Difference Between Supremum and Upper Bounds

An upper bound is any number that lies above every element of a set. A supremum is the least, or smallest, upper bound. Thus, every supremum is an upper bound, but most upper bounds are not the supremum.

The lower-bound distinction is the mirror image: a lower bound lies below every element, while the infimum is the greatest lower bound.

Definitions

Let $S$ be a nonempty subset of $\mathbb{R}$.

A number $u$ is an upper bound of $S$ if

$$x \leq u \quad \text{for every } x\in S.$$

A number $l$ is a lower bound of $S$ if

$$l \leq x \quad \text{for every } x\in S.$$

If $S$ has at least one upper bound, its supremum, written $\sup S$, is the smallest upper bound. Formally, $s=\sup S$ when:

1. $s$ is an upper bound of $S$; and
2. every upper bound $u$ of $S$ satisfies $s\leq u$.

Likewise, $i=\inf S$ when:

1. $i$ is a lower bound of $S$; and
2. every lower bound $l$ of $S$ satisfies $l\leq i$.

The words least and greatest refer to the ordering of real numbers, not to membership in the set.

Worked example: a finite set

Take

$$S=\{2,5,7\}.$$

Every number at least $7$ is an upper bound, so the set of upper bounds is

$$[7,\infty).$$

The smallest of these upper bounds is $7$. Therefore,

$$\sup S=7.$$

Similarly, every number at most $2$ is a lower bound. The set of lower bounds is

$$(-\infty,2],$$

so

$$\inf S=2.$$

Here, both extremal values belong to $S$. Consequently, $7$ is not only the supremum but also the maximum, and $2$ is both the infimum and the minimum.

A supremum need not belong to the set

Consider the open interval

$$S=(0,1).$$

The numbers $1$, $2$, and $10$ are all upper bounds, because every $x\in(0,1)$ satisfies $x<1\leq 2\leq 10$. In fact, every number at least $1$ is an upper bound:

$$\{u\in\mathbb{R}:u\text{ is an upper bound of }S\}=[1,\infty).$$

The least one is $1$, so

$$\sup S=1.$$

But $1\notin(0,1)$. Therefore, $S$ has no maximum. The distinction is:

• $1$ is the supremum because it is the least upper bound.
• $1$ is not a maximum because it is not an element of $S$.

For the lower side, $0$ is the infimum of $(0,1)$, but $0$ is not in the set, so there is no minimum.

Another example: a supremum defined by a square root

Let

$$S=\{x\in\mathbb{R}:x^2<2\text{ and }x\geq 0\}.$$

This is the interval $[0,\sqrt{2})$. Its supremum is $\sqrt{2}$:

• Every $x\in S$ satisfies $x<\sqrt{2}$, so $\sqrt{2}$ is an upper bound.
• No number smaller than $\sqrt{2}$ is an upper bound. If $u<\sqrt{2}$, one can choose a number in $S$ sufficiently close to $\sqrt{2}$ that is greater than $u$.

Thus,

$$\sup S=\sqrt{2}.$$

The supremum is not in $S$, since the definition requires $x^2<2$, whereas $(\sqrt{2})^2=2$. Hence $S$ has no maximum. Its infimum is $0$, and because $0\in S$, its minimum is $0$.

How to recognize the supremum in practice

To prove that a candidate number $s$ is $\sup S$, use two tests:

1. Prove it is an upper bound

Show that

$$x\leq s \quad\text{for every }x\in S.$$

2. Prove no smaller number works

Show that for every $u<s$, there is some $x\in S$ with

$$u<x.$$

This says that every number below $s$ fails to be an upper bound. An equivalent formulation uses the approximation property: for every $\varepsilon>0$, there is an $x\in S$ such that

$$s-\varepsilon<x\leq s.$$

For an infimum, reverse the inequalities. Prove that $i\leq x$ for every $x\in S$, then show that every number greater than $i$ fails to be a lower bound. Equivalently, for every $\varepsilon>0$, some $x\in S$ satisfies

$$i\leq x<i+\varepsilon.$$

Common mistakes

Calling every upper bound the supremum

If $S=(0,1)$, then $2$ is an upper bound but not the supremum. The supremum is $1$, because $1<2$ and $1$ is still an upper bound.

Assuming the supremum belongs to the set

Open intervals provide the standard counterexample. The supremum of $(0,1)$ is $1$, even though $1$ is excluded.

Confusing maximum with supremum

A maximum must be an element of the set and must be at least as large as every other element. A supremum only needs to be the least upper bound; it may be outside the set.

Reversing “least” and “greatest”

The supremum is the least upper bound. The infimum is the greatest lower bound. For upper bounds, move downward until no further move is possible; for lower bounds, move upward.

Ignoring boundedness

In the real numbers, a nonempty set must be bounded above for its supremum to be a real number, and bounded below for its infimum to be a real number. For example, $\mathbb{R}$ has no real supremum or infimum. Some texts use the extended real numbers and write $\sup\mathbb{R}=+\infty$ and $\inf\mathbb{R}=-\infty$, but these are not real numbers.