cauchy sequence

Cauchy sequence

A sequence $(a_n{)}_{n\in \mathbb{N}}\subset \mathbb{C}$ is called a Cauchy sequence if

$$\forall ϵ>0\exists n_0\in \mathbb{N}\forall n,m\ge n_0:|a_n-a_m|<ϵ$$

That is, the terms of a Cauchy sequence are pairwise close to each other for large $n$.

Compare with convergence: convergence says terms get close to a specific value $a$. Cauchy says terms get close to each other. The power of the Cauchy criterion is that you can prove convergence without knowing the limit:

If $|a_{n+1}-a_n|\le C\cdot |a_n-a_{n-1}|$ for some $0<C<1$

Each gap is at most a fixed fraction of the previous gap. By induction: $|a_{n+1}-a_n|\le C^{n-1}|a_2-a_1|$. Since $C<1$, the gaps shrink to zero geometrically, so the terms bunch up: the sequence is Cauchy, hence convergent.

This works for any such sequence. You never need to know the limit.

Cauchy criterion

$$(a_n)\text{ is convergent}⟺(a_n)\text{ is a Cauchy sequence}$$

Proof sketch

$(\Rightarrow )$ If $a_n\to a$, then for large $n,m$:

$$|a_n-a_m|\le |a_n-a|+|a-a_m|<\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon$$

$(\Leftarrow )$ Cauchy $⟹$ bounded (fix $\epsilon =1$, then all terms past some $n_0$ are within $1$ of $a_{n_0}$). Bounded $⟹$ has a convergent subsequence by Bolzano-Weierstrass. Call its limit $a$. Then:

$$|a_n-a|\le |a_n-a_{n_k}|+|a_{n_k}-a|$$

Both terms $<\epsilon /2$ for large enough $n$ and $n_k$ (first by Cauchy, second by subsequence convergence).

Neighbors getting close is not enough

$a_n=\sqrt{n}$: consecutive terms satisfy $|a_{n+1}-a_n|=\sqrt{n+1}-\sqrt{n}\to 0$ (by the conjugate trick: $=\frac{1}{\sqrt{n+1}+\sqrt{n}}\to 0$), but the sequence diverges.

Cauchy requires all pairs $|a_n-a_m|<\epsilon$ for $n,m\ge n_0$, not just neighbors. Here $|a_{4n}-a_n|=\sqrt{4n}-\sqrt{n}=\sqrt{n}\to \infty$.

$a_n=\frac{1}{n}$ is Cauchy

For $m\ge n$: $|a_n-a_m|=\frac{1}{n}-\frac{1}{m}\le \frac{1}{n}$.

Given $\epsilon >0$, pick $n_0>\frac{1}{\epsilon }$. Then for all $n,m\ge n_0$:

$$|a_n-a_m|\le \frac{1}{n}\le \frac{1}{n_0}<\epsilon$$

Completeness

The Cauchy criterion relies on $\mathbb{R}$ being complete. In $\mathbb{Q}$, a Cauchy sequence of rationals can “converge” to an irrational that isn’t in $\mathbb{Q}$, so Cauchy $\nRightarrow$ convergent there.
This is gives us an equivalent way to define completeness: a space is complete iff every Cauchy sequence converges.