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

https://arbital.com/p/reals_as_classes_of_cauchy_sequences_form_a_field

by Patrick Stevens Jul 4 2016 updated Jul 5 2016

The reals are an archetypal example of a field, but if we are to construct them from simpler objects, we need to show that our construction does indeed have the right properties.

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. %%hidden(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 $\epsilon > 0$, 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 $\leq \epsilon$. %%

## 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. %%hidden(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 $\epsilon > 0$ 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 $|a_n - b_n| < \frac{\epsilon}{2 B}$, and similarly for $x_n$ and $y_n$, so the result follows that $|x_n a_n - y_n b_n| < \epsilon$. %%

## 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. %%note:We 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 [todo: this part]

## Additive commutativegroup 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). %%hidden(Show solution): If $(a_n)$ and $(b_n)$ are Cauchy sequences, then let $\epsilon > 0$. We wish to show that there is $N$ such that for all $n, m > N$, we have $|a_n+b_n - a_m - b_m| < \epsilon$.

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. %%

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). %%hidden(Show solution): If $(a_n)$ and $(b_n)$ are Cauchy sequences, then let $\epsilon > 0$. We wish to show that there is $N$ such that for all $n, m > N$, we have $|a_n b_n - a_m b_m| < \epsilon$.

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 $|a_n - a_m| < \frac{\epsilon}{2B}$ and $|b_n - b_m| < \frac{\epsilon}{2B}$ 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 $[a_n] \not = 0$, 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 $[a_n] [b_n] = [1]$. %%hidden(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. %%hidden(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.