# Equivalent condition to sequence being Cauchy

When reading a discussion in lecture notes about Cauchy sequences, the notes mention an equivalent condition for a sequence to be Cauchy that I am having trouble showing is true. The original definition of a Cauchy sequence we have used so far is replicated below.

A sequence $$(a_n)_{n\in\mathbb{N}}$$ is Cauchy if for all $$\epsilon>0$$, there exists $$N\in\mathbb{R}$$ such that $$|a_n-a_m|<\epsilon$$ for all $$n,m>N$$.

## The Condition

Proposition: $$(a_n)_{n\in\mathbb{N}}$$ is Cauchy if and only if for all $$\epsilon>0$$, there exist $$N_\epsilon,x_\epsilon$$ such that $$a_n \in \mathcal O_\epsilon(x_\epsilon)$$ for all $$n>N$$.

## Attempt

Proof: If we have that for all $$\epsilon>0$$ there exists $$N_\epsilon,x_\epsilon$$ such that $$a_n\in\mathcal O_\epsilon(x_\epsilon)$$ then we may say that there exists $$N_\epsilon,x_\epsilon$$ for which $$|a_n-x_\epsilon|<\epsilon/2$$ and $$|a_m-x_\epsilon|<\epsilon/2$$ for all $$n,m>N_\epsilon$$. Hence we have that for all $$\epsilon>0$$, there exists $$N_\epsilon, x_\epsilon$$ such that:

$$|a_m-a_n|=|(a_m-x_\epsilon)-(a_n-x_\epsilon)|\leq |a_m-x_\epsilon|+|a_n-x_\epsilon|<\epsilon/2+\epsilon/2 = \epsilon.$$ Thus the sequence is Cauchy.

Now consider the other direction. We have proven earlier that Cauchy sequences are bounded. Hence there exists a convergent subsequence, call it $$(a_m)$$ such that $$a_m$$ tends to some $$x_\epsilon$$. Then also since $$(a_n)$$ is Cauchy, $$|a_n-a_m|<\epsilon/2$$. Thus $$|a_n-x_\epsilon| = |(a_n-a_m)-(x_\epsilon-a_m)| < \epsilon/2 + \epsilon/2= \epsilon.$$

Is my proof correct?

Edit: as per request in comments I define $$\mathcal O_\epsilon(X):= (X-\epsilon, X+\epsilon)$$.

• As someone whose never seen the $\mathcal{O}$ notation, can you define it for me? Thanks! Commented Sep 27, 2019 at 13:08
• I edited the post for you. Commented Sep 27, 2019 at 13:31

## 1 Answer

Yes it is correct but you can simplify the "other direction" argument :

let $$\epsilon > 0$$. There is $$N$$ such that $$\forall n,m > N, \, |a_n - a_m| < \epsilon$$.
Now set $$x_\epsilon := a_{N+1}$$ and $$N_\epsilon :=N$$. For $$n > N_\epsilon$$, you have $$|a_n - x_\epsilon| < \epsilon$$ as desired.

• This is much simpler, thank you. Commented Sep 27, 2019 at 13:40