# infimum and supremum for real numbers

Suppose $A$ is a non-empty subset of $\mathbbmss{R}$.  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 $m\in\mathbbmss{R}$ such that

1. 1.

$m$ is an upper bound for $A$, that is, $a\leq m$ for all $a\in A$,

2. 2.

if $M$ is another upper bound for $A$, then $m\leq M$.

Such a number $m$ is called the supremum of $A$, and it is denoted by $\sup A$. It is easy to see that there can be only one least upper bound. If $m_{1}$ and $m_{2}$ are two least upper bounds for $A$. Then $m_{1}\leq m_{2}$ and $m_{2}\leq m_{1}$, and $m_{1}=m_{2}$.

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

1. 1.

$M$ is an lower bound for $A$, that is, $M\leq a$ for all $a\in A$,

2. 2.

if $m$ is another lower bound for $A$, then $m\leq M$.

Such a number $M$ is called the infimum of $A$, and it is denoted by $\inf A$. Just as we proved that the supremum is unique, one can also show that the infimum is unique. The next lemma shows that the infimum exists.

###### Lemma 1.

Every non-empty set bounded from below has a greatest lower bound.

###### Proof.

Let $m\in\mathbbmss{R}$ be a lower bound for non-empty set $A$. In other words, $m\leq a$ for all $a\in A$. Let

 $-A=\{-a\in\mathbbmss{R}:a\in A\}.$

Let us recall the following result from this page (http://planetmath.org/InequalityForRealNumbers); 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 $\sup(-A)\leq-m$, or $m\geq-\sup(-A)$. ∎

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. 1.

For example, consider the set of negative real numbers

 $A=\{x\in\mathbbmss{R}:\,\,x<0\}.$

Then  $\sup A=0$. Indeed. First, $a<0$ for all $a\in A$, and if $a for all $a\in A$, then $0\leq b$.

2. 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.

Title infimum and supremum for real numbers InfimumAndSupremumForRealNumbers 2013-03-22 15:41:42 2013-03-22 15:41:42 matte (1858) matte (1858) 6 matte (1858) Topic msc 54C30 msc 26-00 msc 12D99 SetsThatDoNotHaveAnInfimum Infimum Supremum