Cauchy's Criterion

Cauchy's Criterion is a result in analysis that states that a sequence of real numbers $(a_n)$ converges if and only if it is a Cauchy sequence. A sequence $(a_n)$ is considered Cauchy if, for every $\epsilon>0$ , there exists an $N\in\mathbb{N}$ such that for every $m,n\ge N$ that $|a_n-a_m|<\epsilon$ . The criterion is named after Augustine Louis Cauchy, a prominent mathematician known for his results in algebra and analysis.

Proof: Let $(a_n)\to a$ and $\epsilon>0$ be arbitrary. Choose $N\in\mathbb{N}$ such that for $n\ge N$ we have $|a_n-a|<\frac{\epsilon}{2}$ . By the triangle inequality we have $|a_n-a_m|\le |a_n-a|+|a-a_m|< \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$ which proves that this sequence is indeed Cauchy. Now for the reverse direction. We first show that if a sequence is Cauchy it is bounded. Let $\epsilon=1$ such that $|a_n-a_m|<1$ for all $n,m\ge N$ . Then we see that for all $n\in N$ that $|a_n|<|a_N|+1$ which means it is bounded by

$M=\max\{|x_1|,|x_2|,\ldots,|x_{n-1}|,|x_N|+1\}$

so it is therefore bounded. Since $|a_n|\le M$ , by the Bolzano–Weierstrass theorem $(a_n)$ has a convergent subsequence $(a_{n_k})\to a$ . We show that $(a_n)\to a$ . Since $(a_n)$ is assumed to be Cauchy, set $|a_n-a_m|<\frac{\epsilon}{2}$ for $m,n\ge N$ . Additionally, since $(a_{n_k})\to a$ we see that $|a_{n_k}-a|<\frac{\epsilon}{2}$ and we see that by the triangle inequality

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

which proves that $(a_n)$ is convergent.

Page

Toolbox

Search

Cauchy's Criterion