Suppose $A$ is a non-empty subset of . If $A$ is bounded from above, then the axioms of the real numbers imply that there exists a least upper bound for $A$ . That is, there exists an such that

1. $m$ is an upper bound for $A$ , that is, $a\le m$ for all $a\in A$ ,
2. if $M$ is another upper bound for $A$ , then $m\le M$ .

Next, let us consider a set $A$ that is bounded from below. That is, for some we have $m\le a$ for all $a\in A$ . Then we say that is a a greatest lower bound for $A$ if

1. $M$ is an lower bound for $A$ , that is, $M \le a$ for all $a\in A$ ,
2. if $m$ is another lower bound for $A$ , then $m\le M$ .

Proof. Let be a lower bound for non-empty set $A$ . In other words, $m\le a$ for all $a\in A$ . Let

Let us recall the following result from this page; if $m$ is an upper(lower) bound for $A$ , then $-m$ is a lower(upper) bound for $-A$ .

Thus $-A$ is bounded from above by $-m$ .

It follows that $-A$ has a least upper bound $\sup (-A)$ . Now $-\sup (-A)$ is a greatest lower bound for $A$ . First, by the result, it is a lower bound for $A$ . Second, if $m$ is a lower bound for $A$ , then $-m$ is a upper bound for $-A$ , and $\sup (-A)\le -m$ , or $m\ge -\sup (-A)$ .

The proof shows that if $A$ is non-empty and bounded from below, then $$ \inf A = -\sup (-A). $$ In consequence, if $A$ is bounded from above, then $$ \sup A = -\inf (-A). $$

In many respects, the supremum and infimum are similar to the maximum and minimum, or the largest and smallest element in a set. However, it is important to notice that the $\inf A$ and $\sup A$ do not need to belong to $A$ . (See examples below.)

Examples

1. For example, consider the set of negative real numbers
   
   Then $\sup A = 0$ . Indeed. First, $a < 0$ for all $a \in A$ , and if $a < b$ for all $a \in A$ , then $0 \le b$ .
2. The sequence $-(1\!-\!\frac{1}{1}),\,1\!-\!\frac{1}{2},\,-(1\!-\!\frac{1}{3}),\, 1\!-\!\frac{1}{4},\,-(1\!-\!\frac{1}{5}),\,...$ is not convergent. The set $A = \{(-1)^n(1-\frac{1}{n}):\,\, n\in\mathbb{Z}_+\}$ formed by its members has the infimum $-1$ and the supremum 1.