Show a sequence is Cauchy iff $\lim_{N\to\infty} \operatorname{diam} E_N = 0$ 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 $N$?        
 A: 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.
A: $E_N \supset E_{N+1}$ is obvious.
So, $\operatorname{diam} E_N \geq \operatorname{diam} E_{N+1}$ is obvious.  
Let $\{p_n\}$ be a Cauchy sequence.
Let $\epsilon$ be an arbitrary positive real number.  
There is $N \in \{1, 2, \cdots\}$ such that if $n, m \geq N$, then $d(p_n, p_m) < \epsilon$.
So, $\operatorname{diam} E_N \leq \epsilon$.
Because $\operatorname{diam} E_N \geq \operatorname{diam} E_{N+1}$, $\lim_{N \to \infty} \operatorname{diam} E_N = 0$.  
Conversely,
Let $\lim_{N \to \infty} \operatorname{diam} E_N = 0$.
Let $\epsilon$ be an arbitrary positive real number.  
Because $\lim_{N \to \infty} \operatorname{diam} E_N = 0$, there exists $N \in \{1, 2, \cdots\}$ such that $\operatorname{diam} E_N < \epsilon$.
So, if $n, m \geq N$, then $d(x_n, x_m) \leq \operatorname{diam} E_N < \epsilon$.
