Proof of the diameter characterisation of Cauchy sequences

Proof

We prove each implication. Each $T_N$ is nonempty, so its diameter is defined, with a value in $[0, \infty]$.

1. Let $(X, d)$ be a metric space, let $(x_n)_{n \ge 1}$ be a sequence in $X$, and let $T_N = \set{x_n : n \ge N}$.
2. Assume condition 1. Let $\varepsilon > 0$ be real.
3. By condition 1 applied to $\varepsilon / 2$, there is an integer $N$ with $d(x_m, x_n) < \varepsilon / 2$ for all $m, n \ge N$.
4. Let $M \ge N$. Two points of $T_M$ have the form $x_m$ and $x_n$ with $m, n \ge M \ge N$. So their distance is less than $\varepsilon / 2$ by step 3.
5. So $\varepsilon / 2$ is an upper bound of all distances in $T_M$. The supremum is the least upper bound, so $\operatorname{diam}(T_M) \le \varepsilon / 2 < \varepsilon$.
6. Step 5 holds for every $M \ge N$. Since $\varepsilon$ was arbitrary, $\operatorname{diam}(T_N) \to 0$ as $N \to \infty$. So condition 2 holds.
7. Now assume condition 2. Let $\varepsilon > 0$ be real.
8. By condition 2, there is an integer $N$ with $\operatorname{diam}(T_N) < \varepsilon$.
9. Let $m, n \ge N$. Then $x_m$ and $x_n$ lie in $T_N$.
10. The diameter is an upper bound of the distances in $T_N$. So $d(x_m, x_n) \le \operatorname{diam}(T_N) < \varepsilon$.
11. Thus condition 1 holds. By steps 2 to 6 and 7 to 10, the two conditions are equivalent.