Supremum and Infimum

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, particularly a unique property of that order. Specifically, every non-empty subset of $\mathbb{R}$ that is bounded above possesses a least upper bound that remains within $\mathbb{R}$. This property is known as the completeness axiom and serves as a foundational characteristic of the real line. The notions of supremum and infimum provide practical means to apply this axiom.

Upper and lower bounds

Consider a non-empty set $A \subseteq \mathbb{R}$. A real number $M$ is defined as an upper bound of $A$ if:

If such a number exists, the set $A$ is bounded above. Similarly, a real number $m$ is a lower bound of $A$ if:

In this case, $A$ is bounded below. A set is considered bounded if it is both bounded above and below, that is, there exists $K > 0$ such that:

Upper bounds, when they exist, are generally not unique. For example, 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 $A$, so is $m - 1$. 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 defined as 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 says that $s$ is an upper bound of $A$: \(a \leq s \forall a \in A\)

The second condition ensures that any number strictly less than $s$ is exceeded by some element of $A$: \(\forall \epsilon > 0 \exists a \in A : a > s - \epsilon\)

Together, these conditions determine $s$. There can be only one least upper bound. If both $s$ and $s^{‘}$ satisfy the definition, then we have:

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 $a_{n} \rightarrow s$.

The completeness axiom ensures that $sup A$ exists in $\mathbb{R}$ whenever $A$ is non-empty and bounded above. This 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 defined as 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 says that $i$ is a lower bound of $A$:

The second condition ensures that any number strictly greater than $i$ is preceded by some element of $A$:

Together, these conditions uniquely determine $i$. There can be only one greatest lower bound. If both $i$ and $i^{‘}$ satisfy the definition, then we have

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 $a_{n} \rightarrow i$.

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 defined as an element of $A$ that is greater than or equal to every other element. When a maximum exists we have:

The supremum does not necessarily belong to the set $A$. For example, consider $A = ( 0 , 1 )$. Every element of $A$ is strictly less than 1, so $sup A = 1$. Since $1 \notin A$, the set $A$ does not possess a maximum. The number 1 serves as the least upper bound, but it is not an element of $A$. Similarly, $inf A = 0$, yet $0 \notin A$, so $A$ lacks a minimum. In contrast, for $B = [ 0 , 1 ]$ we have:

since the boundary points are included in the set.

In general, the following holds:

\(max A \text{exists} \rightarrow max A = sup A\) \(min A \text{exists} \rightarrow min A = inf A\)

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 $f : D \rightarrow \mathbb{R}$, the supremum of $f$ over $D$ is defined as the supremum of its image:

Similarly, the infimum is defined as:

These quantities represent the least upper bound and greatest lower bound of the values assumed by $f$, without requiring that these bounds are actually attained.

• A real number $s$ is equal to $\underset{x \in D}{sup} f ( x )$ if and only if $f ( x ) \leq s$ for all $x \in D$, and for every $\epsilon > 0$, there exists $x \in D$ such that $f ( x ) > s - \epsilon$.
• Symmetrically, a real number $i$ is equal to $\underset{x \in D}{inf} f ( x )$ if and only if $f ( x ) \geq i$ for all $x \in D$, and for every $\epsilon > 0$, there exists $x \in D$ such that $f ( x ) < i + \epsilon$.

The supremum and infimum of a function are not necessarily attained. For instance, for $f ( x ) = x$ defined on the open interval $( 0 , 1 )$, $\underset{x \in ( 0 , 1 )}{sup} f ( x ) = 1$, yet there is no $x \in ( 0 , 1 )$ such that $f ( x ) = 1$. If the supremum is attained at some point $x_{0} \in D$, meaning $f ( x_{0} ) = \underset{x \in D}{sup} f ( x )$, it coincides with the maximum of $f$ over $D$. The same relationship holds between the infimum and the minimum.

The definitions of supremum and infimum for a function prompt consideration of their distinction from maximum and minimum values. Supremum and infimum represent bounds that the function may approach but does not necessarily attain, whereas maximum and minimum refer to values that the function actually achieves at specific points in $D$.

The approximation property

The $\epsilon$-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. This characterisation is often presented as a standalone property. If $s = sup A$, then for every $\epsilon > 0$ there exists an element $a \in A$ such that

Equivalently, no number strictly less than $s$ serves as an upper bound for $A$. The analogous statement applies to the infimum: if $i = inf A$, then for every $\epsilon > 0$ there exists $a \in A$ such that:

This 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.

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