The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded

Define $d_{\mathbb{R}}:\mathbb{R}\times \mathbb{R}\to \mathbb{R}$ by $d_{\mathbb{R}}(x,y):=\mathrm{\mid }x-y\mathrm{\mid }$ (Absolute value in an ordered field). Then:

1. $d_{\mathbb{R}}$ is a metric on $\mathbb{R}$ (Metric space: $d(x,y)=0$ iff $x=y$, symmetry, and the triangle inequality; pseudometric and ultrametric); it is called the usual metric of $\mathbb{R}$.
2. For $x\in \mathbb{R}$ and $r>0$ the open ball is the bounded open interval (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length, Open ball, closed ball and sphere in a metric space) $B(x,r)=(x-r,x+r),$ and the closed ball is $\bar{B}(x,r)=[x-r,x+r]$.
3. Consequently $U\subseteq \mathbb{R}$ is open in the metric topology of $d_{\mathbb{R}}$ (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement) exactly when for every $x\in U$ there is $r>0$ with $(x-r,x+r)\subseteq U$. This topology is called the usual topology of $\mathbb{R}$.
4. $(\mathbb{R},d_{\mathbb{R}})$ is not a bounded metric space (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space): no ball contains $\mathbb{R}$, so $\mathrm{diam}(\mathbb{R})$ is not defined.

• This is the metric every later ceiling rests on. Every real-line example on the companion page, and every subspace of $\mathbb{R}$ used there, takes its metric from $d_{\mathbb{R}}$ through the subspace construction of Isometry, isometric embedding, and the subspace metric on a subset.
• Unboundedness needs no Archimedean input, and no completeness either. No ordered field is bounded under $d(u,v)=\mathrm{\mid }u-v\mathrm{\mid }$, and the reason is a single element rather than any cofinality property: given a centre $c$ and a radius $r>0$, the element $c+r+1$ satisfies $d(c,c+r+1)=\mathrm{\mid }-(r+1)\mathrm{\mid }=r+1>r$, because $1>0$ and $r+1>0$ (Basic properties of the absolute value, The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities); so it lies outside $B(c,r)$ and no ball contains the field. Step 1.6 above chooses its witness with Every complete ordered field is Archimedean instead, which is a convenience and not a necessity: it delivers a witness that is a canonical natural, and claim 4 needs no such thing. Claim 4 therefore holds verbatim in every ordered field with this $d$, Archimedean or not. Note also that a radius is an element of $\mathbb{R}$, so "a ball of infinite radius" is not something that can be written here.
• The claim that $\mathrm{diam}(\mathbb{R})$ is "not defined" is a claim about the conventions of this development (Conventions: $\sup \mathrm{\varnothing }$, unbounded sets, and the extended reals, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space): suprema here are real numbers and the extended real line, which is introduced on a later page, is not used for them, so an unbounded set has no diameter at all rather than a diameter $+\mathrm{\infty }$.