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

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

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; 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. 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 $a<b$ for all $a\in A$, then $0\leq b$.

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.

## Mathematics Subject Classification

54C30*no label found*26-00

*no label found*12D99

*no label found*

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