Supremum and Infimum
Summarize
The completeness axiom
Although real numbers are frequently introduced via their algebraic properties, the essential distinction between $\mathbb{R}$ and $\mathbb{Q}$ lies in their order structure, and in particular in a unique property of that order. Every non-empty subset of $\mathbb{R}$ that is bounded above has a least upper bound lying within $\mathbb{R}$. The property is known as the completeness axiom and characterises the real line. The notions of supremum and infimum are the practical tools for applying it.
Upper and lower bounds
Consider a non-empty set $A \subseteq \mathbb{R}$. A real number $M$ is an upper bound of $A$ if:
$$ a \leq M \quad \forall \ a \in A $$
If such a number exists, the set $A$ is bounded above. A real number $m$ is a lower bound of $A$ if:
$$ a \geq m \quad \forall \ a \in A $$
In this case, $A$ is bounded below. A set is bounded if it is both bounded above and below, that is, there exists $K > 0$ such that:
$$ |a| \leq K \quad \forall \ a \in A $$
Upper bounds, when they exist, are generally not unique. If $M$ is an upper bound of $A$, then $M + 1$ and $M + 100$ are also upper bounds. The same applies symmetrically to lower bounds. If $m$ is a lower bound of $A$, then $m - 1$ is also a lower bound. The least upper bound is called the supremum, and the greatest lower bound is called the infimum.
- If $A$ is non-empty but not bounded above, the supremum is conventionally defined as $\sup A = +\infty$.
- If $A$ is not bounded below, the infimum is set as $\inf A = -\infty$.
- For the empty set, the conventions $\sup \emptyset = -\infty$ and $\inf \emptyset = +\infty$ are employed.
Supremum
Consider a non-empty subset $A \subseteq \mathbb{R}$ that is bounded above. The supremum of $A$, denoted $\sup A$, is its least upper bound. A real number $s$ is equal to $\sup A$ if and only if both of the following conditions are satisfied. The first condition states that $s$ is an upper bound of $A$:
$$ a \leq s \quad \forall \ a \in A $$
The second condition ensures that any number strictly less than $s$ is exceeded by some element of $A$:
$$ \forall \ \varepsilon > 0 \quad \exists \ a \in A : a > s - \varepsilon $$
Together, these conditions determine $s$ uniquely. There can be only one least upper bound. If both $s$ and $s'$ satisfy the definition, we have:
$$ s \leq s' \ \wedge \ s' \leq s \ \to \ s = s' $$
An equivalent characterisation states that $s = \sup A$ if and only if $s$ is an upper bound of $A$ and there exists a sequence $(a_n) \subseteq A$ such that $a_n \to s$.
The completeness axiom ensures that $\sup A$ exists in $\mathbb{R}$ whenever $A$ is non-empty and bounded above. The property does not hold in $\mathbb{Q}$. For example, the set $\\{q \in \mathbb{Q} : q^2 < 2\\}$ is bounded above in $\mathbb{Q}$, but its least upper bound is $\sqrt{2}$, which is not a rational number. In this case, the supremum exists, but it does not belong to the space. Such a situation cannot occur in $\mathbb{R}$.
Infimum
Consider a non-empty subset $A \subseteq \mathbb{R}$ that is bounded below. The infimum of $A$, denoted $\inf A$, is its greatest lower bound. A real number $i$ is equal to $\inf A$ if and only if both of the following conditions are satisfied. The first condition states that $i$ is a lower bound of $A$:
$$ a \geq i \quad \forall \ a \in A $$
The second condition ensures that any number strictly greater than $i$ is preceded by some element of $A$:
$$ \forall \ \varepsilon > 0 \quad \exists \ a \in A : a < i + \varepsilon $$
Together, these conditions uniquely determine $i$. There can be only one greatest lower bound. If both $i$ and $i'$ satisfy the definition, we have:
$$ i \geq i' \ \wedge \ i' \geq i \ \to \ i = i' $$
An equivalent characterisation states that $i = \inf A$ if and only if $i$ is a lower bound of $A$ and there exists a sequence $(a_n) \subseteq A$ such that $a_n \to i$.
A concrete example is the set $\\{ 1/n : n \in \mathbb{N} \\}.$ Each term is positive, so $0$ is a lower bound, and no positive number can be a lower bound, because the Archimedean property supplies an $n$ with $1/n$ below any proposed positive threshold. Hence $\inf \\{ 1/n : n \in \mathbb{N} \\} = 0.$ The value $0$ is not one of the terms, so the set has an infimum but no minimum.
Supremum and maximum, infimum and minimum
The relationship between supremum and maximum, as well as between infimum and minimum, is frequently misunderstood. The maximum of a set $A$ is an element of $A$ that is greater than or equal to every other element. When a maximum exists, we have:
$$ \max A = \sup A $$
The supremum does not necessarily belong to the set $A$. Consider $A = (0, 1)$. Every element of $A$ is strictly less than $1$, so $\sup A = 1$. Since $1 \notin A$, the set $A$ has no maximum. The number $1$ is the least upper bound, but it is not an element of $A$. Similarly, $\inf A = 0$, yet $0 \notin A$, so $A$ has no minimum. By contrast, for the closed interval $B = [0,1]$ we have:
$$ \sup B = \max B = 1 \qquad \inf B = \min B = 0 $$
since the boundary points are included in the set.
In general, the following implications hold:
$$ \max A \text{ exists} \ \to \ \max A = \sup A $$
$$ \min A \text{ exists} \ \to \ \min A = \inf A $$
The converse does not hold in general. Whether a function actually attains its supremum is a non-trivial question. The Weierstrass theorem gives a sufficient condition: if a function is continuous on a closed and bounded interval, then the supremum and infimum are attained, and the maximum and minimum exist. Outside these conditions, the question must be examined case by case.
Supremum and infimum of functions
The concepts of supremum and infimum extend naturally to functions. For a function $f : D \to \mathbb{R}$, the supremum of $f$ over $D$ is the supremum of its image:
$$ \sup_{x \in D} f(x) = \sup \\{ f(x) : x \in D \\} $$
Similarly, the infimum is defined as:
$$ \inf_{x \in D} f(x) = \inf \\{ f(x) : x \in D \\} $$
These quantities are the least upper bound and greatest lower bound of the values assumed by $f$, and neither is necessarily attained.
- A real number $s$ is equal to $\sup_{x \in D} f(x)$ if and only if $f(x) \leq s$ for all $x \in D$, and for every $\varepsilon > 0$ there exists $x \in D$ such that $f(x) > s - \varepsilon$.
- Symmetrically, a real number $i$ is equal to $\inf_{x \in D} f(x)$ if and only if $f(x) \geq i$ for all $x \in D$, and for every $\varepsilon > 0$ there exists $x \in D$ such that $f(x) < i + \varepsilon$.
The supremum and infimum of a function are not necessarily attained. For the function $f(x) = x$ defined on the open interval $(0, 1)$, $\sup_{x \in (0,1)} f(x) = 1$, yet there is no $x \in (0, 1)$ such that $f(x) = 1$. If the supremum is attained at some point $x_0 \in D$, that is, $f(x_0) = \sup_{x \in D} f(x)$, it coincides with the maximum of $f$ over $D$. The same relationship holds between the infimum and the minimum.
Supremum and infimum are bounds that the function may approach but need not attain, whereas maximum and minimum are values that the function actually reaches at specific points of $D$.
Algebra of suprema and infima
Suprema and infima behave simply under translation and scaling of a set, which often simplifies their computation. For a non-empty bounded set $A \subseteq \mathbb{R}$ and a real number $c,$ write $c + A = \\{ c + a : a \in A \\}$ and $cA = \\{ ca : a \in A \\}.$ Translating a set shifts both bounds by the same amount:
$$ \sup(c + A) = c + \sup A \qquad \inf(c + A) = c + \inf A $$
Scaling by a positive factor scales the bounds while preserving their roles:
$$ \sup(cA) = c \sup A \qquad \inf(cA) = c \inf A \qquad (c > 0) $$
Scaling by a negative factor exchanges the two, since multiplication by a negative number reverses the order:
$$ \sup(cA) = c \inf A \qquad \inf(cA) = c \sup A \qquad (c < 0) $$
Each identity is verified by checking that the right-hand side satisfies the two defining conditions of the bound on the left. The choice $c = -1$ gives the useful special case $\sup(-A) = -\inf A,$ which turns any statement about infima into one about suprema.
A comparison of two sets follows the same reasoning. Suppose $A$ and $B$ are non-empty and every element of $A$ is at most every element of $B,$ that is $a \leq b$ for all $a \in A$ and $b \in B.$ Each $a$ is then a lower bound for $B,$ so $a \leq \inf B,$ and $\inf B$ is in turn an upper bound for $A.$ Taking the least upper bound gives:
$$ \sup A \leq \inf B $$
The conclusion stays non-strict even when the hypothesis is sharpened to $a < b$ for every pair. With $A = \\{ 0 \\}$ and $B = \\{ 1/n : n \in \mathbb{N} \\}$ one has $a < b$ throughout, yet both extrema equal $0,$ so a strict separation $\sup A < \inf B$ cannot be expected.
The approximation property
The $\varepsilon$-characterisation of the supremum and infimum is more than a definitional detail. It is the form in which these concepts most frequently appear in proofs, and is often presented as a standalone property. If $s = \sup A$, then for every $\varepsilon > 0$ there exists an element $a \in A$ such that:
$$ s - \varepsilon < a \leq s $$
Equivalently, no number strictly less than $s$ is an upper bound for $A$. The analogous statement applies to the infimum: if $i = \inf A$, then for every $\varepsilon > 0$ there exists $a \in A$ such that:
$$ i \leq a < i + \varepsilon $$
The property is used throughout analysis whenever one needs to extract elements of a set arbitrarily close to its supremum or infimum, and it appears naturally in existence arguments such as the proof of the Bolzano-Weierstrass theorem and the construction of the Riemann integral.