follows we introduce the meaning of the completeness of \(\mathbb R\), state the completeness axiom in terms of the supremum, and see why this property underlies many of the foundational results of mathematical analysis.
Contents
- An intuitive idea of the completeness of \(\mathbb R\)
- Why \(\mathbb Q\) is not complete
- Review: upper bounds, lower bounds, maximum and minimum
- Review: supremum and infimum
- The completeness axiom for \(\mathbb R\)
- The meaning of the completeness axiom
- Worked examples of the completeness axiom
- The completeness of \(\mathbb R\) and sequences
- Fundamental consequences of the completeness of \(\mathbb R\)
- Final summary
An intuitive idea of the completeness of \(\mathbb R\)
To say that \(\mathbb R\) is complete means, intuitively, that the real line has no gaps.
This statement should be read with care. The rational numbers \(\mathbb Q\) lie very densely on the line: between any two distinct rationals there is always at least one more rational. Yet, despite this density, \(\mathbb Q\) does not fill the line completely.
Consider, for instance, the number
\[ \sqrt{2}. \]
It is not rational. Even so, one can construct rational numbers that lie closer and closer to \(\sqrt{2}\), both from below and from above.
In other words, within \(\mathbb Q\) there are approximation processes that “aim at” a number which does not belong to \(\mathbb Q\). From the point of view of the rationals, that point is a gap.
The set \(\mathbb R\), on the other hand, is built precisely to fill in these gaps. Every quantity that can be singled out as a limit, as a supremum, or as a point separating two classes of numbers belongs to the real line.
The completeness of \(\mathbb R\) makes this idea precise: every nonempty set of real numbers that is bounded above has a supremum in \(\mathbb R\).
Why \(\mathbb Q\) is not complete
To understand the completeness of \(\mathbb R\), it helps first to see why \(\mathbb Q\) is not complete.
Consider the set
\[ A=\{q\in\mathbb Q:q^2<2\}. \]
The set \(A\) consists of those rational numbers whose square is less than \(2\).
For example,
\[ 1\in A, \]
since
\[ 1^2=1<2. \]
Moreover, \(A\) is bounded above in \(\mathbb Q\). For instance, \(2\) is a rational upper bound of \(A\), because every rational \(q\) with \(q^2<2\) is certainly less than \(2\).
Nevertheless, \(A\) has no supremum in \(\mathbb Q\).
Indeed, reasoning inside \(\mathbb R\), its supremum is
\[ \sup A=\sqrt{2}. \]
But
\[ \sqrt{2}\notin\mathbb Q. \]
Hence, viewing the set only within \(\mathbb Q\), the number that ought to be the least upper bound is missing.
This is the crucial point: \(\mathbb Q\) contains a great many numbers and is dense on the line, yet it is not complete. There exist nonempty, bounded-above subsets of \(\mathbb Q\) that have no rational supremum.
The gap corresponding to \(\sqrt{2}\)
The set
\[ A=\{q\in\mathbb Q:q^2<2\} \]
describes all the rationals that lie, intuitively, to the left of \(\sqrt{2}\).
Within \(\mathbb Q\), however, the number \(\sqrt{2}\) does not exist. So the set \(A\) approaches indefinitely a threshold that does not belong to the rationals.
In the reals, by contrast, that threshold does exist, and it is precisely \(\sqrt{2}\). For this reason \(\mathbb R\) is complete, whereas \(\mathbb Q\) is not.
Review: upper bounds, lower bounds, maximum and minimum
Before stating the completeness axiom, let us recall a few basic definitions.
Let \(A\subseteq\mathbb R\) be a nonempty set.
A number \(M\in\mathbb R\) is called an upper bound of \(A\) if every element of \(A\) is less than or equal to \(M\). In symbols:
\[ x\leq M \qquad \text{for every } x\in A. \]
If \(A\) has at least one upper bound, then \(A\) is said to be bounded above.
Likewise, a number \(m\in\mathbb R\) is called a lower bound of \(A\) if every element of \(A\) is greater than or equal to \(m\). In symbols:
\[ m\leq x \qquad \text{for every } x\in A. \]
If \(A\) has at least one lower bound, then \(A\) is said to be bounded below.
Maximum and minimum
An element \(M\in A\) is called the maximum of \(A\) if it is greater than or equal to every element of \(A\):
\[ x\leq M \qquad \text{for every } x\in A. \]
In this case one writes
\[ M=\max A. \]
An element \(m\in A\) is called the minimum of \(A\) if it is less than or equal to every element of \(A\):
\[ m\leq x \qquad \text{for every } x\in A. \]
In this case one writes
\[ m=\min A. \]
It is important to note that the maximum and the minimum, when they exist, must belong to the set.
For example, the interval
\[ (0,1) \]
is bounded above and below, yet it has neither a maximum nor a minimum. Indeed, \(1\) and \(0\) are its supremum and infimum, respectively, but they do not belong to the interval.
Review: supremum and infimum
The notions of maximum and minimum are not enough to describe every bounded set. There are, in fact, sets that have no maximum yet still possess a smallest upper bound.
Let \(A\subseteq\mathbb R\) be a nonempty set that is bounded above. A number \(s\in\mathbb R\) is called the supremum (or least upper bound) of \(A\) if it satisfies two properties:
- \(s\) is an upper bound of \(A\);
- \(s\) is the smallest among all upper bounds of \(A\).
In this case one writes
\[ s=\sup A. \]
To say that \(s=\sup A\) therefore means that
\[ x\leq s \qquad \text{for every } x\in A, \]
and that no number smaller than \(s\) is still an upper bound of \(A\).
Equivalently, \(s=\sup A\) if and only if \(s\) is an upper bound of \(A\) and, for every \(\varepsilon>0\), there exists at least one element \(x\in A\) such that
\[ s-\varepsilon<x\leq s. \]
This second characterization is very important: it says that the elements of \(A\) can come arbitrarily close to \(\sup A\) from below.
Infimum
In an analogous way, let \(A\subseteq\mathbb R\) be a nonempty set that is bounded below. A number \(i\in\mathbb R\) is called the infimum (or greatest lower bound) of \(A\) if it satisfies two properties:
- \(i\) is a lower bound of \(A\);
- \(i\) is the greatest among all lower bounds of \(A\).
In this case one writes
\[ i=\inf A. \]
To say that \(i=\inf A\) therefore means that
\[ i\leq x \qquad \text{for every } x\in A, \]
and that no number greater than \(i\) is still a lower bound of \(A\).
Equivalently, \(i=\inf A\) if and only if \(i\) is a lower bound of \(A\) and, for every \(\varepsilon>0\), there exists at least one element \(x\in A\) such that
\[ i\leq x<i+\varepsilon. \]
The difference between maximum and supremum
The maximum, when it exists, is an element of the set. The supremum, on the other hand, need not belong to the set at all.
For example, for the interval
\[ A=(0,1) \]
one has
\[ \sup A=1, \qquad \inf A=0. \]
However,
\[ 1\notin A \qquad \text{and} \qquad 0\notin A. \]
Hence \(A\) has neither a maximum nor a minimum.
This example shows that the supremum and the infimum are more general concepts than the maximum and the minimum.
The completeness axiom for \(\mathbb R\)
We can now state the fundamental property of the real numbers.
Completeness axiom for \(\mathbb R\). Every nonempty subset of \(\mathbb R\) that is bounded above has a supremum in \(\mathbb R\).
In symbols, if \(A\subseteq\mathbb R\), \(A\neq\varnothing\), and \(A\) is bounded above, then there exists a real number \(s\in\mathbb R\) such that
\[ s=\sup A. \]
This axiom asserts that, on the real line, every nonempty set possessing at least one upper bound also possesses the smallest of its upper bounds.
The completeness of \(\mathbb R\) can therefore be expressed by saying that, in \(\mathbb R\), the suprema of nonempty, bounded-above sets are never missing.
The equivalent formulation in terms of the infimum
The completeness axiom can also be stated in terms of the infimum.
Every nonempty subset of \(\mathbb R\) that is bounded below has an infimum in \(\mathbb R\).
In symbols, if \(A\subseteq\mathbb R\), \(A\neq\varnothing\), and \(A\) is bounded below, then there exists a real number \(i\in\mathbb R\) such that
\[ i=\inf A. \]
The two formulations are equivalent. Indeed, the existence of suprema yields the existence of infima by applying the axiom to the reflected set
\[ -A=\{-x:x\in A\}. \]
In particular, if \(A\) is bounded below, then
\[ \inf A=-\sup(-A). \]
Why is it called an axiom?
It is called an axiom because completeness cannot be deduced from the algebraic and order properties alone — the very properties that the rational numbers already enjoy.
\(\mathbb Q\) is also an ordered field: one can add, multiply and compare rational numbers. Yet \(\mathbb Q\) is not complete.
The property that distinguishes \(\mathbb R\) from \(\mathbb Q\) is precisely the existence of the supremum for every nonempty, bounded-above set of reals.
The meaning of the completeness axiom
The completeness axiom asserts that if a set of reals is nonempty and cannot exceed a certain threshold, then there exists a smallest threshold that bounds it from above.
This smallest threshold is the supremum.
To grasp the meaning of the axiom, picture a set \(A\subseteq\mathbb R\) made up of points placed on the real line. If \(A\) is bounded above, then all of its points lie to the left of at least one real number.
The completeness axiom guarantees that among all these upper bounds there is a smallest one. In other words, there exists a real number that marks exactly the upper edge of the set.
This edge may or may not belong to the set.
If it belongs to the set, then it coincides with the maximum. If it does not belong to the set, it is nonetheless present on the real line as the supremum.
Example: a closed interval
Consider the interval
\[ A=[0,1]. \]
The set \(A\) is nonempty and bounded above.
Its supremum is
\[ \sup A=1. \]
In this case \(1\in A\), so the supremum is also the maximum:
\[ \max A=1. \]
Example: an open interval
Now consider the interval
\[ A=(0,1). \]
This set, too, is nonempty and bounded above.
Its supremum is again
\[ \sup A=1. \]
However, \(1\notin A\), so \(A\) has no maximum.
The completeness axiom nonetheless guarantees the existence of the supremum, even when that supremum does not belong to the set.
The essential point
The essential point is this: completeness does not say that every bounded set has a maximum or a minimum.
It says instead that every nonempty, bounded-above set has a supremum, and every nonempty, bounded-below set has an infimum.
This distinction is fundamental. The maximum must belong to the set; the supremum, by contrast, need not belong to the set at all.
Worked examples of the completeness axiom
Let us now look at some examples showing how the completeness axiom guarantees the existence of the supremum even when the set has no maximum.
Example 1: an open interval
Consider the set
\[ A=(0,1). \]
The set \(A\) is nonempty and bounded above. For instance, \(1\) is an upper bound of \(A\), since
\[ x\leq 1 \qquad \text{for every } x\in A. \]
By the completeness axiom, \(A\) has a supremum in \(\mathbb R\). In this case
\[ \sup A=1. \]
Yet \(1\notin A\), so \(A\) has no maximum.
This example shows that the completeness axiom does not guarantee the existence of a maximum, but it does guarantee the existence of the supremum.
Example 2: the set of numbers whose square is less than \(2\)
Consider the set
\[ A=\{x\in\mathbb R:x^2<2\}. \]
This set is nonempty, since \(0\in A\). It is also bounded above: for instance, \(2\) is an upper bound of \(A\).
By the completeness axiom, the supremum
\[ \sup A \]
exists. In this case one has
\[ \sup A=\sqrt{2}. \]
Indeed, \(\sqrt{2}\) is an upper bound of \(A\), because if \(x^2<2\), then \(-\sqrt{2}<x<\sqrt{2}\), and in particular \(x<\sqrt{2}\). Moreover, no number smaller than \(\sqrt{2}\) can be an upper bound, since there are elements of \(A\) arbitrarily close to \(\sqrt{2}\) from the left.
This example highlights the essential role of the real numbers: the number \(\sqrt{2}\), which is missing from \(\mathbb Q\), exists in \(\mathbb R\) and can be recognized as the supremum of a set.
Example 3: a discrete set with no maximum
Consider the set
\[ A=\left\{1-\frac{1}{n}:n\in\mathbb N,\ n\geq 1\right\}. \]
The first elements of the set are
\[ 0,\frac{1}{2},\frac{2}{3},\frac{3}{4},\ldots \]
The set is nonempty and bounded above. Indeed, every element of it is less than \(1\).
By the completeness axiom, \(A\) has a supremum. In this case
\[ \sup A=1. \]
But \(1\notin A\), because there is no \(n\in\mathbb N\) for which
\[ 1-\frac{1}{n}=1. \]
Hence \(A\) has no maximum.
This example is important because it exhibits a discrete set, made up of infinitely many isolated points, that approaches indefinitely a value lying outside the set.
The completeness of \(\mathbb R\) and sequences
The completeness of \(\mathbb R\) can also be expressed through sequences. One of its most important formulations is the Cauchy criterion.
A sequence \((x_n)\) of real numbers is called a Cauchy sequence if its terms become arbitrarily close to one another as the indices grow.
In symbols, \((x_n)\) is Cauchy if, for every \(\varepsilon>0\), there exists \(N\in\mathbb N\) such that, for all \(m,n\geq N\),
\[ |x_n-x_m|<\varepsilon. \]
The idea is that a Cauchy sequence requires no advance knowledge of its limit: it describes a sequence whose terms settle ever more closely toward one another.
Completeness via Cauchy sequences
The completeness of \(\mathbb R\) can be stated as follows:
Every Cauchy sequence of real numbers converges to a real number.
In symbols, if \((x_n)\subseteq\mathbb R\) is a Cauchy sequence, then there exists \(x\in\mathbb R\) such that
\[ x_n\to x. \]
This property is another form of the completeness of \(\mathbb R\). It says that, if a real sequence behaves as though it ought to converge, then its limit really does exist within \(\mathbb R\).
Why \(\mathbb Q\) is not sequentially complete
In the rationals this fails. There are sequences of rational numbers that are Cauchy yet converge to no rational number.
For example, we may take a sequence of rational approximations of \(\sqrt{2}\):
\[ 1,\ 1.4,\ 1.41,\ 1.414,\ 1.4142,\ldots \]
This sequence is made up of rational numbers, and its terms grow closer and closer to one another. In \(\mathbb R\) it converges to
\[ \sqrt{2}. \]
However,
\[ \sqrt{2}\notin\mathbb Q. \]
So, regarded as a sequence in \(\mathbb Q\), it does not converge to a rational number.
This shows that \(\mathbb Q\) is not complete: it contains sequences that ought to converge, yet whose limit falls outside \(\mathbb Q\).
Connection with the supremum axiom
The supremum axiom and completeness via Cauchy sequences are two different formulations of the same fundamental property of the real numbers.
The supremum axiom asserts that nonempty, bounded-above sets of reals have a real upper edge.
Completeness via Cauchy sequences asserts instead that every approximation process taking place within the reals converges to a real number.
Both formulations express the same idea: on the real line, the limit points needed to complete approximation processes are never missing.
Fundamental consequences of the completeness of \(\mathbb R\)
The completeness of \(\mathbb R\) is not an isolated property. Many of the foundational theorems of real analysis depend precisely on the fact that the real line has no gaps.
Let us look at some of the most important consequences.
Existence of suprema and infima
The most immediate consequence is the existence of suprema and infima.
If \(A\subseteq\mathbb R\) is nonempty and bounded above, then
\[ \sup A\in\mathbb R \]
exists. If \(A\subseteq\mathbb R\) is nonempty and bounded below, then
\[ \inf A\in\mathbb R \]
exists.
This property lets us work with sets that have no maximum or minimum but that nonetheless have a well-defined upper or lower edge.
The Nested Interval Theorem
Another consequence of completeness is the Nested Interval Theorem.
If
\[ I_n=[a_n,b_n], \qquad n\in\mathbb N, \]
is a sequence of closed, bounded, nested intervals — that is,
\[ I_1\supseteq I_2\supseteq I_3\supseteq \cdots, \]
then their intersection is nonempty:
\[ \bigcap_{n=1}^{+\infty} I_n\neq\varnothing. \]
If, in addition, the lengths of the intervals tend to \(0\), that is,
\[ b_n-a_n\to 0, \]
then the intersection contains exactly one point.
This result depends on completeness: in a non-complete set, a sequence of nested intervals can “close in” around a missing point.
The Bolzano–Weierstrass Theorem
Completeness also underlies the Bolzano–Weierstrass Theorem.
This theorem states that every bounded sequence of reals has a convergent subsequence.
In symbols, if \((x_n)\) is a bounded sequence of real numbers, then there exist a subsequence \((x_{n_k})\) and a real number \(x\in\mathbb R\) such that
\[ x_{n_k}\to x. \]
The crucial point is that the limit of the subsequence still belongs to \(\mathbb R\). This is possible because \(\mathbb R\) is complete.
The Cauchy convergence criterion
A fundamental consequence of completeness is the Cauchy convergence criterion.
A real sequence converges if and only if it is a Cauchy sequence.
In symbols:
\[ (x_n) \text{ converges in } \mathbb R \quad \Longleftrightarrow \quad (x_n) \text{ is Cauchy}. \]
The left-to-right implication holds in every metric space: every convergent sequence is Cauchy.
The reverse implication, by contrast, is a completeness property: in \(\mathbb R\), every Cauchy sequence converges to a real number.
Existence theorems in analysis
Many existence results in analysis rest, directly or indirectly, on the completeness of \(\mathbb R\).
For example, completeness underlies the Extreme Value Theorem (Weierstrass’ theorem), the existence-of-zeros theorem (Bolzano’s theorem), the Intermediate Value Theorem, and several results on the convergence of sequences and series.
In all of these cases the underlying idea is the same: one constructs an object by means of successive approximations, and completeness guarantees that the limiting object truly exists in \(\mathbb R\).
Final summary
The completeness of \(\mathbb R\) is the property that distinguishes the real numbers from the rational numbers. It expresses the fact that the real line has no gaps.
The most classical formulation of completeness is the supremum axiom:
every nonempty, bounded-above subset of \(\mathbb R\) has a supremum in \(\mathbb R\).
In symbols, if \(A\subseteq\mathbb R\), \(A\neq\varnothing\), and \(A\) is bounded above, then
\[ \sup A\in\mathbb R \]
exists.
Equivalently, every nonempty, bounded-below subset of \(\mathbb R\) has an infimum in \(\mathbb R\).
Completeness does not assert that every bounded set has a maximum or a minimum. It asserts instead that every nonempty, bounded-above set has a supremum, even when that supremum does not belong to the set.
The set of rational numbers \(\mathbb Q\) is not complete: there exist nonempty, bounded-above subsets of \(\mathbb Q\) that have no rational supremum. A key example is the set of rationals \(q\) with \(q^2<2\), whose real supremum is \(\sqrt{2}\), which does not belong to \(\mathbb Q\).
The completeness of \(\mathbb R\) can also be expressed in sequential form: every Cauchy sequence of real numbers converges to a real number.
This property underlies many of the foundational results of mathematical analysis, among them the Nested Interval Theorem, the Bolzano–Weierstrass Theorem, the Cauchy convergence criterion and numerous existence theorems.
In short, completeness is what makes \(\mathbb R\) the natural setting for analysis: every well-defined approximation process finds its own limit inside the real line.