The Axiom of Completeness
The Axiom of Completeness
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.
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 \(\sup (c+A)=c+\sup A\).
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 \(s \le b\).