Supremum and Infimum: Definition, Properties and Examples

The supremum and infimum generalize the notions of maximum and minimum of a set, and they make it possible to describe the behaviour of bounded sets in a rigorous way.

Unlike the maximum and the minimum, the supremum and the infimum may exist even when the corresponding extreme values do not belong to the set.

In the following sections we introduce the definitions of upper bound, lower bound, supremum and infimum, study their main properties and work through several significant examples.

Contents

• Upper bounds and lower bounds
• Supremum
• Infimum
• Uniqueness of the supremum and the infimum
• Characterization of the supremum
• Characterization of the infimum
• Relationship with maximum and minimum
• Examples
• Completeness of the real numbers

To introduce the notions of supremum and infimum we must start from two fundamental concepts: those of upper bound and lower bound.

Let \(A\subseteq\mathbb{R}\) be a non-empty set.

Definition. A real number \(M\) is called an upper bound of \(A\) if:

\[ x\leq M \qquad \forall x\in A. \]

In other words, an upper bound is a number greater than or equal to every element of the set.

Likewise, a real number \(m\) is called a lower bound of \(A\) if:

\[ m\leq x \qquad \forall x\in A. \]

A lower bound is therefore a number less than or equal to every element of the set.

Example 1. Consider the interval \( A=(1,5). \)

Since every element is less than \(5\), the number \(5\) is an upper bound of \(A\). The numbers \(6\), \(10\), \(100\) — and, more generally, every real number greater than or equal to \(5\) — are upper bounds of the set as well.

Similarly, the number \(1\) is a lower bound of \(A\). So are \(0\), \(-3\), \(-100\) and, more generally, every real number less than or equal to \(1\).

We thus observe that one and the same set may possess infinitely many upper bounds and infinitely many lower bounds.

Sets bounded above and below

Definition. A set that possesses at least one upper bound is said to be bounded above, whereas a set that possesses at least one lower bound is said to be bounded below.

A set that is bounded both above and below is simply said to be bounded.

It is worth noting that the term bounded bears no relation to the notion of limit of a sequence or of a function. To say that a set is bounded simply means that all of its elements lie between a suitable lower bound and a suitable upper bound.

Example 2. The interval

\[ (1,5) \]

is bounded above and below. For instance, \(5\) is an upper bound and \(1\) is a lower bound.

Example 3. The set \( [0,+\infty) \) is bounded below but not bounded above. Indeed, \(0\) is a lower bound, whereas no real number is greater than or equal to every element of the set.

Example 4. The set \( \mathbb{R} \) is neither bounded above nor bounded below.

The notions of upper bound and lower bound are the starting point for introducing the supremum and the infimum, which will be defined in the following sections.

Let \(A\subseteq\mathbb{R}\) be a non-empty set that is bounded above. The set of its upper bounds is therefore non-empty, and among them there is a distinguished one: the smallest.

Definition. The supremum of \(A\), denoted by \(\sup A\), is the smallest of the upper bounds of \(A\).

Equivalently, a real number \(s\) is the supremum of \(A\) if it satisfies the following two conditions:

• \(s\) is an upper bound of \(A\), that is, \(x\leq s\) for every \(x\in A\);
• if \(M\) is an upper bound of \(A\), then \(s\leq M\).

The first condition states that \(s\) "lies above" every element of \(A\); the second, that no number smaller than \(s\) enjoys the same property.

The existence of the supremum for every non-empty set that is bounded above is by no means obvious: it is guaranteed by a fundamental property of the real numbers, the completeness axiom, which we shall discuss in the final section.

Example 5. We return to the interval:

\[ A=(1,5). \]

Its upper bounds are precisely the real numbers greater than or equal to \(5\), that is, the set \([5,+\infty)\). The smallest of these is \(5\), hence:

\[ \sup A=5. \]

Note that \(5\notin A\): the supremum need not belong to the set.

The infimum is introduced in a perfectly symmetric way. Let \(A\subseteq\mathbb{R}\) be a non-empty set that is bounded below: the set of its lower bounds is non-empty and contains a distinguished element, the largest.

Definition. The infimum of \(A\), denoted by \(\inf A\), is the largest of the lower bounds of \(A\).

Equivalently, a real number \(i\) is the infimum of \(A\) if it satisfies the following two conditions:

• \(i\) is a lower bound of \(A\), that is, \(i\leq x\) for every \(x\in A\);
• if \(m\) is a lower bound of \(A\), then \(m\leq i\).

Here too, the existence of the infimum for every non-empty set that is bounded below follows from the completeness axiom.

Example 6. We again consider:

\[ A=(1,5). \]

Its lower bounds are precisely the real numbers less than or equal to \(1\), that is, the set \((-\infty,1]\). The largest of these is \(1\), hence:

\[ \inf A=1. \]

As with the supremum, we observe that \(1\notin A\).

Remark (unbounded sets). The definitions above concern sets that are bounded above or below. In order to handle the unbounded case uniformly as well, one often adopts the following convention: if \(A\) is not bounded above one sets

\[ \sup A=+\infty, \]

while if \(A\) is not bounded below one sets

\[ \inf A=-\infty. \]

The symbols \(+\infty\) and \(-\infty\) are not real numbers: the equality \(\sup A=+\infty\) is merely a concise way of stating that \(A\) has no upper bound. Under this convention, every non-empty subset of \(\mathbb{R}\) is endowed with a supremum and an infimum, whether finite or infinite. For example, \(\sup\mathbb{R}=+\infty\) and \(\inf\mathbb{R}=-\infty\), while for the set \([0,+\infty)\) one has \(\inf=0\) and \(\sup=+\infty\).

When they exist, the supremum and the infimum of a set are unique.

Proposition. If a set \(A\subseteq\mathbb{R}\) possesses a supremum, then it is unique.

Proof. Suppose that \(s_1\) and \(s_2\) are two suprema of \(A\).

Since \(s_1\) is a supremum, every upper bound of \(A\) is greater than or equal to \(s_1\). In particular, as \(s_2\) is an upper bound of \(A\), we have:

\[ s_1\leq s_2. \]

Similarly, since \(s_2\) is a supremum and \(s_1\) is an upper bound of \(A\), it follows that:

\[ s_2\leq s_1. \]

From the two inequalities we deduce:

\[ s_1=s_2. \]

Therefore the supremum is unique.

By an entirely analogous argument one proves that the infimum, when it exists, is likewise unique.

Monotonicity with respect to inclusion

A further useful property concerns the behaviour of these extrema when a set is enlarged: adding elements can only increase (or leave unchanged) the supremum and decrease (or leave unchanged) the infimum.

Proposition. Let \(A,B\subseteq\mathbb{R}\) be non-empty with \(A\subseteq B\). If \(B\) is bounded above, then so is \(A\), and

\[ \sup A\leq\sup B. \]

Likewise, if \(B\) is bounded below, then so is \(A\), and

\[ \inf A\geq\inf B. \]

Proof. Suppose \(B\) is bounded above and set \(s=\sup B\). For every \(x\in A\) we have \(x\in B\), since \(A\subseteq B\), and therefore \(x\leq s\). Hence \(s\) is an upper bound of \(A\): in particular, \(A\) is bounded above and admits a supremum. Since \(\sup A\) is the smallest of the upper bounds of \(A\) and \(s\) is one of them, it follows that:

\[ \sup A\leq s=\sup B. \]

The case of the infimum is entirely analogous. Setting \(i=\inf B\), we have \(i\leq x\) for every \(x\in B\), and hence for every \(x\in A\); thus \(i\) is a lower bound of \(A\) and \(\inf A\) exists. Since \(\inf A\) is the largest of the lower bounds of \(A\) and \(i\) is one of them, we obtain \(\inf A\geq i=\inf B\).

Remark. With the convention introduced in the previous section, the inequalities \(\sup A\leq\sup B\) and \(\inf A\geq\inf B\) remain valid for any pair of non-empty sets with \(A\subseteq B\), including unbounded ones.

The definition of supremum states that \(\sup A\) is the smallest of all the upper bounds of \(A\). Verifying this property directly would, in principle, require comparing \(s\) with the entire collection of upper bounds of the set.

There is, however, a far more workable characterization, which allows one to recognize a supremum by examining only the elements of \(A\).

Proposition. Let \(A\subseteq\mathbb{R}\) be a non-empty set that is bounded above, and let \(s\in\mathbb{R}\). Then:

\[ s=\sup A \]

if and only if both of the following conditions hold:

• \(x\leq s\) for every \(x\in A\);
• for every \(\varepsilon>0\) there exists an element \(x\in A\) such that \[ s-\varepsilon<x. \]

The first condition states that \(s\) is an upper bound of \(A\).

The second condition, on the other hand, guarantees that no number strictly less than \(s\) can be an upper bound of the set.

Indeed, for any arbitrarily fixed \(\varepsilon>0\), there always exists an element of \(A\) strictly greater than \(s-\varepsilon\). Consequently, \(s-\varepsilon\) cannot be an upper bound of \(A\).

The equivalence becomes clear once we consider the meaning of the two conditions. If the first held but not the second, there would exist an \(\varepsilon>0\) such that no element of \(A\) exceeds \(s-\varepsilon\): then \(s-\varepsilon\) would itself be an upper bound, strictly less than \(s\), so that \(s\) could not be the smallest of the upper bounds. Conversely, if both conditions hold, then \(s\) is an upper bound and no number smaller than \(s\) is: \(s\) is therefore the least of the upper bounds, that is, \(\sup A\).

Example 7. Consider the interval:

\[ A=(1,5). \]

Let us verify, by means of the characterization, that:

\[ \sup A=5. \]

First condition. Every element of \(A\) satisfies \(x<5\), and hence, a fortiori, \(x\leq 5\). Thus \(5\) is an upper bound of \(A\).

Second condition. Let \(\varepsilon>0\). We must exhibit an element of \(A\) greater than \(5-\varepsilon\).

If \(\varepsilon\geq 4\), then:

\[ 5-\varepsilon\leq 1, \]

and hence every element of \(A\) is already greater than \(5-\varepsilon\): the condition is trivially satisfied.

If, instead, \(0<\varepsilon<4\), consider the number:

\[ x=5-\frac{\varepsilon}{2}. \]

Since \(0<\dfrac{\varepsilon}{2}<2\), we have:

\[ 3<x<5, \]

and therefore \(x\in(1,5)=A\). Moreover, since \(\dfrac{\varepsilon}{2}<\varepsilon\), it follows that:

\[ 5-\varepsilon<x. \]

In both cases we have found an element of \(A\) greater than \(5-\varepsilon\). The second condition is therefore satisfied for every \(\varepsilon>0\).

Since both conditions hold, we conclude that:

\[ \sup A=5, \]

even though \(5\notin A\). This example brings out the nature of the supremum: a value that the elements of the set can approach arbitrarily closely, without it necessarily having to belong to the set.

Just as with the supremum, the infimum too admits an equivalent characterization that is particularly useful in applications.

Proposition. Let \(A\subseteq\mathbb{R}\) be a non-empty set that is bounded below, and let \(i\in\mathbb{R}\). Then:

\[ i=\inf A \]

if and only if both of the following conditions hold:

• \(i\leq x\) for every \(x\in A\);
• for every \(\varepsilon>0\) there exists an element \(x\in A\) such that \[ x<i+\varepsilon. \]

The first condition states that \(i\) is a lower bound of \(A\).

The second condition, on the other hand, guarantees that no number strictly greater than \(i\) can be a lower bound of the set.

Indeed, for any arbitrarily fixed \(\varepsilon>0\), there always exists an element of \(A\) strictly less than \(i+\varepsilon\). Consequently, \(i+\varepsilon\) cannot be a lower bound of \(A\).

The equivalence becomes clear once we consider the meaning of the two conditions. If the first held but not the second, there would exist an \(\varepsilon>0\) such that:

\[ x\geq i+\varepsilon \qquad \forall x\in A. \]

In that case \(i+\varepsilon\) would be a lower bound of \(A\) strictly greater than \(i\), contradicting the fact that \(i\) is the largest of the lower bounds.

Conversely, if both conditions hold, then \(i\) is a lower bound and no number greater than \(i\) is a lower bound. Hence \(i\) coincides with the greatest of the lower bounds, that is, with the infimum of \(A\).

Example 8. Consider the interval:

\[ A=(1,5). \]

Let us verify, by means of the previous characterization, that:

\[ \inf A=1. \]

First condition. Every element of \(A\) satisfies \(1<x\), and hence, a fortiori, \(1\leq x\). It follows that \(1\) is a lower bound of \(A\).

Second condition. Let \(\varepsilon>0\).

If \(\varepsilon\geq 4\), then:

\[ 1+\varepsilon\geq 5, \]

and hence every element of \(A\) is less than \(1+\varepsilon\).

If, instead, \(0<\varepsilon<4\), consider:

\[ x=1+\frac{\varepsilon}{2}. \]

Since:

\[ 0<\frac{\varepsilon}{2}<2, \]

we obtain:

\[ 1<x<3<5. \]

Therefore:

\[ x\in(1,5)=A. \]

Moreover:

\[ x=1+\frac{\varepsilon}{2} < 1+\varepsilon. \]

In both cases there exists an element of \(A\) less than \(1+\varepsilon\). The second condition is therefore verified.

Since both conditions hold, we conclude that:

\[ \inf(1,5)=1. \]

Remark. Since:

\[ 1\notin(1,5), \]

the infimum is not necessarily an element of the set.

The supremum and infimum are closely related to the maximum and minimum, of which they are a generalization. There is just one essential difference: the maximum and the minimum must belong to the set, whereas the supremum and the infimum need not.

Proposition. Let \(A\subseteq\mathbb{R}\) be non-empty and bounded above. Then \(A\) possesses a maximum if and only if:

\[ \sup A\in A, \]

and in that case:

\[ \max A=\sup A. \]

Proof. Suppose first that \(A\) possesses a maximum and set \(m=\max A\).

By the definition of maximum, \(m\in A\) and \(x\leq m\) for every \(x\in A\), so \(m\) is an upper bound of \(A\). Moreover, if \(M\) is any upper bound of \(A\), then \(M\geq x\) for every \(x\in A\), and in particular, since \(m\in A\):

\[ M\geq m. \]

Hence \(m\) is the smallest of the upper bounds, that is, \(m=\sup A\); in particular \(\sup A=m\in A\).

Conversely, suppose that \(\sup A\in A\) and set \(s=\sup A\). Then \(s\) is an upper bound, so \(x\leq s\) for every \(x\in A\); moreover \(s\in A\). By definition, \(s\) is therefore the maximum of \(A\), and \(\max A=s=\sup A\).

In an entirely analogous way one proves that \(A\), non-empty and bounded below, possesses a minimum if and only if \(\inf A\in A\), and in that case:

\[ \min A=\inf A. \]

To summarize: the supremum always exists (for a non-empty set that is bounded above), whereas the maximum exists only when the supremum belongs to the set. The same holds, symmetrically, for the infimum and the minimum.

Example 9. For the closed interval:

\[ A=[1,5], \]

one has \(\sup A=5\) and \(\inf A=1\); since \(5\in A\) and \(1\in A\), both extrema belong to the set, and therefore:

\[ \max A=5,\qquad \min A=1. \]

For the open interval:

\[ A=(1,5), \]

one still has \(\sup A=5\) and \(\inf A=1\), but now \(5\notin A\) and \(1\notin A\): the set possesses neither a maximum nor a minimum, even though it is endowed with a supremum and an infimum.

We now apply the definitions and results above to a few notable sets, determining for each its supremum, infimum and, where they exist, its maximum and minimum.

Example 10. \[ A=[-2,3). \]

The infimum is \(-2\), which belongs to the set; hence:

\[ \inf A=\min A=-2. \]

The supremum is \(3\), which, on the other hand, does not belong to the set:

\[ \sup A=3, \]

while the maximum does not exist.

Example 11. \[ A=\left\{\frac1n:n\in\mathbb{N},\ n\geq 1\right\}. \]

The elements are \(1,\ \frac12,\ \frac13,\ldots\) The largest value is \(1\), attained at \(n=1\), and it belongs to the set:

\[ \sup A=\max A=1. \]

The elements decrease, approaching \(0\) without ever reaching it. The number \(0\) is a lower bound and, for every \(\varepsilon>0\), by choosing \(n\) such that \(\frac1n<\varepsilon\) one obtains an element less than \(0+\varepsilon\). Hence:

\[ \inf A=0, \]

while the minimum does not exist, since \(0\notin A\).

Example 12. \[ A=\left\{\frac{n}{n+1}:n\in\mathbb{N},\ n\geq 1\right\}. \]

Since \(\frac{n}{n+1}=1-\frac1{n+1}\), the sequence is increasing. The first element, at \(n=1\), is \(\frac12\), and it belongs to the set:

\[ \inf A=\min A=\frac12. \]

The elements increase, approaching \(1\) without ever reaching it, hence:

\[ \sup A=1, \]

while the maximum does not exist, since \(1\notin A\).

Example 13. \[ A=\left\{(-1)^n+\frac1n:n\in\mathbb{N},\ n\geq 1\right\}. \]

It is convenient to distinguish the terms of even and odd index.

For even \(n\) the element equals \(1+\frac1n\); these terms decrease, and the largest is obtained at \(n=2\):

\[ 1+\frac12=\frac32. \]

All the other elements of the set are less than \(\frac32\), which belongs to \(A\). Therefore:

\[ \sup A=\max A=\frac32. \]

For odd \(n\) the element equals \(-1+\frac1n\); these terms decrease, approaching \(-1\) without ever reaching it. The number \(-1\) is a lower bound of \(A\) and, for every \(\varepsilon>0\), by choosing an odd \(n\) with \(\frac1n<\varepsilon\) one obtains an element less than \(-1+\varepsilon\). Hence:

\[ \inf A=-1, \]

while the minimum does not exist, since no element of the set equals \(-1\).

We have stated several times that every non-empty set that is bounded above possesses a supremum. This property is not a consequence of the algebraic rules or of the ordering: it is a structural property of the real numbers, taken as an axiom.

Completeness axiom. Every non-empty subset of \(\mathbb{R}\) that is bounded above admits a supremum in \(\mathbb{R}\).

From this axiom one immediately deduces the symmetric property for the infimum: every non-empty subset of \(\mathbb{R}\) that is bounded below admits an infimum in \(\mathbb{R}\). It suffices to observe that, setting \(-A=\{-x:x\in A\}\), one has:

\[ \inf A=-\sup(-A). \]

The importance of the completeness axiom emerges clearly when we compare \(\mathbb{R}\) with the field of rational numbers \(\mathbb{Q}\), which does not enjoy this property.

A rational set with no supremum in \(\mathbb{Q}\)

Consider the subset of \(\mathbb{Q}\):

\[ B=\left\{x\in\mathbb{Q}:x>0,\ x^2<2\right\}. \]

The set \(B\) is non-empty, since \(1\in B\), and it is bounded above in \(\mathbb{Q}\): if \(x\in B\) then \(x^2<2<4\), whence \(x<2\), and so \(2\) is an upper bound.

We show, however, that \(B\) has no supremum within \(\mathbb{Q}\). Suppose, for the sake of contradiction, that there exists \(s\in\mathbb{Q}\) with \(s=\sup B\); since \(\sqrt2\) is not rational, we must have \(s^2\neq 2\), so that either \(s^2<2\) or \(s^2>2\).

If \(s^2<2\), we choose a rational \(h\) with \(0<h<1\) and:

\[ h<\frac{2-s^2}{2s+1}. \]

Then, using \(h^2<h\):

\[ (s+h)^2=s^2+2sh+h^2<s^2+h(2s+1)<s^2+(2-s^2)=2, \]

so that \(s+h\in B\) and \(s+h>s\): this contradicts the fact that \(s\) is an upper bound.

If, instead, \(s^2>2\), we choose a rational \(h\) with:

\[ 0<h<\frac{s^2-2}{2s}. \]

Then:

\[ (s-h)^2=s^2-2sh+h^2>s^2-2sh>s^2-(s^2-2)=2. \]

It follows that \(s-h\) is still an upper bound of \(B\) (every element \(x\in B\) satisfies \(x^2<2<(s-h)^2\), whence \(x<s-h\)), but \(s-h<s\): this contradicts the fact that \(s\) is the smallest of the upper bounds.

In both cases we reach a contradiction. Hence \(B\) has no supremum in \(\mathbb{Q}\).

In the set of real numbers, by contrast, the supremum does exist, and it is:

\[ \sup B=\sqrt2. \]

This example shows that \(\mathbb{Q}\) has "holes": there exist rational sets that are bounded above and cluster around a value without that value belonging to \(\mathbb{Q}\). The completeness axiom asserts precisely that in \(\mathbb{R}\) these holes do not exist: the real numbers form a continuum with no gaps.

It is this property that makes possible the rigorous development of the notions of limit, continuity, derivative and integral, which constitute the foundation of mathematical analysis.