Epsilon characterisation of Cauchy sequences

Statement

Let $(X, d)$ be a metric space and let $(x_n)_{n \ge 1}$ be a sequence in $X$. For each $N \ge 1$, let $T_N = \{x_n : n \ge N\}$. Then the following two conditions are equivalent:

1. For each real $\varepsilon > 0$ there is an integer $N$ such that $d(x_m, x_n) < \varepsilon$ for all $m, n \ge N$.
2. The diameters $\operatorname{diam}(T_N)$ tend to $0$ as $N \to \infty$.

The sequence $(x_n)$ is a Cauchy sequence when it satisfies these conditions.

Remarks

Condition 1 is the usual definition of a Cauchy sequence. Each convergent sequence is a Cauchy sequence: if $x_n \to x$, then $d(x_m, x_n) \le d(x_m, x) + d(x, x_n)$. In $\mathbb{Q}$ with the usual metric, the decimal approximations $1, 1.4, 1.41, 1.414, \ldots$ of $\sqrt{2}$ form a Cauchy sequence that does not converge in $\mathbb{Q}$. Each Cauchy sequence is bounded.

Explanations

L1 human

The terms of a Cauchy sequence get closer and closer to each other. You can check this without knowing a limit, and this is the point. A Cauchy sequence looks as if it converges. But the space can have no point for it to converge to.