What Are the Minimum, Maximum, Infimum, and Supremum?

Minimum and maximum describe elements that a set actually contains. Infimum and supremum describe the sharpest lower and upper bounds, which may or may not be in the set. This membership distinction is the key to using the four terms correctly.

Definitions

Let $A$ be a nonempty subset of the real numbers.

• A minimum of $A$ is an element $m\in A$ such that $m\leq a$ for every $a\in A$. It is also written $\min A$.
• A maximum of $A$ is an element $M\in A$ such that $a\leq M$ for every $a\in A$. It is also written $\max A$.
• A lower bound of $A$ is a number $l$ such that $l\leq a$ for every $a\in A$. The infimum $\inf A$ is the greatest lower bound: it is a lower bound, and every other lower bound is at most it.
• An upper bound of $A$ is a number $u$ such that $a\leq u$ for every $a\in A$. The supremum $\sup A$ is the least upper bound: it is an upper bound, and it is at most every other upper bound.

The infimum and supremum are unique when they exist. Unlike a minimum or maximum, they do not have to be members of $A$.

How the four concepts relate

If $A$ has a minimum, then that minimum is also its infimum. If $A$ has a maximum, then that maximum is also its supremum. The reverse implications need not hold: a set can have an infimum or supremum that it never reaches.

For example, consider the open interval $A=(0,1)$. Every element of $A$ is greater than $0$ and less than $1$, so $0$ is its greatest lower bound and $1$ is its least upper bound. But neither endpoint belongs to the interval. Therefore,

$$\inf A=0,\qquad \sup A=1,$$

while $A$ has no minimum and no maximum.

By contrast, for $B=\{-2,1,4\}$, the endpoints are included:

$$\min B=\inf B=-2,\qquad \max B=\sup B=4.$$

A practical test for infimum and supremum

For a nonempty set $A\subseteq\mathbb{R}$ that is bounded above, a number $s$ is $\sup A$ exactly when both conditions hold:

1. $s$ is an upper bound: $a\leq s$ for every $a\in A$.
2. No smaller number is an upper bound. Equivalently, for every $\varepsilon>0$, some $a\in A$ satisfies $s-\varepsilon<a$.

The second condition says that members of $A$ can be found arbitrarily close to $s$, on or below it. It does not say that any member equals $s$. There is a corresponding test for an infimum: $i=\inf A$ exactly when $i$ is a lower bound and, for every $\varepsilon>0$, some $a\in A$ satisfies $a<i+\varepsilon$.

For a quick application, let $C=\{x\in\mathbb{R}:x^2<4\}=(-2,2)$. The number $2$ is an upper bound. Given any $\varepsilon>0$, choose a positive number $\delta<\min(\varepsilon,2)$; then $2-\delta\in C$ and $2-\varepsilon<2-\delta$. Thus elements of $C$ occur arbitrarily close to $2$, so $\sup C=2$. Since $2\notin C$, there is no maximum. The same reasoning at the other endpoint gives $\inf C=-2$, with no minimum.

Common mistakes

• Assuming a bound must be included. An upper bound can lie outside the set; in fact, most upper bounds need not be the supremum. The supremum is the least one.
• Calling an infimum a minimum without checking membership. First find the greatest lower bound, then check whether it belongs to the set.
• Confusing an infimum with any lower bound. For $(0,1)$, every number at most $0$ is a lower bound, but only $0$ is the greatest one.
• Assuming every set has all four values. A nonempty set can be unbounded above or below, in which case it has no real supremum or infimum in that direction. The real-number completeness theorem guarantees a real supremum for every nonempty set that is bounded above, and a real infimum for every nonempty set that is bounded below.

For the empty set, there is no minimum or maximum. In the real numbers, its infimum and supremum are usually left undefined; in the extended real numbers, a common convention is $\inf\varnothing=+\infty$ and $\sup\varnothing=-\infty$. Check the convention being used before applying these formulas.