The Cauchy property

is a useful idea that
describes sequences that seem to converge
without mentioning any limit. It is a modification of the usual
definition of convergence except that we cannot compare the values of
the sequence to $l$; instead we have to compare such values to
each other.

Definition.

We say that a sequence $\left({a}_{n}\right)$ has the Cauchy property if the following statement holds:

Convergent sequences do have the Cauchy property. This is quite easy to prove.

Proposition.

Suppose ${a}_{n}\to l$ as $n\to \infty $. Then $\left({a}_{n}\right)$ has the Cauchy property.

**Proof.**

Let $\epsilon >0$ be arbitrary, and let $N\in \mathbb{N}$ be such that $\forall n\u2a7eN\left|{a}_{n}-l\right|\frac{\epsilon}{2}$, from the fact that ${a}_{n}\to l$ as $n\to \infty $. Then given arbitrary $n,k\u2a7eN$ we have

as required.

We're going to prove the converse to this result now, that every Cauchy sequence (i.e., one with the Cauchy property) converges to some limit. Once again, the completeness of the real numbers is the fact that makes this work. First we need a useful lemma, the proof of which is almost identical to the theorem that says every convergent sequence is bounded.

Lemma.

Suppose that $\left({a}_{n}\right)$ has the Cauchy property. Then the sequence $\left({a}_{n}\right)$ is bounded.

**Proof.**

Let $\epsilon =1$. From the Cauchy property (as written above, in the spacial case when $k=N$) there is $N\in \mathbb{N}$ be such that $\forall n\u2a7eN\left|{a}_{n}-{a}_{N}\right|1$. Then for all $n\u2a7eN$ we have $\left|{a}_{n}\right|<1+\left|{a}_{N}\right|$. Therefore

is a bound on the whole sequence $\left({a}_{n}\right)$.

Theorem.

Suppose that $\left({a}_{n}\right)$ has the Cauchy property. Then there is $l\in \mathbb{R}$ such that ${a}_{n}\to l$ as $n\to \infty $.

**Proof.**

By the lemma, $\left({a}_{n}\right)$ is bounded, so by the Bolzano-Weierstrass theorem (which used the completeness of the reals) there is $l\in \mathbb{R}$ and a subsequence ${b}_{n}={a}_{f\left(n\right)}\to l$ as $n\to \infty $. We shall use the Cauchy property again to prove that the whole sequence $\left({a}_{n}\right)$ converges to $l$.

Let $\epsilon >0$ be arbitrary. From the convergence of the subsequence
there is ${N}_{1}\in \mathbb{N}$ such that
$\forall n\u2a7e{N}_{1}\left|{b}_{n}-l\right|\frac{\epsilon}{2}$
i.e.,
$\forall n\u2a7e{N}_{1}\left|{a}_{f\left(n\right)}-l\right|\frac{\epsilon}{2}$.
From the Cauchy property there is ${N}_{2}\in \mathbb{N}$ such that
$\forall n\u2a7e{N}_{2}\forall k\u2a7e{N}_{2}\left|{a}_{n}-{a}_{k}\right|\frac{\epsilon}{2}$.
Let $N=max({N}_{2},f({N}_{1}\left)\right)$. Now let $n\u2a7eN$ be arbitrary.
Then there is $k\u2a7e{N}_{1}$ such that $f\left(k\right)\u2a7eN$. This is because
$f$ is increasing, i.e., because the subsequence ${b}_{n}$ selects
*infinitely* many terms in ${a}_{n}$ and the function $f$ enumerates them.
Then
$\left|{a}_{f\left(k\right)}-l\right|<\frac{\epsilon}{2}$ as $k\u2a7e{N}_{1}$
and
$\left|{a}_{n}-{a}_{f\left(k\right)}\right|<\frac{\epsilon}{2}$ as $f\left(k\right)\u2a7e{N}_{2}$
so, using the triangle inequality,

as required.

The real importance of this result is that it enables us to state the completeness
axiom for the reals in a way that uses the distance function only, and not
using the order relation on the reals. Indeed many people prefer to take
as their completeness axiom the statement Every Cauchy sequence has a limit.

(Note that the monotone convergence theorem needs the order relation $<$
to define what it means to be a monotonic sequence.)

The main result just presented (that every Cauchy sequence has a limit) is another version of the completeness property for the fields. Because it doesn't require the order relation, $<$, it is a useful axiom to consider for other fields orther than ordered fields: all that is required is the distance function $d(x,y)$ to have meaning in the field. In paticular, a sequence $\left({a}_{n}\right)$ has the Cauchy property in such a field if: $\forall \epsilon >0\exists N\in \mathbb{N}\forall n\u2a7eN\forall k\u2a7eNd({a}_{n},{a}_{k})\epsilon $. It is equally easy to formulate the notion of convergence of a sequence $\left({a}_{n}\right)$ to a limit $l$ in terms of the distance function: $\forall \epsilon >0\exists N\in \mathbb{N}\forall n\u2a7eNd({a}_{n},l)\epsilon $. This gives the following idea of completeness.

Completeness, Cauchy form.

A field $F$ with a distance function $d(x,y)$ is complete if for every sequence $\left({a}_{n}\right)$ of elements of $F$ with the Cauchy property there is $l\in F$ such that ${a}_{n}\to l$ as $n\to \infty $.

An example of a field with a distance function which isn't ordered is the field of complex numbers, $\u2102$. Here we may take $d(x,y)=\left|x-y\right|$ and it turns out that in this sense $\u2102$ is a complete field.

The two approaches to completeness for Archimedean ordered fields
such as the reals (via monotonic sequences and via Cauchy sequences)
are equivalent: it is possible to show directly from the Archimedean
Property of the reals that every bounded monotonic sequence is Cauchy,
and hence the above theorem implies that every bounded monotonic
sequence has a limit. So for the reals it is entirely a matter of
choice (or taste) which approach one prefers. Personally, I find the
monotone convergence theorem more obviously true

and therefore
preferable as an *axiom* for the reals. However, as mentioned
already, there are other situations in which the Cauchy sequence
approach is the only one possible.