Real numbers/Cauchy sequence/Monotonic/Introduction/Section

A problem with the concept of convergence is that in its very formulation already the limit is used, which in many cases is not known in advance. The Babylonian method to construct a sequence ${}{\left(x_{n}\right)}_{n\in \mathbb {N} }$ (for the computation of ${}{\sqrt {5}}$ say) starting with a rational number gives a sequence of rational numbers. If we consider this sequence inside the real numbers ${}\mathbb {R}$ where ${}{\sqrt {5}}$ exists, this sequence converges. However, within the rational numbers, this sequence does not converge. We would like to formulate within the rational numbers alone the property that the members of the sequence are getting closer and closer without referring to a limit point. This purpose fulfills the notion of Cauchy sequence.

Augustin Louis Cauchy (1789-1857)

Definition

A real sequence ${}{\left(x_{n}\right)}_{n\in \mathbb {N} }$ is called a Cauchy sequence, if the following condition holds.

For every ${}\epsilon >0$, there exists an ${}n_{0}\in \mathbb {N}$, such that for all ${}n,m\geq n_{0}$, the estimate

${}\vert {x_{n}-x_{m}}\vert \leq \epsilon \,$

holds.

Lemma

Every convergent sequence is a Cauchy sequence.

Proof

Let ${}{\left(x_{n}\right)}_{n\in \mathbb {N} }$ be a convergent sequence with limit ${}x$. Let ${}\epsilon >0$ be given. We apply the convergence property for ${}\epsilon /2$. Therefore there exists an ${}n_{0}$ with

$\vert {x_{n}-x}\vert \leq \epsilon /2{\text{ for all }}n\geq n_{0}.$

For arbitrary ${}n,m\geq n_{0}$ we then have due to the triangle inequality

${}\vert {x_{n}-x_{m}}\vert \leq \vert {x_{n}-x}\vert +\vert {x-x_{m}}\vert \leq \epsilon /2+\epsilon /2=\epsilon \,.$

  Hence we have a Cauchy sequence.

$\Box$

Definition

Let ${}{\left(x_{n}\right)}_{n\in \mathbb {N} }$ be a real sequence. For any strictly increasing mapping ${}\mathbb {N} \rightarrow \mathbb {N} ,i\mapsto n_{i}$, the sequence

$i\mapsto x_{n_{i}}$

is called a subsequence of the sequence.

Definition

A real sequence ${}{\left(x_{n}\right)}_{n\in \mathbb {N} }$ is called increasing, if ${}x_{n+1}\geq x_{n}$ holds for all ${}n\in \mathbb {N}$, and strictly increasing, if ${}x_{n+1}>x_{n}$ holds for all ${}n\in \mathbb {N}$.

A sequence ${}{\left(x_{n}\right)}_{n\in \mathbb {N} }$ is called decreasing if ${}x_{n+1}\leq x_{n}$ holds for all ${}n\in \mathbb {N}$, and strictly decreasing, if ${}x_{n+1}<x_{n}$ holds for all

${}n\in \mathbb {N}$.

Lemma

Let ${}{\left(x_{n}\right)}_{n\in \mathbb {N} }$ be a real increasing sequence which is bounded from above. Then ${}{\left(x_{n}\right)}_{n\in \mathbb {N} }$ is a Cauchy sequence.

Proof

Let ${}b\in \mathbb {R}$ denote a bound from above, so that ${}x_{n}\leq b$ holds for all ${}x_{n}$. We assume that ${}{\left(x_{n}\right)}_{n\in \mathbb {N} }$ is not a Cauchy sequence. Then there exists some ${}\epsilon >0$ such that for every ${}n_{0}$, there exist indices ${}n>m\geq n_{0}$ fulfilling ${}x_{n}-x_{m}\geq \epsilon$. Because of the monotonicity, there is also for every ${}n_{0}$ an ${}n>n_{0}$ with ${}x_{n}-x_{n_{0}}\geq \epsilon$. Hence, we can define inductively an increasing sequence of natural numbers satisfying

$n_{1}>n_{0}{\text{ such that }}x_{n_{1}}-x_{n_{0}}\geq \epsilon ,$

$n_{2}>n_{1}{\text{ such that }}x_{n_{2}}-x_{n_{1}}\geq \epsilon ,$

and so on. On the other hand, there exists, due to the axiom of Archimedes, some ${}k\in \mathbb {N}$ with

${}k\epsilon >b-x_{n_{0}}\,.$

The sum of the first ${}k$ differences of the subsequence ${}x_{n_{j}}$, ${}j\in \mathbb {N}$, is

${}{\begin{aligned}x_{n_{k}}-x_{n_{0}}&={\left(x_{n_{k}}-x_{n_{k-1}}\right)}+{\left(x_{n_{k-1}}-x_{n_{k-2}}\right)}+\cdots +{\left(x_{n_{2}}-x_{n_{1}}\right)}+{\left(x_{n_{1}}-x_{n_{0}}\right)}\\&\geq k\epsilon \\&>b-x_{n_{0}}.\end{aligned}}$

This implies ${}x_{n_{k}}>b$, contradicting the condition that ${}b$ is an upper bound for the sequence.

$\Box$