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LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-02 (claude-opus-5)
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The supremum metric d(f,g)=supxf(x)g(x)d_\infty(f,g) = \sup_x |f(x) - g(x)| is a metric on the bounded real-valued functions on a nonempty set

Statement

Let SS be a nonempty set. Call a function f:SRf : S \to \mathbb{R} bounded when its range f[S]={f(s):sS}f[S] = \{f(s) : s \in S\} is a bounded subset of R\mathbb{R} (Lower bound, bounded below, bounded set), and write

B(S):={f:f is a bounded function SR}.\mathcal{B}(S) := \{\, f : f \text{ is a bounded function } S \to \mathbb{R} \,\}.

For f,gB(S)f, g \in \mathcal{B}(S) put D(f,g):={f(s)g(s):sS}D(f,g) := \{\, |f(s) - g(s)| : s \in S \,\} and

d(f,g):=supD(f,g).d_\infty(f,g) := \sup D(f,g).

This is well defined: D(f,g)D(f,g) is nonempty because SS is, and it is bounded above (step 1.1 below), so its least upper bound exists (Complete ordered field (least-upper-bound property)) and is unique (Suprema and infima are unique).

Then dd_\infty is a metric on B(S)\mathcal{B}(S) (Metric space: d(x,y)=0d(x,y) = 0 iff x=yx = y, symmetry, and the triangle inequality; pseudometric and ultrametric), the supremum metric (also called the uniform metric).

The hypotheses ensure that the formula is a finite real-valued metric for every pair in the stated function space. Boundedness of ff and gg makes D(f,g)D(f,g) bounded above, and nonemptiness of SS makes it nonempty. Some unbounded pairs can still have a finite supremum, but allowing all real-valued functions would not give a finite-valued metric: for example, on S=RS=\mathbb{R} the functions f(s)=sf(s)=s and g(s)=0g(s)=0 make D(f,g)D(f,g) unbounded above (Conventions: sup\sup \emptyset, unbounded sets, and the extended reals).

Facts & Assumptions

Given: A nonempty set SS and bounded functions f,g,hB(S)f, g, h \in \mathcal{B}(S), with ff(s)uf\ell_f \le f(s) \le u_f, gg(s)ug\ell_g \le g(s) \le u_g and hh(s)uh\ell_h \le h(s) \le u_h for all sSs \in S; a fixed s0Ss_0 \in S.

[L1]

Bounded subset of R\mathbb{R}: TT is bounded when there are ,uR\ell, u \in \mathbb{R} with tu\ell \le t \le u for every tTt \in T (Lower bound, bounded below, bounded set).

[L2]

Least-upper-bound property: a nonempty subset of R\mathbb{R} that is bounded above has a least upper bound, that is an upper bound below every upper bound; it is unique (Complete ordered field (least-upper-bound property), Suprema and infima are unique).

[L3]

Absolute value: u0|u| \ge 0; u=0|u| = 0 if and only if u=0u = 0; u=u|-u| = |u|; and u|u| equals uu or u-u (Basic properties of the absolute value, Absolute value in an ordered field).

[L4]

Two-term triangle inequality: u+vu+v|u + v| \le |u| + |v| (The triangle inequality).

[L5]

A two-element subset of R\mathbb{R} has a maximum, which is one of the two elements and bounds both (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L6]

Order arithmetic: inequalities may be added and a constant added to both sides, in the strict form of Order is preserved by adding a constant and by adding inequalities and, together with the case of equality settled by totality (Ordered field, Complete ordered field (least-upper-bound property)), in the nonstrict form used below; and by trichotomy a0a \le 0 together with a0a \ge 0 gives a=0a = 0.

Proof

technique · direct
1.1

For every sSs \in S the value f(s)g(s)|f(s) - g(s)| is f(s)g(s)f(s) - g(s) or g(s)f(s)g(s) - f(s), and f(s)g(s)ufgf(s) - g(s) \le u_f - \ell_g while g(s)f(s)ugfg(s) - f(s) \le u_g - \ell_f; so M:=max{ufg, ugf}M := \max\{u_f - \ell_g,\ u_g - \ell_f\} bounds D(f,g)D(f,g) above, and since s0Ss_0 \in S makes D(f,g)D(f,g) nonempty, d(f,g)=supD(f,g)d_\infty(f,g) = \sup D(f,g) exists and is unique.

givenL1L2L3L5L6
1.2

Symmetry (M2): g(s)f(s)=(f(s)g(s))=f(s)g(s)|g(s) - f(s)| = |-(f(s) - g(s))| = |f(s) - g(s)| for every sSs \in S, so D(g,f)D(g,f) and D(f,g)D(f,g) are the same subset of R\mathbb{R} and therefore have the same supremum.

L2L3
2.1

Separation (M1): d(f,g)d_\infty(f,g) bounds D(f,g)D(f,g) above, so d(f,g)f(s0)g(s0)0d_\infty(f,g) \ge |f(s_0) - g(s_0)| \ge 0; if d(f,g)=0d_\infty(f,g) = 0 then f(s)g(s)0|f(s) - g(s)| \le 0 and f(s)g(s)0|f(s) - g(s)| \ge 0 for every ss, hence f(s)=g(s)f(s) = g(s) for every ss and f=gf = g; conversely if f=gf = g then D(f,g)={0}D(f,g) = \{0\}, whose least upper bound is 00.

step 1.1L2L3L6
2.2

For every sSs \in S: f(s)h(s)=(f(s)g(s))+(g(s)h(s))f(s)g(s)+g(s)h(s)d(f,g)+d(g,h)|f(s) - h(s)| = |(f(s) - g(s)) + (g(s) - h(s))| \le |f(s) - g(s)| + |g(s) - h(s)| \le d_\infty(f,g) + d_\infty(g,h), the last inequality because each supremum bounds its own set above.

step 1.1L2L4L6
3.1

Triangle inequality (M3): step 2.2 says the real number d(f,g)+d(g,h)d_\infty(f,g) + d_\infty(g,h) is an upper bound of D(f,h)D(f,h), and d(f,h)d_\infty(f,h) is the least upper bound of that set, so d(f,h)d(f,g)+d(g,h)d_\infty(f,h) \le d_\infty(f,g) + d_\infty(g,h).

step 2.2L2
4.1

The function dd_\infty therefore satisfies (M1) by step 2.1, (M2) by step 1.2 and (M3) by step 3.1, so it is a metric on B(S)\mathcal{B}(S).

step 1.2step 2.1step 3.1

Remarks

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