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Section 3.4 Completeness and the real numbers

Definition 3.4.1. Completeness.

A totally ordered set \((S,\lt)\) is complete if and only if, for every subset \(T \subseteq S\) with an upper bound in \(S\text{,}\) there is a supremum of \(T\) in \(S\text{.}\)

Equivalently, \(S\) is complete if and only if every subset of \(S\) that has a lower bound has an infimum. We have seen above that \(\QQ\) is not complete. Nevertheless, we can complete \(\QQ\text{.}\)

We won't prove this theorem, as the proof is quite long and tedious, and there's not much to be learned from it. So there is a complete ordered field, which, up to a simple renaming of the elements, is unique. We call it \(\RR\) or the real line, and we call its elements real numbers.

This \(\RR\) contains a copy of the natural numbers: \(0\) and \(1\) are in there because of the field axioms. Note that \(0 \lt 1\text{.}\) (Why?) for every natural number \(n\text{,}\) if we can regard \(n\) as a real number, then we can regard the successor of \(n\) as the real number \(n+1\text{.}\)

Since \(\RR\) is a field, we can take the additive inverses of the natural numbers to get a copy of \(\ZZ\) lying in \(\RR\text{.}\) We can also take products of integers and multiplicative inverses of integers to get a copy of \(\QQ\) lying in \(\RR\text{.}\)

We are happy to pretend that we have literal inclusions

\begin{equation*} \NN \subset \NN_0 \subset \ZZ \subset \QQ \subset \RR\text{.} \end{equation*}

The following theorem implies that there are enough natural numbers. This will be used to show that in fact every real number can be arbitrarily well approximated by rational numbers.

Suppose not. Then \(x\) is an upper bound for \(\NN_0 \subseteq\RR\text{.}\) By completeness, it follows that there is a real number \(N = \sup\NN_0\text{.}\) Now consider \(N-1\text{;}\) for any natural number \(n\text{,}\) the real number \(n+1\) is also a natural number, and therefore \(n+1\leq N\text{,}\) so \(n\leq N-1\text{.}\) Thus \(N-1\) is also an upper bound for \(\NN_0\text{.}\) But this contradicts the fact that \(N\) is a least upper bound.

By the previous theorem, choose a natural number \(n\gt x/y\text{.}\)

Let's prove the existence first. Consider the set \(S = \{z\in \RR : (z \geq 0)\wedge (z^2\geq x)\}\text{.}\) Observe that \(S\) has a lower bound, namely \(0\text{.}\) Now by completeness, we may define \(y = \inf S\text{.}\) We claim that \(y^2=x\text{.}\) Suppose not; then either \(y^2 \gt x\) or \(y^2 \lt x\text{.}\)

If \(y^2\gt x\text{,}\) then use the Archimedean property to find a natural number \(n\) such that

\begin{equation*} n(y^2-x) \gt 2y \end{equation*}

Thus \(y^2-(2/n)y\gt x\text{.}\) Now let \(u=y-1/n\text{.}\) We have

\begin{equation*} u^2 = (y-1/n)^2 = y^2-(2/n)y+1/n^2 \gt y^2 - (2/n)y \gt x\text{.} \end{equation*}

Thus \(u\in S\text{,}\) and this contradicts our assumption that \(y\) is a lower bound.

If on the other hand \(y^2\lt x\text{,}\) then use the Archimedean property to find a natural number \(n\) such that

\begin{equation*} n(x-y^2) \gt 2y+1 \end{equation*}

Thus \(y^2+(2/n)y+1/n^2 \lt y^2+(2/n)y+1/n \lt x\text{.}\) Now let \(u=y+1/n\text{.}\) We have

\begin{equation*} u^2 = (y+1/n)^2 = y^2+(2/n)y+1/n^2 \lt x\text{.} \end{equation*}

Now if \(z\in S\text{,}\) then \(z^2>u^2\text{.}\) This implies that \(u\lt z\text{.}\) (Indeed if \(u \geq z\text{,}\) then \(u^2\geq z^2\text{.}\)) Thus \(u \) is a lower bound for \(S\text{,}\) which contradicts our assumption that \(y\) is the greatest lower bound.

This shows that \(y^2=x\text{.}\) Now, to prove uniqueness, suppose \(v\geq0\) is a positive real number; if \(v\lt y\text{,}\) then \(v^2 \lt y^2 = x\text{,}\) and if \(y\lt v\text{,}\) then \(y^2 \lt v^2\text{.}\) Hence \(v^2=y^2\) if and only if \(v=y\text{.}\)