infimum and supremum for real numbers

Suppose $A$ is a non-empty subset of $ℝ$.  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 $m\in ℝ$ 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 $m\in ℝ$ we have $m\le a$ for all $a\in A$.  Then we say that $M\in ℝ$ 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 non-empty set bounded from below has a greatest lower bound.

Proof.

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

$$-A=\{-a\in ℝ:a\in A\}.$$

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 non-empty 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	2013-03-22 15:41:42
Last modified on	2013-03-22 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 26-00
Classification	msc 12D99
Related topic	SetsThatDoNotHaveAnInfimum
Related topic	Infimum
Related topic	Supremum