Equivalence of Cauchy sequence definitions

Question

Let $(x_n)$ be a sequence in a metric space $E$. Show that the following are equivalent:

• $(x_n)$ is Cauchy;
• $\forall \epsilon > 0 \;\; \exists N \in \mathbb{N} \;\; \forall n \ge N, \;\; d(x_N, x_n) < \epsilon $.

Similarly, show TFAE:

• $(x_n)$ is not Cauchy;
• $(x_n)$ has a subsequence $(x_{n_k})$ satisfying $$ \text{inf}\{d(x_{n_k}, x_{n_{k+1}}):\; k \in \mathbb{N}\} > 0 .$$

We are given that $(x_n)$ is Cauchy if $$\lim_{N\to\infty} \sup_{p\in \mathbb{N}} d(x_N, x_{N+p}) = 0.$$ Intuitively, I can see the equivalence of the first part, but I'm stuck as to how to formalise a proof using $(\epsilon,\delta)$.

Please help!

Answer 1

I'll prove one direction of the equivalence between the first two bullets, hopefully you will understand how to prove the other direction.

$(\Rightarrow)$ Assume that $(x_n)_n$ is Cauchy, that is $$\lim_{N\rightarrow \infty}\sup_{p\in\mathbb{N}}d(x_N,x_{N+p})=0.$$ Let $\varepsilon >0$. We want to find $K$ such that for all $n \geq K, \;\; d(x_K, x_n) < \varepsilon$, or equivalently, such that for all $ p\in \mathbb{N},\;\; d(x_K, x_{K+p})<\varepsilon$.

$K$ exists by the very definition of limit: $\lim_{N\rightarrow \infty}\sup_{p\in\mathbb{N}}d(x_N,x_{N+p})=0$ implies that there exists a number $K$ such that $$\sup_{p\in\mathbb{N}}d(x_K,x_{K+p})<\varepsilon.$$ Of course this implies that for all $ p\in \mathbb{N},\;\; d(x_K, x_{K+p})<\varepsilon$.

Answer 2

$\implies $ is obvious. For the converse, let $\varepsilon>0$ and let $N$ s.t $d(x_N,x_n)<\varepsilon/2$ if $n\geq N$. If $n,m\geq N$, then $$d(x_n,x_m)\leq d(x_N,x_n)+d(x_N,x_m) <\varepsilon.$$

For TFAE, it's just the negation.