# Construction of the reals via Cauchy sequences

This section discusses the construction of the real numbers from the rationals via the idea of a Cauchy sequence.

This approach is perhaps the most natural one from the point of view of Cauchy's version of the completeness axiom, and has the advantage that it generalizes to other settings where there is a distance function $d ( x , y )$ (like $d ( x , y ) = | x - y |$) but no order relation $x < y$ available in the number-system. This includes the case of building the complex numbers from the rationals as limits of Cauchy sequences of numbers of the form $p + i q$ with $p , q ∈ ℚ$. It also provides an introduction to the construction more advanced topic of a completion of metric space.

This web page is still under construction. The final version will contain some (but not all) of the proofs omitted here at present.

## 1. Defining the reals from the rationals

We start by recalling the definition of a Cauchy sequence.

Definition.

A sequence $( a n )$ of rationals is said to be Cauchy (or: has the Cauchy Property) if . Note that I have modified the definition of Cauchy sequence given previously by replacing the quantifier with the equivalent . (This is equivalent because of the Archimedean property, of course.)

We have seen that we expect all Cauchy sequences to converge, and all reals to be the limit of a Cauchy sequence of rationals. The Cauchy property is particularly useful as it doesn't mention the limit $l$, so you don't need to know what the limit is to talk about convergence.

Definition.

We let $R$ be the set of all Cauchy sequences of rationals.

Once again, a real number may have more than one rational Cauchy sequence converging to it, so we must factor out by an equivalence relation to obtain the true version of the reals.

Definition.

For Cauchy sequences of rationals $( a n ) , ( b n ) ∈ R$, we say $( a n ) ∼ ( b m )$ if .

Proposition.

The relation $∼$ is an equivalence relation on $R$.

Proof.

Reflexivity of $∼$, that $( a n ) ∼ ( a n )$ is just the statement that the sequence $( a n )$ is Cauchy. Symmetry of $∼$ is obvious, as $| a n - b m | = | b m - a n |$.

To prove transitivity, we assume and and argue as follows.

Subproof.

Let $k ∈ ℕ$ be arbitrary.

Let $K ∈ ℕ$ and $M ∈ ℕ$ satisfy

and

Subproof.

Let $N = max ( M , K )$.

Subproof.

Let $n , m ⩾ N$ be arbitrary. Then

$| a n - c m | = | a n - b n + b n - c m | ⩽ | a n - b n | + | b n - c m | < 1 2 ( k +1 ) + 1 2 ( k +1 ) = 1 k +1$

by the triangle inequality, as required.

So .

So .

So .

Definition.

The set $ℝ$ is the set of equivalence classes $[ a n ]$ of elements $( a n )$ of $R$.

For a rational, $q ∈ ℚ$, the constant sequence with value $q$ is a Cauchy sequence. You can check that two distinct constant sequences $q 1$ and $q 2$ are inequivalent. Therefore we can identify each $q ∈ ℚ$ with the equivalence class $[ q ]$ of the constant sequence $q$.

We expect the arithmetic operations of $+,·$, etc., to be continuous. This suggests the following definition.

Definition.

For sequences $( a n )$ and $( b n )$ in $R$ and their equvalence classes $[ a n ]$, $[ b n ]$ we define

1. $[ a n ] + [ b n ] = [ a n + b n ]$
2. $[ a n ] · [ b n ] = [ a n b n ]$

This completes the definition part, as we have now defined $ℝ$ with its arithmetic operations, and order. All that remains it to check the axioms.

Theorem.

The set $ℝ$ with $+,· , ⩽$ is an ordered field. It also satisfies the Archimedean property and is a complete ordered field.

Proof.

Another long exercise (sigh).