The Axiom of Completeness

 

The Axiom of Completeness

Definition: Upper bound

A real number $\alpha$ is called an upper bound for $S$ if $x\le \alpha$ for all $x \in S$. The set $S$ is said to be bounded above if it has an upper bound.

 

Definition: Lower bound

A real number $\alpha$ is called an lower bound for $S$ if $\alpha \le x$ for all $x \in S$. The set $S$ is said to be bounded below if it has an lower bound.

 

Definition: Least upper bound

A real number $\alpha$ is called the least upper bound (supremum of sup) of $S$, if (i) $\alpha$ is an upper bound for $S$; and (ii) if $\beta$ is any upper bound for $S$, then $\alpha \le \beta$. The supremum, if it exists, is unique, and is denoted by $\sup S$.

 

Definition: Greatest lower bound

A real number $\alpha$ is called the greatest lower bound (infimum of inf) of $S$, if (i) $\alpha$ is an lower bound for $S$; and (ii) if $\beta$ is any lower bound for $S$, then $\beta \le \alpha$. The infimum, if it exists, is unique, and is denoted by $\inf S$.

 

Axiom: Axiom of Completeness

Every nonempty set of real number that is bounded above has a least upper bound.

 

Example: Let $$A = \left\{ {\frac{1}{n}|n \in \mathbb{N}} \right\} = \left\{1, \frac{1}{2},\frac{1}{3},\dots \right\}$$ The set $A$ is bounded above and below. For the least upper bound, we claim $\sup A = 1$. We observe that $1\ge \frac{1}{n}$ for all choice of $n\in \mathbb{N}$. We begin by assuming we are in possession of some other upper bound $b$. Because $1\in A$ and $b$ is an upper bound for A, we must have $1\le b$. For the greatest lower bound, we claim $\inf A = 0$. We observe that $0\le \frac{1}{n}$ for all choice of $n \in \mathbb{N}$. We begin by assuming we are in possession of some other lower bound $c$.

 

An important lesson to take from Example is that $\sup A$ and $\inf A$ may or may not be elements of the set $A$. This issue is tied to understanding the crucial difference between the maximum and the supremum (or minimum and the infimum) of a given set.

Definition: Maximum

A real number $a_0$ is a maximum of the set $A$ if $a_0$ is an element of $A$ and $a_0 \ge a$ for all $a\in A$. Similarly, a number $a_1$ is a minimum of $A$ if $a_1 \in A$ and $a_1 \le a$ for every $a \in A$.

 

Example: To belabor the point, consider the open interval $$(0,2) = \left\{x\in \mathbb{R}| 0<x<2\right\},$$ And the closed interval $$[0,2] = \left\{x\in \mathbb{R}| 0\le x\le 2\right\}.$$ Both sets are bounded above (and below), and both have the same least upper bound, namely $2$. It is not the case, however, that both sets hae a maximum. A maximum is a specific type of upper bound that is required to be an element of the set in question, and the open interval $(0,2)$ does not possess such an element. Thus, the supremum can exist and not be a maximum, but when a maximum exists, then it is also the supremum.

 

Although we can see now that not every nonempty bounded set contains a maximum, the Axiom of Completeness asserts that every such set does have a least upper bound.

 

Example: Consider again the set $$S = \left\{r\in \mathbb{Q} | r^2 < 2\right\},$$ and pretend for the moment that our world consists only of rational numbers. The set $S$ is certainly bounded above. Taking $b=2$ works, as does $b=\frac{3}{2}$. But notice what happens as we go in search of the least upper bound. We might try $b=\frac{142}{100}$, which is indeed an upper bound, but then we discover that $b=\frac{1415}{1000}$ is an upper bound that is smaller still. Is there a smallest one? In the rational numbers, there is not. In the real numbers, there is. Back in $\mathbb{R}$, the Axiom of Completeness states that we may set $\alpha = \sup S$ and be confident that such a number exists. If we are restricting our attention to only rational numbers, then $\alpha$ is not an allowable option for $\sup S$, and the search for a least upper bound goes on indefinitely. Whatever rational upper bound is discovered, it is always possible to find one smaller.

 

Example: Let $A\subseteq \mathbb{R}$ be nonempty and bounded above, and let $c\in \mathbb{R}$. Define the set $c+A$ by $$c+A = \left\{c+a|a\in A\right\}.$$ Then $\sup (c+A) = c + \sup A$. Setting $s= \sup A$, we see that $a \le s$ for all $a \in A$, which implies $c+a \le c+s$ for all $a\in A$. Thus $c+s$ is an upper bound for $c+A$. Let $b$ be an arbitrary upper bound for $c+A$; i.e., $c+a\le b$ for all $a \in A$. This is equivalent to $a\le b -c$ for all $a \in A$, from which we conclude that $b-c$ is an upper bound for $A$. Because $s$ is the least upper bound of $A$, $s \le b-c$, which can be rewritten as $c+s \le b$. We conclude $\sup (c+A)=c+\sup A$.

 

Lemma:

Assume $s \in \mathbb{R}$ is an upper bound for a set $A \subseteq \mathbb{R}$. Then $s = \sup A$ if and only if, for every choice of $\epsilon > 0$ there exists an element $a \in A$ satisfying $s - \epsilon<a$.

 

Proof. ($\Rightarrow$) For the forward direction, we assume $s = \sup A$ and consider $s-\epsilon$ where $\epsilon>0$ has been arbitrarily chosen. Because $s - \epsilon<s$, $s - \epsilon$ is not an upper bound for $A$. If this is the case, then there must be some element $a \in A$ for which $s - \epsilon < a$. This proves the lemma in one direction.

($\Leftarrow$) Conversely, assume $s$ is an upper bound with the property that no matter how $\epsilon$ is chosen, $s-\epsilon$ is no longer an upper bound for $A$. Notice that what this implies is that if $b$ is any number less than $s$, then $b$ is not an upper bound. To prove that $s = \sup A$, we have just argued that any number smaller than $s$ cannot be an upper bound, it follows that if b is some other upper bound for $A$, than $s \le b$.

 

Lemma:

Assume $l \in \mathbb{R}$ is an lower bound for a set $A \subseteq \mathbb{R}$. Then $l = \inf A$ if and only if, for every choice of $\epsilon > 0$ there exists an element $a \in A$ satisfying $l + \epsilon>a$.