Cauchy Sequences

A Cauchy sequence is a sequence whose terms become very close to each other as the sequence progresses. Formally, the sequence $\{a_n\}_{n=0}^{\infty}$ is a Cauchy sequence if, for every $\epsilon>0,$ there is an $N>0$ such that $$n,m>N\implies |a_n-a_m|<\epsilon.$$ Translating the symbols, this means that for any small distance, there is a certain index past which any two terms are within that distance of each other, which captures the intuitive idea of the terms becoming close. This can also be written as $$\limsup_{m,n} |a_m-a_n|=0,$$ where the limit superior is being taken.

Cauchy sequences are useful because they give rise to the notion of a complete field, which is a field in which every Cauchy sequence converges. Because the Cauchy sequences are the sequences whose terms grow close together, the fields where all Cauchy sequences converge are the fields that are not ``missing" any numbers. The canonical complete field is $\mathbb{R}$, so understanding Cauchy sequences is essential to understanding the properties and structure of $\mathbb{R}$.

The definition of Cauchy sequences given above can be used to identify sequences as Cauchy sequences.

Is the sequence $a_n=\frac{1}{2^n}$ a Cauchy sequence?

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Yes. Take any $\epsilon>0$, and choose $N$ so large that $2^{-N}<\epsilon$. Then, if $n,m>N$, we have $$|a_n-a_m|=\left|\frac{1}{2^n}-\frac{1}{2^m}\right|\leq \frac{1}{2^n}+\frac{1}{2^m}\leq \frac{1}{2^N}+\frac{1}{2^N}=\epsilon,$$ so this sequence is Cauchy. $_\square$

Showing that a sequence is not Cauchy is slightly trickier. For a sequence not to be Cauchy, there needs to be some $N>0$ such that for any $\epsilon>0$, there are $m,n>N$ with $|a_n-a_m|>\epsilon$. In other words, no matter how far out into the sequence the terms are, there is no guarantee they will be close together.

Is the sequence $a_n=n$ a Cauchy sequence?

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No. Take $\epsilon=1$. Then, for any $N$, if we take $n=N+3$ and $m=N+1$, we have that $|a_m-a_n|=2>1$, so there is never any $N$ that works for this $\epsilon.$ Thus, the sequence is not Cauchy. $_\square$

When attempting to determine whether or not a sequence is Cauchy, it is easiest to use the intuition of the terms growing close together to decide whether or not it is, and then prove it using the definition.

The ideas from the previous sections can be used to consider Cauchy sequences in a general metric space $(X,d).$ In this context, a sequence $\{a_n\}$ is said to be Cauchy if, for every $\epsilon>0$, there exists $N>0$ such that $$m,n>n\implies d(a_m,a_n)<\epsilon.$$ On an intuitive level, nothing has changed except the notion of "distance" being used.

Consider the metric space of continuous functions on $[0,1]$ with the metric $$d(f,g)=\int_0^1 |f(x)-g(x)|\, dx.$$ Is the sequence $f_n(x)=nx$ a Cauchy sequence in this space?

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Note that $$d(f_m,f_n)=\int_0^1 |mx-nx|\, dx =\left[|m-n|\frac{x^2}{2}\right]_0^1=\frac{|m-n|}{2}.$$ By taking $m=n+1$, we can always make this $\frac12$, so there are always terms at least $\frac12$ apart, and thus this sequence is not Cauchy. $_\square$

Cauchy sequences are intimately tied up with convergent sequences. For example, every convergent sequence is Cauchy, because if $a_n\to x$, then $$|a_m-a_n|\leq |a_m-x|+|x-a_n|,$$ both of which must go to zero. The converse of this question, whether every Cauchy sequence is convergent, gives rise to the following definition:

A field is complete if every Cauchy sequence in the field converges to an element of the field.

Is $\mathbb{Q}$ a complete field?

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No. Take a sequence given by $a_0=1$ and satisfying $a_n=\frac{a_{n-1}}{2}+\frac{1}{a_{n}}$. This sequence has limit $\sqrt{2}$, so it is Cauchy, but this limit is not in $\mathbb{Q},$ so $\mathbb{Q}$ is not a complete field. $_\square$

In particular, $\mathbb{R}$ is a complete field, and this fact forms the basis for much of real analysis: to show a sequence of real numbers converges, one only need show that it is Cauchy. Similarly, given a Cauchy sequence, it automatically has a limit, a fact that is widely applicable.