The real numbers, when constructed as equivalence classes of Cauchy sequences of rationals, form a totally ordered field, with the inherited field structure given by

$[a_{n}] \times [b_{n}] = [a_{n} \times b_{n}]$
$[a_{n}] \leq [b_{n}]$ if and only if either $[a_{n}] = [b_{n}]$ or for sufficiently large $n$ , $a_{n} \leq b_{n}$ .

Proof

Firstly, we need to show that those operations are even well-defined: that is, if we pick two different representatives $(x_{n})_{n = 1}^{\infty}$ and $(y_{n})_{n = 1}^{\infty}$ of the same equivalence class $[x_{n}] = [y_{n}]$ , we don't somehow get different answers.

Well-definedness of $+$

We wish to show that $[x_{n}] + [a_{n}] = [y_{n}] + [b_{n}]$ whenever $[x_{n}] = [y_{n}]$ and $[a_{n}] = [b_{n}]$ ; this is an exercise.

Show solution

Since $[x_{n}] = [y_{n}]$ , it must be the case that both $(x_{n})$ and $(y_{n})$ are Cauchy sequences such that $x_{n} - y_{n} \to 0$ as $n \to \infty$ . Similarly, $a_{n} - b_{n} \to 0$ as $n \to \infty$ .

We require $[x_{n} + a_{n}] = [y_{n} + b_{n}]$ ; that is, we require $x_{n} + a_{n} - y_{n} - b_{n} \to 0$ as $n \to \infty$ .

But this is true: if we fix rational $ϵ > 0$ , we can find $N_{1}$ such that for all $n > N_{1}$ , we have $| x_{n} - y_{n} | < \frac{ϵ}{2}$ ; and we can find $N_{2}$ such that for all $n > N_{2}$ , we have $| a_{n} - b_{n} | < \frac{ϵ}{2}$ . Letting $N$ be the maximum of the two $N_{1}, N_{2}$ , we have that for all $n > N$ , $| x_{n} + a_{n} - y_{n} - b_{n} | \leq | x_{n} - y_{n} | + | a_{n} - b_{n} |$ by the triangle_inequality, and hence $\leq ϵ$ .

Well-definedness of $\times$

We wish to show that $[x_{n}] \times [a_{n}] = [y_{n}] \times [b_{n}]$ whenever $[x_{n}] = [y_{n}]$ and $[a_{n}] = [b_{n}]$ ; this is also an exercise.

Show solution

We require $[x_{n} a_{n}] = [y_{n} b_{n}]$ ; that is, $x_{n} a_{n} - y_{n} b_{n} \to 0$ as $n \to \infty$ .

Let $ϵ > 0$ be rational. Then $| x_{n} a_{n} - y_{n} b_{n} | = | x_{n} (a_{n} - b_{n}) + b_{n} (x_{n} - y_{n}) |$ using the very handy trick of adding the expression $x_{n} b_{n} - x_{n} b_{n}$ .

By the triangle inequality, this is $\leq | x_{n} | | a_{n} - b_{n} | + | b_{n} | | x_{n} - y_{n} |$ .

We now use the fact that Cauchy_sequences_are_bounded, to extract some $B$ such that $| x_{n} | < B$ and $| b_{n} | < B$ for all $n$ ; then our expression is less than $B (| a_{n} - b_{n} | + | x_{n} - y_{n} |)$ .

Finally, for $n$ sufficiently large we have $| a_{n} - b_{n} | < \frac{ϵ}{2 B}$ , and similarly for $x_{n}$ and $y_{n}$ , so the result follows that $| x_{n} a_{n} - y_{n} b_{n} | < ϵ$ .

Well-definedness of $\leq$

We wish to show that if $[a_{n}] = [c_{n}]$ and $[b_{n}] = [d_{n}]$ , then $[a_{n}] \leq [b_{n}]$ implies $[c_{n}] \leq [d_{n}]$ .

Suppose $[a_{n}] \leq [b_{n}]$ , but suppose for contradiction that $[c_{n}]$ is not $\leq [d_{n}]$ : that is, $[c_{n}] \neq [d_{n}]$ and there are arbitrarily large $n$ such that $c_{n} > d_{n}$ . Then there are two cases.

If $[a_{n}] = [b_{n}]$ then $[d_{n}] = [b_{n}] = [a_{n}] = [c_{n}]$ , so the result follows immediately. ^[1]
If for all sufficiently large $n$ we have $a_{n} \leq b_{n}$ , then
this part

Additive commutative group structure on $R$

The additive identity is $[0]$ (formally, the equivalence class of the sequence $(0, 0, \dots)$ ). Indeed, $[a_{n}] + [0] = [a_{n} + 0] = [a_{n}]$ .

The additive inverse of the element $[a_{n}]$ is $[- a_{n}]$ , because $[a_{n}] + [- a_{n}] = [a_{n} - a_{n}] = [0]$ .

The operation is commutative: $[a_{n}] + [b_{n}] = [a_{n} + b_{n}] = [b_{n} + a_{n}] = [b_{n}] + [a_{n}]$ .

The operation is closed, because the sum of two Cauchy sequences is a Cauchy sequence (exercise).

Show solution

If $(a_{n})$ and $(b_{n})$ are Cauchy sequences, then let $ϵ > 0$ . We wish to show that there is $N$ such that for all $n, m > N$ , we have $| a_{n} + b_{n} - a_{m} - b_{m} | < ϵ$ .

But $| a_{n} + b_{n} - a_{m} - b_{m} | \leq | a_{n} - a_{m} | + | b_{n} - b_{m} |$ by the triangle inequality; so picking $N$ so that $| a_{n} - a_{m} | < \frac{ϵ}{2}$ and $| b_{n} - b_{m} | < \frac{ϵ}{2}$ for all $n, m > N$ , the result follows.

The operation is associative: $[a_{n}] + ([b_{n}] + [c_{n}]) = [a_{n}] + [b_{n} + c_{n}] = [a_{n} + b_{n} + c_{n}] = [a_{n} + b_{n}] + [c_{n}] = ([a_{n}] + [b_{n}]) + [c_{n}]$

Ring structure

The multiplicative identity is $[1]$ (formally, the equivalence class of the sequence $(1, 1, \dots)$ ). Indeed, $[a_{n}] \times [1] = [a_{n} \times 1] = [a_{n}]$ .

$\times$ is closed, because the product of two Cauchy sequences is a Cauchy sequence (exercise).

Show solution

If $(a_{n})$ and $(b_{n})$ are Cauchy sequences, then let $ϵ > 0$ . We wish to show that there is $N$ such that for all $n, m > N$ , we have $| a_{n} b_{n} - a_{m} b_{m} | < ϵ$ .

But $| a_{n} b_{n} - a_{m} b_{m} | = | a_{n} (b_{n} - b_{m}) + b_{m} (a_{n} - a_{m}) | \leq | b_{m} | | a_{n} - a_{m} | + | a_{n} | | b_{n} - b_{m} |$ by the triangle inequality.

Cauchy sequences are bounded, so there is $B$ such that $| a_{n} |$ and $| b_{m} |$ are both less than $B$ for all $n$ and $m$ .

So picking $N$ so that $| a_{n} - a_{m} | < \frac{ϵ}{2 B}$ and $| b_{n} - b_{m} | < \frac{ϵ}{2 B}$ for all $n, m > N$ , the result follows.

$\times$ is clearly commutative: $[a_{n}] \times [b_{n}] = [a_{n} \times b_{n}] = [b_{n} \times a_{n}] = [b_{n}] \times [a_{n}]$ .

$\times$ is associative: $[a_{n}] \times ([b_{n}] \times [c_{n}]) = [a_{n}] \times [b_{n} \times c_{n}] = [a_{n} \times b_{n} \times c_{n}] = [a_{n} \times b_{n}] \times [c_{n}] = ([a_{n}] \times [b_{n}]) \times [c_{n}]$

$\times$ distributes over $+$ : we need to show that $[x_{n}] \times ([a_{n}] + [b_{n}]) = [x_{n}] \times [a_{n}] + [x_{n}] \times [b_{n}]$ . But this is true: $[x_{n}] \times ([a_{n}] + [b_{n}]) = [x_{n}] \times [a_{n} + b_{n}] = [x_{n} \times (a_{n} + b_{n})] = [x_{n} \times a_{n} + x_{n} \times b_{n}] = [x_{n} \times a_{n}] + [x_{n} \times b_{n}] = [x_{n}] \times [a_{n}] + [x_{n}] \times [b_{n}]$

Field structure

To get from a ring to a field, it is necessary and sufficient to find a multiplicative inverse for any $[a_{n}]$ not equal to $[0]$ .

Since $[a_{n}] \neq 0$ , there is some $N$ such that for all $n > N$ , $a_{n} \neq 0$ . Then defining the sequence $b_{i} = 1$ for $i \leq N$ , and $b_{i} = \frac{1}{a_{i}}$ for $i > N$ , we obtain a sequence which induces an element $[b_{n}]$ of $R$ ; and it is easy to check that $[a_{n}] [b_{n}] = [1]$ .

Show solution

$[a_{n}] [b_{n}] = [a_{n} b_{n}]$ ; but the sequence $(a_{n} b_{n})$ is $1$ for all $n > N$ , and so it lies in the same equivalence class as the sequence $(1, 1, \dots)$ .

Ordering on the field

We need to show that:

if $[a_{n}] \leq [b_{n}]$ , then for every $[c_{n}]$ we have $[a_{n}] + [c_{n}] \leq [b_{n}] + [c_{n}]$ ;
if $[0] \leq [a_{n}]$ and $[0] \leq [b_{n}]$ , then $[0] \leq [a_{n}] \times [b_{n}]$ .

We may assume that the inequalities are strict, because if equality holds in the assumption then everything is obvious.

Show obvious bits

If $[a_{n}] = [b_{n}]$ , then for every $[c_{n}]$ we have $[a_{n}] + [c_{n}] = [b_{n}] + [c_{n}]$ by well-definedness of addition. Therefore $[a_{n}] + [c_{n}] \leq [b_{n}] + [c_{n}]$ .

If $[0] = [a_{n}]$ and $[0] \leq [b_{n}]$ , then $[0] = [0] \times [b_{n}] = [a_{n}] \times [b_{n}]$ , so it is certainly true that $[0] \leq [a_{n}] \times [b_{n}]$ .

For the former: suppose $[a_{n}] < [b_{n}]$ , and let $[c_{n}]$ be an arbitrary equivalence class. Then $[a_{n}] + [c_{n}] = [a_{n} + c_{n}]$ ; $[b_{n}] + [c_{n}] = [b_{n} + c_{n}]$ ; but we have $a_{n} + c_{n} \leq b_{n} + c_{n}$ for all sufficiently large $n$ , because $a_{n} \leq b_{n}$ for sufficiently large $n$ . Therefore $[a_{n}] + [c_{n}] \leq [b_{n}] + [c_{n}]$ , as required.

For the latter: suppose $[0] < [a_{n}]$ and $[0] < [b_{n}]$ . Then for sufficiently large $n$ , we have both $a_{n}$ and $b_{n}$ are positive; so for sufficiently large $n$ , we have $a_{n} b_{n} \geq 0$ . But that is just saying that $[0] \leq [a_{n}] \times [b_{n}]$ , as required.

^{^︎}
We didn't need the extra assumption that \(C_n \not \leq D_n\) here.

The reals (constructed as classes of Cauchy sequences of rationals) form a field