The supremum metric $d_\infty(f,g) = \sup_x |f(x) - g(x)|$ on the bounded real-valued functions on a set

Let $S$ be a nonempty set, let $\mathcal{B}(S)$ be the set of bounded functions $S \to \mathbb{R}$ and let $d_\infty(f,g) = \sup\{|f(s)-g(s)| : s \in S\}$ be the supremum metric; that this is a metric is The supremum metric $d_\infty(f,g) = \sup_x |f(x) - g(x)|$ is a metric on the bounded real-valued functions on a nonempty set and is quoted here rather than reproved. This example records two things about it.

1. The constants form an isometric copy of the real line. For $c \in \mathbb{R}$ let $\hat c \in \mathcal{B}(S)$ be the constant function with value $c$. Then $c \mapsto \hat c$ is an isometric embedding of $(\mathbb{R}, d_{\mathbb{R}})$ into $(\mathcal{B}(S), d_\infty)$ (Isometry, isometric embedding, and the subspace metric on a subset, The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$, and it is unbounded): $d_\infty(\hat c, \hat c\,') = |c - c\,'| \qquad (c, c\,' \in \mathbb{R}).$
2. The supremum need not be attained. Take $S = \mathbb{N}$, so that $\mathcal{B}(\mathbb{N})$ is the set of bounded sequences of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences), and put $f(k) := 1 - \frac{1}{k+1} \quad (k \in \mathbb{N}), \qquad g := \hat 0 .$ Then $f, g \in \mathcal{B}(\mathbb{N})$, $d_\infty(f,g) = 1$, and $|f(k) - g(k)| < 1$ for every $k \in \mathbb{N}$. So $d_\infty(f,g)$ is a supremum that is not a maximum, and no single point of $S$ realises the distance.

The index shift in $1/(k+1)$ is forced: $\mathbb{N}$ contains $0$ (The natural numbers $\mathbb{N}$ (von Neumann)) and sequences here are indexed from $0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), so $1/k$ would be undefined at $k = 0$.

• Why the constants matter. The isometric copy of $\mathbb{R}$ inside $\mathcal{B}(S)$ shows that $(\mathcal{B}(S), d_\infty)$ is unbounded whenever $S \ne \emptyset$, since $\mathbb{R}$ is (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$, and it is unbounded, An isometric embedding is injective and carries the metric topology of the source onto the subspace topology of its image).
• With $S = \mathbb{N}$ this is the space usually written $\ell^\infty$, the bounded real sequences with the supremum metric. Its completeness and its separability are questions for later pages and are not touched here.
• Attainment is a genuinely different question from existence. The supremum exists because the set is nonempty and bounded above (Complete ordered field (least-upper-bound property)); whether it lies in the set is exactly the question of whether the sup is a maximum, and claim 2 answers it negatively for a specific pair.