infimum and supremum for real numbers
Suppose $A$ is a nonempty subset of $\mathbb{R}$. 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 $m\in \mathbb{R}$ 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 $m\le M$.
Such a number $m$ is called the supremum of $A$, and it is denoted by $supA$. 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 ${m}_{1}\le {m}_{2}$ and ${m}_{2}\le {m}_{1}$, and ${m}_{1}={m}_{2}$.
Next, let us consider a set $A$ that is bounded from below. That is, for some $m\in \mathbb{R}$ we have $m\le a$ for all $a\in A$. Then we say that $M\in \mathbb{R}$ 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 $m\le M$.
Such a number $M$ is called the infimum^{} of $A$, and it is denoted by $infA$. 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 nonempty set bounded from below has a greatest lower bound.
Proof.
Let $m\in \mathbb{R}$ be a lower bound for nonempty set $A$. In other words, $m\le a$ for all $a\in A$. Let
$$A=\{a\in \mathbb{R}:a\in A\}.$$ 
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 $sup(A)\le m$, or $m\ge sup(A)$. ∎
The proof shows that if $A$ is nonempty and bounded from below, then
$$infA=sup(A).$$ 
In consequence, if $A$ is bounded from above, then
$$supA=inf(A).$$ 
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 $infA$ and $supA$ do not need to belong to $A$. (See examples below.)
Examples

1.
For example, consider the set of negative real numbers
$$ Then $supA=0$. Indeed. First, $$ for all $a\in A$, and if $$ for all $a\in A$, then $0\le b$.

2.
The sequence^{} $(1\frac{1}{1}),\mathrm{\hspace{0.17em}1}\frac{1}{2},(1\frac{1}{3}),\mathrm{\hspace{0.17em}1}\frac{1}{4},(1\frac{1}{5}),\mathrm{\dots}$ 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 

Canonical name  InfimumAndSupremumForRealNumbers 
Date of creation  20130322 15:41:42 
Last modified on  20130322 15:41:42 
Owner  matte (1858) 
Last modified by  matte (1858) 
Numerical id  6 
Author  matte (1858) 
Entry type  Topic 
Classification  msc 54C30 
Classification  msc 2600 
Classification  msc 12D99 
Related topic  SetsThatDoNotHaveAnInfimum 
Related topic  Infimum 
Related topic  Supremum 