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

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

\([a_n] + [b_n] = [a_n+b_n]\)

\([a_n] \times [b_n] = [a_n \times b_n]\)

\([a_n] \leq [b_n]\) 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.

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 \(\epsilon > 0\), we can find \(N_1\) such that for all \(n > N_1\), we have \(|x_n - y_n| < \frac{\epsilon}{2}\); and we can find \(N_2\) such that for all \(n > N_2\), we have \(|a_n - b_n| < \frac{\epsilon}{2}\). 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 \(\leq \epsilon\). <div><div>

## 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.

Let \(\epsilon > 0\) be rational. Then

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 \(|a_n - b_n| < \frac{\epsilon}{2 B}\), and similarly for \(x_n\) and \(y_n\), so the result follows that \(|x_n a_n - y_n b_n| < \epsilon\). <div><div>

## 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, \([c_n] \not = [d_n]\) 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. noteWe didn’t need the extra assumption that \([c_n] \not \leq [d_n]\) here.

If for all sufficiently large \(n\) we have \(a_n \leq b_n\), then this part

## Additive commutative group structure on \(\mathbb{R}\)

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).

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 \(|a_n - a_m| < \frac{\epsilon}{2}\) and \(|b_n - b_m| < \frac{\epsilon}{2}\) for all \(n, m > N\), the result follows. <div><div>

The operation is associative:

## 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).

But

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 \(|a_n - a_m| < \frac{\epsilon}{2B}\) and \(|b_n - b_m| < \frac{\epsilon}{2B}\) for all \(n, m > N\), the result follows. <div><div>

\(\times\) 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:

\(\times\) 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:

## 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 \([a_n] \not = 0\), there is some \(N\) such that for all \(n > N\), \(a_n \not = 0\). 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 \(\mathbb{R}\); and it is easy to check that \([a_n] [b_n] = [1]\).

## 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.

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 \([0] \leq [a_n] \times [b_n]\). <div><div>

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.

Parents:

- Real number (as Cauchy sequence)
There are several ways to construct real numbers; this is the most natural way to use them in computations.