Maximum Versus Supremum

Attained Edges and Unattained Edges

The supremum describes the least upper bound of a set. A maximum describes an upper edge that the set actually contains. These ideas can give the same number, but they are not interchangeable: a set may approach its supremum without ever reaching it.

Recall from “Upper Bounds” that an upper bound for $S\subseteq\mathbb{R}$ is a number $U$ such that $x\leq U$ for every $x\in S$. The supremum is the least upper bound. A maximum, when it exists, must be an element of the set and must be at least as large as every element.

Recall (Maximum). A number $M$ is the maximum of $S\subseteq\mathbb{R}$ if $M\in S$ and $x\leq M$ for every $x\in S$. It is denoted by $\max S$. The supremum $\sup S$, when it exists, is the least upper bound of $S$; it need not belong to $S$.

Theorem (A Supremum Is a Maximum Exactly When It Is Attained), from “Supremum,” gives the precise connection: for a nonempty set bounded above, its supremum is a maximum if and only if the supremum belongs to the set. We will use this earlier result rather than re-prove it. The practical distinction is membership: every maximum is attained, while a supremum may only mark a boundary the set gets arbitrarily close to.

Question	Maximum	Supremum
Must it be an upper bound?	Yes	Yes
Must it belong to the set?	Yes	No
Must it be at least as large as every other upper bound?	No	Yes; it is the least upper bound
Can it fail to exist for a nonempty set bounded above?	Yes	No, by completeness of $\mathbb{R}$

Worked Example: A Supremum That Is Not a Maximum

Let $S=(1,9)$. Every $x\in S$ satisfies $x<9$, so $9$ is an upper bound. To see that it is the least upper bound, use the Approximation Characterization of the Supremum. Given $\varepsilon>0$, let $\delta=\min\{\varepsilon,8\}$ and choose $x=9-\delta/2$. Since $0<\delta\leq8$, we have $x\geq5>1$ and $x<9$, so $x\in S$. Also $\delta/2<\varepsilon$, and therefore $x=9-\delta/2>9-\varepsilon$. Thus, for every positive $\varepsilon$, there is an element of $S$ greater than $9-\varepsilon$. The approximation characterization gives $\sup S=9$.

But $9\notin S$. No element of $S$ can be its maximum: if $m\in S$, then $m<9$, and $(m+9)/2$ is also in $S$ and is strictly greater than $m$. Hence $S$ has a supremum but no maximum. The endpoint $9$ is an upper edge, not an attained value.

Worked Example: A Maximum and a Supremum That Agree

Let $T=\{-5,\,2,\,11/3,\,8\}$. The number $8$ belongs to $T$, and $-5\leq8$, $2\leq8$, $11/3\leq8$, and $8\leq8$. Thus $8$ is the maximum of $T$. It is also an upper bound. Every upper bound for $T$ must be at least $8$, because $8\in T$. Therefore $8$ is the least upper bound as well: $\max T=\sup T=8$.

Here the supremum is attained. Its membership in $T$ is what makes it a maximum; simply being the least upper bound would not, by itself, guarantee membership.

Points Approaching a Supremum

When a set has a supremum, its elements can be found as close to that supremum as desired, from below or at the supremum. The Approximation Characterization of the Supremum states this using an arbitrary positive tolerance. Choosing a particular sequence of tolerances gives a useful way to record the approximation.

Theorem (Sequential Approximation to the Supremum). Let $S\subseteq\mathbb{R}$ be nonempty, and let $\alpha\in\mathbb{R}$. Then $\alpha=\sup S$ if and only if $\alpha$ is an upper bound for $S$ and there is a sequence $(s_n)_{n\geq1}$ of elements of $S$ such that $$ \alpha-\frac{1}{n}<s_n\leq\alpha $$ for every positive integer $n$.

Proof. First suppose $\alpha=\sup S$. Then $\alpha$ is an upper bound. For each positive integer $n$, apply the Approximation Characterization of the Supremum with $\varepsilon=1/n$. It supplies an $s_n\in S$ such that $s_n>\alpha-1/n$. Since $\alpha$ is an upper bound, $s_n\leq\alpha$. This gives the required sequence.

Conversely, suppose $\alpha$ is an upper bound and there are elements $s_n\in S$ satisfying the displayed inequalities for every $n\geq1$. Let $U$ be any upper bound for $S$. We show that $\alpha\leq U$. If instead $U<\alpha$, choose a positive integer $n$ such that $1/n<\alpha-U$. Such an integer exists by the Archimedean property of the real numbers. Then $$ s_n>\alpha-\frac{1}{n}>U. $$ But $U$ is an upper bound and $s_n\in S$, so $s_n\leq U$, a contradiction. Thus every upper bound $U$ satisfies $\alpha\leq U$. Since $\alpha$ is itself an upper bound, it is the least upper bound, so $\alpha=\sup S$. $\square$

The inequalities explain why this approximation does not assert that the supremum is attained. Every $s_n$ may be strictly below $\alpha$. If one of the selected elements equals $\alpha$, then $\alpha\in S$, and the earlier attainment criterion says that $\alpha$ is a maximum. If none does, the sequence still records how closely the set comes to its unattained upper edge.

Worked Example: Choosing Elements Near an Unattained Supremum

Let $A=(0,5)$, and set $\alpha=5$. Every element of $A$ is less than $5$, so $5$ is an upper bound. For each positive integer $n$, define $s_n=5-1/(n+1)$. Since $n\geq1$, we have $0<1/(n+1)\leq1/2$, and hence $9/2\leq s_n<5$. Thus $s_n\in A$. Moreover, $$ 5-\frac{1}{n}<5-\frac{1}{n+1}=s_n<5, $$ because $1/(n+1)<1/n$. The Sequential Approximation Theorem gives $\sup A=5$. No $s_n$ equals $5$, and $5\notin A$, so this supremum is not a maximum.

This example also shows why one should not confuse “there are elements arbitrarily close to the supremum” with “there is an element equal to the supremum.” The former is guaranteed by the least-upper-bound property; the latter requires attainment.

Removing an Element Without a Maximum

A set without a maximum does not have one particular element at its upper edge. In fact, removing any single element from such a set leaves its supremum unchanged. This gives a useful contrast with a set whose maximum is an essential attained endpoint: removing that maximum can lower the supremum.

Theorem (Deleting One Element from a Set Without a Maximum). Let $S\subseteq\mathbb{R}$ be nonempty and bounded above, and suppose $S$ has no maximum. For every $x\in S$, $$ \sup\bigl(S\setminus\{x\}\bigr)=\sup S. $$

Proof. Write $\alpha=\sup S$. By Theorem (A Supremum Is a Maximum Exactly When It Is Attained), $\alpha\notin S$, since otherwise $\alpha$ would be a maximum. Fix $x\in S$. In particular, $x<\alpha$. The set $S\setminus\{x\}$ is nonempty: if it were empty, then $S=\{x\}$, and $x$ would be a maximum.

Because $S\setminus\{x\}\subseteq S$, $\alpha$ is an upper bound for $S\setminus\{x\}$. Now let $\varepsilon>0$, and define $$ \delta=\frac{\min\{\varepsilon,\alpha-x\}}{2}. $$ Both numbers in the minimum are positive, so $\delta>0$. Also $\delta<\varepsilon$ and $\delta<\alpha-x$. By the Approximation Characterization of the Supremum, there is an $s\in S$ such that $s>\alpha-\delta$. Since $\delta<\alpha-x$, we have $\alpha-\delta>x$, and consequently $s>x$. Thus $s\neq x$, so $s\in S\setminus\{x\}$. Finally, $\delta<\varepsilon$ gives $$ s>\alpha-\delta>\alpha-\varepsilon. $$ For every positive $\varepsilon$, the set after deletion therefore contains an element greater than $\alpha-\varepsilon$. Applying the Approximation Characterization again shows that its supremum is $\alpha$. $\square$

Worked Example: Deleting a Point Does Not Change the Supremum

Let $B=(2,6)$ and delete the point $4$, forming $B\setminus\{4\}$. The number $6$ is an upper bound for this remaining set. Given $\varepsilon>0$, let $\delta=\min\{\varepsilon,4\}/2$ and choose $x=6-\delta$. We have $0<\delta\leq2$, so $4\leq x<6$, and $x\neq4$ unless $x=4$. In that boundary case, choose instead $x=6-\delta/2$; this satisfies $x>4$, $x<6$, and $x>6-\varepsilon$. In either case there is an element of $B\setminus\{4\}$ greater than $6-\varepsilon$. Thus $\sup(B\setminus\{4\})=6$. The set still approaches $6$, even though one of its interior points is gone.

Contrast this with $C=\{2,5\}$. Its maximum and supremum are both $5$. After deleting $5$, the remaining set is $\{2\}$, whose supremum is $2$. Here the deleted point was the attained upper edge. In the preceding example, there was no maximum point to remove.

What the Distinction Is Good For

When deciding whether a proposed number is a maximum, check membership as well as order: the number must belong to the set and dominate every element. To verify a supremum, membership is unnecessary. Instead, show that the number is an upper bound and that elements of the set occur arbitrarily close below it, using the Approximation Characterization of the Supremum.

A frequent error is to conclude that every bounded-above set has a maximum. Completeness guarantees a supremum for every nonempty set bounded above, but that supremum can lie outside the set. The interval $(1,9)$ is a direct example. Conversely, when the supremum is in the set, the attainment criterion immediately identifies it as the maximum. Keeping these conditions separate prevents confusing an edge that is approached with one that is reached.

Key takeaway. A maximum is an attained upper edge; a supremum is the least upper bound, whether or not it is attained. For a nonempty set that is bounded above and has no maximum, deleting any one element leaves its supremum unchanged.

Check Your Understanding

Use the definitions and results in this tutorial to answer each question.

What additional condition must a maximum satisfy that a supremum need not satisfy?
State the criterion that tells when the supremum of a nonempty set bounded above is also its maximum.
What inequalities can be used to select elements approximating a supremum for each positive integer $n$?
Why does deleting one element from a set without a maximum leave its supremum unchanged?
What happens to the supremum of $\{2,5\}$ when $5$ is deleted, and how does this differ from deleting an element of $(2,6)$?

What You'll Learn