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

Written by Patrick Stevens, et al. last updated

Summaries

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

  • if and only if either or for sufficiently large , .

Proof

Firstly, we need to show that those operations are even well-defined: that is, if we pick two different representatives and of the same equivalence class , we don't somehow get different answers.

Well-definedness of

We wish to show that whenever and ; this is an exercise.

Show solution

Since , it must be the case that both and are Cauchy sequences such that as . Similarly, as .

We require ; that is, we require as .

But this is true: if we fix rational , we can find such that for all , we have ; and we can find such that for all , we have . Letting be the maximum of the two , we have that for all , by the triangle_inequality, and hence .

Well-definedness of

We wish to show that whenever and ; this is also an exercise.

Show solution

We require ; that is, as .

Let be rational. Then using the very handy trick of adding the expression .

By the triangle inequality, this is .

We now use the fact that Cauchy_sequences_are_bounded, to extract some such that and for all ; then our expression is less than .

Finally, for sufficiently large we have , and similarly for and , so the result follows that .

Well-definedness of

We wish to show that if and , then implies .

Suppose , but suppose for contradiction that is not : that is, and there are arbitrarily large such that . Then there are two cases.

  • If then , so the result follows immediately. [1]
  • If for all sufficiently large we have , then

Additive commutative group structure on

The additive identity is (formally, the equivalence class of the sequence ). Indeed, .

The additive inverse of the element is , because .

The operation is commutative: .

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

Show solution

If and are Cauchy sequences, then let . We wish to show that there is such that for all , we have .

But by the triangle inequality; so picking so that and for all , the result follows.

The operation is associative:

Ring structure

The multiplicative identity is (formally, the equivalence class of the sequence ). Indeed, .

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

Show solution

If and are Cauchy sequences, then let . We wish to show that there is such that for all , we have .

But by the triangle inequality.

Cauchy sequences are bounded, so there is such that and are both less than for all and .

So picking so that and for all , the result follows.

is clearly commutative: .

is associative:

distributes over : we need to show that . 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 not equal to .

Since , there is some such that for all , . Then defining the sequence for , and for , we obtain a sequence which induces an element of ; and it is easy to check that .

Show solution

; but the sequence is for all , and so it lies in the same equivalence class as the sequence .

Ordering on the field

We need to show that:

  • if , then for every we have ;
  • if and , then .

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

Show obvious bits

If , then for every we have by well-definedness of addition. Therefore .

If and , then , so it is certainly true that .

For the former: suppose , and let be an arbitrary equivalence class. Then ; ; but we have for all sufficiently large , because for sufficiently large . Therefore , as required.

For the latter: suppose and . Then for sufficiently large , we have both and are positive; so for sufficiently large , we have . But that is just saying that , as required.

  1. ^︎

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

1.
^︎

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