Supremum

Beyond Maximum and Minimum

A maximum is an upper bound that belongs to the set. But, as the previous tutorial showed, a set can have an upper bound without having a maximum. For example, the numbers in an open interval can approach its right endpoint without reaching it. The supremum records the least upper bound in both situations: whether or not that bound is an element of the set.

Recall from “Upper Bounds” that \(U\) is an upper bound for \(S\subseteq\mathbb{R}\) when \(x\leq U\) for every \(x\in S\). The key additional idea is to compare all of a set’s upper bounds. The supremum is the smallest one.

The definition has two parts: \(\alpha\) must first be an upper bound, and it must then be no larger than any other upper bound. A maximum, by contrast, must belong to \(S\). Thus a supremum need not be a maximum, though every maximum is also the supremum.

The Completeness Property of the Real Numbers

In the real numbers, every nonempty set that is bounded above has a supremum. This guarantee is the least-upper-bound property, also called the completeness property, of \(\mathbb{R}\). It is a fundamental property of the real numbers, not a consequence of the field and order rules alone. In the development of real analysis, it is what ensures that a least upper bound exists when a set is nonempty and has an upper bound.

Both conditions matter. The empty set has no supremum in \(\mathbb{R}\): every real number is an upper bound for it, but there is no least real number. A set that is unbounded above has no upper bounds, so it cannot have a supremum as defined here. For a nonempty set bounded above, however, the Completeness Principle guarantees that the supremum exists.

An Approximation Test for the Supremum

The definition says that the supremum is no larger than any other upper bound. A useful equivalent test describes how elements of the set must come arbitrarily close to it from below. The test makes precise the idea that a supremum cannot be lowered, even by a small amount, while remaining an upper bound.

Proof. First suppose \(\alpha=\sup S\), and let \(\varepsilon>0\). If no \(s\in S\) satisfied \(\alpha-\varepsilon<s\), then every \(s\in S\) would satisfy \(s\leq\alpha-\varepsilon\). This would make \(\alpha-\varepsilon\) an upper bound for \(S\). But \(\alpha-\varepsilon<\alpha\), contradicting that \(\alpha\) is the least upper bound. Therefore there is an \(s\in S\) with \(\alpha-\varepsilon<s\).

Conversely, suppose that for every \(\varepsilon>0\) there is an \(s\in S\) with \(\alpha-\varepsilon<s\). We already know that \(\alpha\) is an upper bound. Let \(U\) be any upper bound for \(S\). If \(U<\alpha\), then \(\varepsilon=\alpha-U>0\). The assumed property gives some \(s\in S\) such that \(\alpha-\varepsilon<s\). Since \(\alpha-\varepsilon=U\), this says \(U<s\), contradicting that \(U\) is an upper bound. Thus no upper bound is less than \(\alpha\), so \(\alpha\leq U\) for every upper bound \(U\). Hence \(\alpha=\sup S\). \(\square\)

When Is the Supremum a Maximum?

The Approximation Characterization also clarifies the relationship between a supremum and a maximum. If the supremum is in the set, its upper-bound property makes it the maximum. Conversely, if a maximum exists, every upper bound must be at least that maximum, so the maximum is the least upper bound.

Proof. Write \(\alpha=\sup S\). By definition, \(\alpha\) is an upper bound for \(S\). If \(\alpha\in S\), then \(\alpha\) belongs to the set and every \(s\in S\) satisfies \(s\leq\alpha\). These are exactly the conditions for \(\alpha\) to be a maximum.

Conversely, suppose \(M\) is a maximum of \(S\). Then \(M\in S\), and \(M\) is an upper bound. Since \(\sup S\) is the least upper bound, \(\sup S\leq M\). But \(M\in S\) and \(\sup S\) is an upper bound, so \(M\leq\sup S\). Antisymmetry gives \(M=\sup S\), and therefore the supremum belongs to \(S\). \(\square\)

A common mistake is to conclude that a supremum is a maximum merely because it bounds the set. The membership condition is essential. The open interval example has supremum \(4\), but \(4\notin(1,4)\), so it has no maximum.

Suprema of Unions

The upper-bound viewpoint also makes it possible to find the supremum of a union from the suprema of its two parts. The larger of the two suprema bounds both sets. It is the least such bound because any upper bound for the union must bound each set separately.

Proof. By the Completeness Principle, \(\sup S\) and \(\sup T\) exist. Let \(\alpha\) be the larger of these two numbers. Since \(\sup S\) is an upper bound for \(S\) and \(\sup S\leq\alpha\), every element of \(S\) is at most \(\alpha\). Similarly, every element of \(T\) is at most \(\alpha\). Thus \(\alpha\) is an upper bound for \(S\cup T\).

Now let \(U\) be any upper bound for \(S\cup T\). It is an upper bound for \(S\) and for \(T\), so the least-upper-bound definitions give \(\sup S\leq U\) and \(\sup T\leq U\). Hence their larger value \(\alpha\) also satisfies \(\alpha\leq U\). Therefore \(\alpha\) is the least upper bound of \(S\cup T\), proving the result. \(\square\)

Suprema allow us to describe the precise upper edge of a set without requiring that the set contain an element at that edge. The Completeness Principle guarantees that this edge exists for every nonempty set bounded above; the approximation test verifies a proposed value; and membership determines whether that supremum is also a maximum.