Every Cauchy sequence in a metric space is bounded

Let $(X,d)$ be a metric space (Metric space: $d(x,y) = 0$ iff $x = y$, symmetry, and the triangle inequality; pseudometric and ultrametric) and let $(x_k)$ be a Cauchy sequence in $X$ (Cauchy sequence in a metric space). Then its range $R := \{\, x_k : k \in \mathbb{N} \,\}$ is a bounded subset of $X$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space): there are a point $x_0 \in X$ and a real $r > 0$ with $R \subseteq B(x_0, r)$ (Open ball, closed ball and sphere in a metric space).

Consequently $R$ is nonempty and bounded, so $\operatorname{diam}(R)$ exists (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

• The maximum is taken over $j \le N$ and not over $j < N$, and the extra element $1$ is in the set as well. Both are deliberate. Indices run from $0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), so $N = 0$ is possible and $\{\, d(x_j,x_N) : j < N \,\}$ would then be empty, which has no maximum (Maximum and minimum of a set); adjoining $1$ makes the set nonempty in every case and simultaneously covers the tail bound.
• Boundedness of the range is strictly weaker than Cauchyness. The sequence $x_k = (-1)^k$ in $\mathbb{R}$ has bounded range and is not Cauchy, since two consecutive terms are always at distance $2$. So this lemma cannot be reversed, and nothing on this page reverses it.
• What the lemma is for. It is what makes $\operatorname{diam}$ available for the tails of a Cauchy sequence, which is the form in which Cauchyness enters the converse half of In a complete metric space nested nonempty closed sets whose diameters tend to $0$ meet in exactly one point, and this property characterises completeness.