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 $\mathcal O_\epsilon(X):= (X-\epsilon, X+\epsilon)$.
Best 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.