Does Every Nonempty Bounded-Above Set of Real Numbers Have a Supremum?

Yes—provided that $B$ is a nonempty subset of the real numbers. If $B\subseteq\mathbb{R}$ is bounded above, then there is a real number $s$ that is the supremum of $B$. This follows from the completeness property of $\mathbb{R}$.

There is one small qualification: the empty set is bounded above under the usual definition, but it has no supremum in $\mathbb{R}$. Also, boundedness alone does not guarantee a supremum if the set is in a number system that is not complete, such as $\mathbb{Q}$.

What “supremum” means

A number $u$ is an upper bound for $B$ if every element of $B$ is at most $u:

$$\forall x\in B,\quad x\le u.$$

The supremum $s=\sup B$ is the least upper bound: it is an upper bound, and it is no greater than any other upper bound. In symbols,

$$(\forall x\in B,\ x\le s)\quad\text{and}\quad (\forall u\in\mathbb{R},\ [\forall x\in B,\ x\le u]\Rightarrow s\le u).$$

“Least” refers to the numerical order, not to membership. The supremum may or may not be in $B$. If it is in $B$, it is also the maximum of $B$.

Why boundedness gives a supremum in $\mathbb{R}$

The real numbers satisfy the least-upper-bound property, also called completeness: every nonempty subset of $\mathbb{R}$ that has an upper bound has a least upper bound in $\mathbb{R}$.

So, once you know that $B$ is nonempty and bounded above, the existence of $\sup B$ is precisely an application of this property. It is not necessary to produce a formula for the supremum or to show that the set contains it. A proof that a particular number is the supremum must, however, verify that it is an upper bound and that no smaller number is an upper bound.

Worked example: a supremum that is not in the set

Let

$$B=\left\{1-\frac{1}{n}:n\in\mathbb{N},\ n\ge1\right\} =\left\{0,\frac12,\frac23,\frac34,\ldots\right\}.$$

Every element is less than $1$, so $1$ is an upper bound. To see that it is the least upper bound, take any real number $u<1$. Choose a positive integer $n$ large enough that $1/n<1-u$. Then

$$1-\frac1n>u,$$

so $u$ is not an upper bound for $B$. Thus no number less than $1$ is an upper bound, and $\sup B=1$. But $1$ is not in $B$: for every finite positive integer $n$, $1-1/n<1$. The set has a supremum but no maximum.

Why the qualifications matter

The empty set. The statement “$B$ is bounded above” means that some $M\in\mathbb{R}$ satisfies $x\le M$ for every $x\in B$. For $B=\varnothing$, this condition is true for every $M$, because there are no elements that could violate it. But there is no least real upper bound: if a real number $s$ were the least upper bound, then $s-1$ would also be an upper bound and would be smaller. Some treatments of the extended real numbers define $\sup\varnothing=-\infty$, but that is not a real supremum.

A different ambient number system. Completeness is special to $\mathbb{R}$; $\mathbb{Q}$ does not have the least-upper-bound property. For example,

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

is nonempty and bounded above in $\mathbb{Q}$. Its least upper bound in $\mathbb{R}$ is $\sqrt2$, which is irrational. It has no supremum in $\mathbb{Q}$. Thus, a claim about existence must specify the ambient ordered set; boundedness in the rationals is not enough.

Common mistakes

• Forgetting nonemptiness: Under the usual definition, the empty set is bounded above, but it has no supremum in $\mathbb{R}$.
• Assuming the supremum must be attained: A supremum need not belong to the set. In the example, $1$ is approached by elements of $B$ but is never reached.
• Confusing boundedness with completeness: Boundedness says an upper bound exists. Completeness is what guarantees that there is a least upper bound in the ambient number system.
• Switching number systems without checking: A set can have a real supremum that is not rational, so it need not have a supremum when considered as a subset of $\mathbb{Q}$.