Why Does a Nonempty Bounded Set of Reals Have a Supremum?

A supremum is the smallest number that is at least as large as every element of a set. The existence question is answered by a defining property of the real numbers: every nonempty subset of $\mathbb{R}$ that is bounded above has a supremum in $\mathbb{R}$. This property is called completeness or the least-upper-bound property.

The existence theorem

Let $A\subseteq\mathbb{R}$ be nonempty and bounded above. Then there is a real number $s$ such that:

1. $s$ is an upper bound for $A$: $a\le s$ for every $a\in A$.
2. $s$ is no larger than any other upper bound for $A$.

That number is written $s=\sup A$. The second condition is what makes it the least upper bound.

In a standard real-analysis course, this is usually taken as an axiom or established earlier in the construction of the real numbers. It cannot be derived just from the usual order and field rules: the rational numbers satisfy those rules, but are not complete.

How the definition verifies a supremum

To prove that a proposed number $s$ is $\sup A$, show both of the following:

• Upper-bound condition: every $a\in A$ satisfies $a\le s$.
• Arbitrarily-close condition: for every $\varepsilon>0$, some $a\in A$ satisfies $s-\varepsilon<a$.

The second condition says that no number strictly below $s$ can still be an upper bound. Indeed, if $u<s$, take $\varepsilon=s-u>0$. Then there is an $a\in A$ with $a>s-\varepsilon=u$, so $u$ is not an upper bound.

This is often the most practical way to identify a supremum once existence is guaranteed.

Worked example: the interval $(0,1)$

Let $A=(0,1)$. Every element of $A$ is less than $1$, so $1$ is an upper bound. Now let $u<1$ be any smaller real number. Choose $0<\varepsilon<\min\{1,1-u\}$. The number $a=1-\varepsilon$ belongs to $(0,1)$ and is greater than $u$. Thus $u$ is not an upper bound.

Therefore $\sup A=1$. But $1\notin A$, so $A$ has no maximum. A maximum must be an element of the set; a supremum does not have to be.

Why completeness matters

Consider the set of rational numbers

$$B=\{q\in\mathbb{Q}: q>0\text{ and }q^2<2\}.$$

It is nonempty and bounded above in $\mathbb{Q}$. In $\mathbb{R}$, its supremum is $\sqrt{2}$. But $\sqrt{2}$ is irrational, so it is not in $\mathbb{Q}$. In fact, $B$ has no supremum that is a rational number: any rational upper bound below $\sqrt{2}$ is not an upper bound, while any rational upper bound above $\sqrt{2}$ is not the least one.

So the theorem is specific to a complete number system such as $\mathbb{R}$. The same assertion is false for arbitrary ordered fields, including $\mathbb{Q}$.

If the real numbers are defined using Dedekind cuts, the existence theorem can be seen directly. A real number is represented by the rationals below it. For a nonempty family of real numbers bounded above, take the union of their corresponding sets of lower rationals. That union is itself a Dedekind cut and represents the least upper bound. This is one construction that builds completeness into $\mathbb{R}$.

Common mistakes

• Forgetting nonemptiness: Under the usual real-valued definition, the empty set has no supremum in $\mathbb{R}$; every real number is an upper bound, and there is no least real number.
• Forgetting boundedness above: A set unbounded above has no real supremum. In the extended real numbers, one may assign it supremum $+\infty$ instead.
• Confusing supremum with maximum: The supremum may lie outside the set, as $1$ does for $(0,1)$.
• Trying to prove completeness from order rules alone: Completeness is an additional property distinguishing $\mathbb{R}$ from $\mathbb{Q}$.

When working within real analysis, use the theorem once you have checked that the set is nonempty and bounded above. Then identify the candidate and verify that it is an upper bound that can be approached from within the set.