The Supremum Strategy

A Sharp Bound Requires Two Arguments

The Compactness Strategy used finite subcovers or convergent subsequences to turn information about many points into a global conclusion. The Supremum Strategy also turns local information into a global statement, but it is driven by the least-upper-bound property of the real numbers. It is especially useful when a proof asks for the best possible upper bound, or when a set has no largest element but its elements approach a boundary.

To prove that a number $s$ is the supremum of a set $A$, it is not enough to show that every element of $A$ is at most $s$. That proves only that $s$ is an upper bound. One must also show that no smaller number is an upper bound. The practical way to do this is to show that for every $\varepsilon>0$, some element of $A$ lies above $s-\varepsilon$. This says that elements of $A$ can be found as close to $s$ as desired from below.

Definition (Supremum): Let $A$ be a nonempty subset of $\mathbb{R}$ that is bounded above. A number $s\in\mathbb{R}$ is the supremum of $A$, written $s=\sup A$, if $s$ is an upper bound for $A$ and $s\leq u$ for every upper bound $u$ of $A$.

The least-upper-bound property of $\mathbb{R}$ guarantees that every nonempty set bounded above has a supremum. In a proof, that property supplies existence; the work that remains is to identify the supremum. The following characterization turns that task into two explicit checks.

Theorem (Epsilon Characterization of the Supremum): Let $A\subseteq\mathbb{R}$ be nonempty and bounded above. A number $s$ is $\sup A$ if and only if (1) $a\leq s$ for every $a\in A$, and (2) for every $\varepsilon>0$, there exists $a\in A$ such that $s-\varepsilon<a$.

Proof. First suppose $s=\sup A$. By definition, $s$ is an upper bound, so $a\leq s$ for every $a\in A$. Let $\varepsilon>0$. If no $a\in A$ satisfied $s-\varepsilon<a$, then every $a\in A$ would satisfy $a\leq s-\varepsilon$. Thus $s-\varepsilon$ would be an upper bound for $A$. This contradicts the fact that $s$ is the least upper bound, since $s-\varepsilon<s$. Hence an $a\in A$ with $s-\varepsilon<a$ exists.

Conversely, suppose both conditions hold. The first says that $s$ is an upper bound. Let $u$ be any upper bound for $A$. We show $s\leq u$. If instead $u<s$, choose $\varepsilon=s-u$, which is positive. By the second condition, there is an $a\in A$ with

u=s-\varepsilon<a.

This contradicts that $u$ is an upper bound. Therefore every upper bound $u$ satisfies $s\leq u$, and $s=\sup A$. $\square$

The proof gives a useful witness principle: if $s=\sup A$ and $c<s$, there must be an element $a\in A$ with $c<a$. Indeed, take $\varepsilon=s-c$ in the characterization. This is often the exact step needed to contradict a proposed smaller upper bound.

Applying the Characterization

When identifying a supremum, keep the two parts of the characterization separate. First prove that the candidate is an upper bound. Then, given an arbitrary positive $\varepsilon$, produce an element of the set that lies above the candidate minus $\varepsilon$. The element may depend on $\varepsilon$; that dependence is expected.

Worked Example: A Set That Approaches Its Supremum Without Attaining It

Let $A=\{2-1/n:n\text{ is a positive integer}\}$. We claim that $\sup A=2$. For every positive integer $n$, $1/n>0$, so $2-1/n<2$. Thus $2$ is an upper bound for $A$.

Now let $\varepsilon>0$. Choose a positive integer $n$ with $n>1/\varepsilon$, using the Archimedean property of the real numbers. Then $1/n<\varepsilon$, and hence

2-\varepsilon<2-\frac{1}{n}<2.

So $A$ contains an element above $2-\varepsilon$ for every $\varepsilon>0$. The epsilon characterization gives $\sup A=2$. No element of $A$ equals $2$, because $2-1/n=2$ would imply $1/n=0$, which is impossible for a positive integer $n$. Thus the supremum need not be a member of the set.

A supremum is therefore not the same as a maximum. A maximum must belong to the set and be at least as large as every other element. A supremum is the least upper bound whether or not it belongs to the set. The epsilon condition handles both situations: it requires approximation from below, not attainment.

Worked Example: The Supremum of a Set Defined by an Inequality

Consider $C=\{x\in\mathbb{R}:x\geq0\text{ and }x^2<16\}$. We show that $\sup C=4$. If $x\in C$ and $x\geq4$, then $x\geq0$ and

x^2-16=(x-4)(x+4)\geq0,

so $x^2\geq16$, contrary to $x^2<16$. Therefore every $x\in C$ satisfies $x<4$, and $4$ is an upper bound.

Given $\varepsilon>0$, set $\delta=\min(\varepsilon/2,1)$ and $x=4-\delta$. Since $0<\delta\leq1$, we have $3\leq x<4$, so $x\geq0$ and $x^2<16$; hence $x\in C$. Also $\delta<\varepsilon$: if $\varepsilon\leq2$, then $\delta=\varepsilon/2<\varepsilon$, and if $\varepsilon>2$, then $\delta=1<\varepsilon$. Consequently,

4-\varepsilon<4-\delta=x<4.

The epsilon characterization now gives $\sup C=4$. The explicit choice of $x$ verifies the approximation condition even when $\varepsilon$ is large; the set constraint $x\geq0$ has not been left implicit.

Combining Supremum Arguments

The same approximation method can prove formulas for sets built from other sets. For example, let $A+B=\{a+b:a\in A,\ b\in B\}$. If $A$ and $B$ are nonempty and bounded above, their individual suprema provide an upper bound for every sum. To show that this bound is least, approximate each supremum separately and add the resulting inequalities.

Theorem (Supremum of a Sum Set): Let $A,B\subseteq\mathbb{R}$ be nonempty and bounded above. Then $A+B$ is nonempty and bounded above, and $$ \sup(A+B)=\sup A+\sup B. $$

Proof. Write $\alpha=\sup A$ and $\beta=\sup B$. Since $A$ and $B$ are nonempty, choose $a_0\in A$ and $b_0\in B$; then $a_0+b_0\in A+B$, so $A+B$ is nonempty. For any $a\in A$ and $b\in B$, the upper-bound properties give $a\leq\alpha$ and $b\leq\beta$, whence $a+b\leq\alpha+\beta$. Thus $A+B$ is bounded above by $\alpha+\beta$.

It remains to show that elements of $A+B$ approach $\alpha+\beta$ from below. Let $\varepsilon>0$. The epsilon characterization applied to $A$ gives $a\in A$ with $\alpha-\varepsilon/2<a$, and applied to $B$ gives $b\in B$ with $\beta-\varepsilon/2<b$. Adding these inequalities yields

\alpha+\beta-\varepsilon<a+b.

Since $a+b\in A+B$, the epsilon characterization, now applied to the upper bound $\alpha+\beta$, shows that $\sup(A+B)=\alpha+\beta$. $\square$

The proof depends on being able to choose one element from each set within half the required error. It does not require either set to contain its supremum. Nor is it enough to say that each set has elements “near” its supremum: the error must be specified so that the two errors add to less than the target $\varepsilon$.

Worked Example: Computing a Supremum of Sums

Let $A=\{2-1/n:n\geq1\}$ and $B=\{5-3/m:m\geq1\}$, where $n$ and $m$ range over positive integers. The first worked example established $\sup A=2$. For $m\geq1$, $5-3/m\leq5$, and for every $\varepsilon>0$, choosing $m>3/\varepsilon$ gives $5-\varepsilon<5-3/m$. Thus $\sup B=5$. The Supremum of a Sum Set Theorem gives

\sup(A+B)=2+5=7.

The elements of the sum set have the form

\left(2-\frac{1}{n}\right)+\left(5-\frac{3}{m}\right) =7-\frac{1}{n}-\frac{3}{m}<7.

To verify directly that these sums approach $7$, let $\varepsilon>0$. Choose positive integers $n>2/\varepsilon$ and $m>6/\varepsilon$. Then $1/n<\varepsilon/2$ and $3/m<\varepsilon/2$, so

7-\varepsilon<7-\frac{1}{n}-\frac{3}{m}<7.

This confirms both the upper-bound calculation and the approximation step. In this example the supremum $7$ is not attained: attaining it would require $1/n+3/m=0$, although both terms are positive.

A Supremum Workflow and a Common Pitfall

The strategy is most reliable when the proof identifies the candidate and then treats its two roles separately. For a set defined through inequalities, first derive a bound valid for every element. For a set described by a formula, then choose an element whose parameters make it close to the candidate. In a contradiction proof, the witness principle can be used directly: a number strictly below the supremum cannot still be an upper bound.

Check existence conditions.
Verify that the set is nonempty and bounded above, so its supremum is defined.

Prove the candidate is an upper bound.
Show the claimed inequality for every element of the set, including any boundary cases allowed by its definition.

Choose an epsilon-dependent witness.
For an arbitrary $\varepsilon>0$, construct an element strictly above the candidate minus $\varepsilon$.

Conclude leastness.
Apply the epsilon characterization, or show directly that every smaller proposed upper bound is contradicted by a witness.

A common pitfall is to prove only that $s$ is an upper bound and then announce $s=\sup A$. Many larger numbers are also upper bounds; the defining feature is that no smaller number is an upper bound. Another pitfall is to assume the supremum belongs to the set. If the proof needs an actual element equal to the bound, it must establish attainment separately. The epsilon characterization supplies elements arbitrarily close from below, but it does not supply an element equal to the supremum.

Check Your Understanding

Use the epsilon characterization and the examples in this tutorial to answer these questions.

What two conditions must be proved to identify $s$ as the supremum of a nonempty set bounded above?
If $s=\sup A$ and $c<s$, how does the epsilon characterization produce an element of $A$ greater than $c$?
Why does $\sup\{2-1/n:n\geq1\}=2$ not imply that $2$ belongs to the set?
In the proof for $C=\{x\geq0:x^2<16\}$, why is it necessary to choose an element depending on $\varepsilon$?
In the sum-set theorem, why are the errors chosen as $\varepsilon/2$ for each set?
What additional fact must be proved before a supremum can be called a maximum?

What You'll Learn