Show a sequence is Cauchy iff $\lim_{N\to\infty} \operatorname{diam} E_N = 0$

Question

So this is right out of Rudin, but doesn't offer a proof or anything, just simply says that "given the definition of a Cauchy sequence and the definition of a diameter, the proof is clear." The sequence in question is $E_N = {p_N, p_{N+1}..}$. But I just don't quite understand. From the text, the diameter is the supremum of the distance of the points in the set. How does the supremum of $E_N$ approach $0$ for large $N$?

Answer 1

Given $\epsilon > 0$, we want to find $N$ so that for all indices $m,n \geq N$, $d(p_m,p_n) < \epsilon$. By definition, $$ \operatorname{diam} E_N \equiv \sup\{d(p,q)\mid p,q \in E\} = \sup\{d(p_m,p_n)\mid m,n \geq N \} $$ We can choose $N$ so that $\operatorname{diam} E_N < \epsilon$. Since $\sup\{d(p_m,p_n)\mid m,n \geq N \}$ is the upper bound of $d(p_m,p_n)$, $(p_n)_{n=0}^\infty$ is Cauchy.

The converse is similar.