How to Calculate the Supremum of a Set

The supremum of a set is its least upper bound. To calculate it, you need to establish two facts: your candidate is an upper bound, and no smaller number is an upper bound. This second part is why simply spotting a number that bounds the set is not enough.

A supremum does not have to be an element of the set. When it is an element, it is also the set’s maximum. The distinction matters especially for open intervals and sequences that approach a limit without reaching it.

What the supremum means

Let $A$ be a nonempty set of real numbers. A number $u$ is an upper bound of $A$ if $a\leq u$ for every $a\in A$. The supremum, written $\sup A$, is the smallest of all such upper bounds.

Equivalently, a real number $s$ is $\sup A$ precisely when:

1. $a\leq s$ for every $a\in A$; and
2. for every $\varepsilon>0$, some $a\in A$ satisfies $s-\varepsilon<a$.

The first condition says that $s$ is an upper bound. The second says that elements of $A$ get as close to $s$ as you like from below. Together, they rule out any smaller upper bound.

In the real numbers, every nonempty set that is bounded above has a supremum. If a set is unbounded above, it has no real supremum. Some contexts use the extended real numbers and write its supremum as $+\infty$.

A step-by-step method

1. Identify the set and its domain. For a sequence, note which indices are allowed; for an interval or function, check whether endpoints are included.
2. Guess a candidate. Use the pattern of values, inequalities, or endpoint behavior to find a possible least upper bound.
3. Prove it is an upper bound. Verify that every element of the set is at most the candidate.
4. Prove it is the least upper bound. Show that for every $\varepsilon>0$, at least one element exceeds the candidate minus $\varepsilon$.
5. Check whether the bound is attained. If the candidate is in the set, it is the maximum. If not, the set has no maximum at that value.

For a finite, nonempty set, this process is simple: the supremum is the largest element. For infinite sets, the last two proof steps are often the main work.

Worked example: a sequence approaching 3

Let

$$A=\left\{3-\frac{1}{n}:n\in\mathbb{N},\ n\geq 1\right\}.$$

The first few elements are $2$, $2.5$, and approximately $2.667$. We will show that $\sup A=3$.

First, show 3 is an upper bound. Since $n\geq1$, we have $1/n>0$, so

$$3-\frac{1}{n}<3.$$

Thus every element of $A$ is less than 3.

Next, show that no smaller number is an upper bound. Let $\varepsilon>0$. Choose a positive integer $n$ large enough that $n>1/\varepsilon$. Then $1/n<\varepsilon$, and therefore

$$3-\frac{1}{n}>3-\varepsilon.$$

So however close to 3 you choose a number from below, some element of $A$ lies above it. By the supremum criterion, $\sup A=3$.

But $3\notin A$: no finite positive integer $n$ makes $1/n=0$. Therefore $A$ has no maximum. Its supremum is 3 even though it never reaches 3.

Maximum versus supremum

Consider the interval $B=(1,4]$. Every element is at most 4, and 4 belongs to $B$. Hence $\sup B=4$, and 4 is also the maximum.

For $C=(1,4)$, every element is still less than 4, and elements can be chosen arbitrarily close to 4. Thus $\sup C=4$, but $C$ has no maximum because 4 is not included. The supremum describes the upper edge of a set, not necessarily an element in the set.

Common mistakes

• Giving an upper bound without proving it is least. For the sequence above, 10 is an upper bound, but it is not the supremum. You must show that values approach the proposed bound closely enough to exclude every smaller upper bound.
• Assuming the supremum must belong to the set. That is required for a maximum, not for a supremum.
• Confusing an unbounded set with one whose supremum is not attained. A set can approach a finite supremum without attaining it. An unbounded-above set has no finite real upper bound at all.
• Ignoring endpoint inclusion. The supremum of $(a,b)$ and $[a,b]$ is $b$ in both cases, but only the latter includes its supremum and has a maximum.

When the set is defined by a function, the same logic applies: find a candidate upper bound for all allowed inputs, then show that function values can approach it as closely as desired. Whether the function actually takes that value depends on the domain and whether the relevant endpoint or input is included.