ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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The bounded real-valued functions on a set, with the supremum metric, form a complete metric space

Example

Let $S$ be a nonempty set, let $B (S) : = {f : f is a bounded function S \to R}$ , and let $d_{\infty} (f, g) : = \sup {∣ f (s) - g (s) ∣ : s \in S}$ be the supremum metric, which is a metric on $B (S)$ (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, Lower bound, bounded below, bounded set).

Then $(B (S), d_{\infty})$ is a complete metric space (Complete metric space: every Cauchy sequence converges in the space).

The limit is produced pointwise and then shown to be bounded and to be approached uniformly; that order is the content of the proof.

Facts & Assumptions

Given: A nonempty set $S$ ; the space $B (S)$ with the supremum metric $d_{\infty}$ ; a Cauchy sequence $(f_{k})$ in $(B (S), d_{\infty})$ ; a point $s \in S$ ; a real $ε > 0$ .

[A1]

Cauchyness of $(f_{k})$ : for every real $ε > 0$ there is $K$ with $d_{\infty} (f_{k}, f_{l}) < ε$ for all $k, l \geq K$ (Cauchy sequence in a metric space, The rationals embed densely in the reals).

[L1]

$d_{\infty}$ is a metric on $B (S)$ , and $d_{\infty} (f, g)$ is the least upper bound of ${∣ f (s) - g (s) ∣ : s \in S}$ , so it dominates each of those numbers and is dominated by every upper bound of them (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, Complete ordered field (least-upper-bound property), Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L2]

A function $h : S \to R$ is bounded when its range is a bounded subset of $R$ , that is when there is a real $M \geq 0$ with $∣ h (s) ∣ \leq M$ for every $s$ ; the passage from a pair of bounds to a single $M$ is the maximum of two absolute values (Lower bound, bounded below, bounded set, Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set, Basic properties of the absolute value).

[L3]

Every Cauchy sequence of reals converges, and the limit of a real sequence is unique, which licenses $\lim_{k} a_{k}$ for a sequence already known to converge (The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges, A sequence has at most one limit, Limits and Cauchy sequences of reals).

[L4]

Limits of reals preserve non-strict inequalities holding eventually, and behave additively (Limits preserve non-strict inequalities, Algebra of limits: sums, scalar multiples, products and quotients).

[L5]

$∣ ∣ a ∣ - ∣ b ∣ ∣ \leq ∣ a - b ∣$ for reals, the reverse triangle inequality of the usual metric of $R$ with third point $0$ ; hence $a_{k} \to a$ gives $∣ a_{k} ∣ \to ∣ a ∣$ (The reverse triangle inequality $∣ d (x, z) - d (y, z) ∣ \leq d (x, y)$ in any metric space, Basic properties of the absolute value, The absolute value makes $R$ 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).

[L6]

Convergence in a metric space may be tested with real $ε > 0$ (Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ , The rationals embed densely in the reals).

Verification

technique · direct

1.1

For every $s \in S$ and all $k, l$ the number $∣ f_{k} (s) - f_{l} (s) ∣$ belongs to the set whose supremum is $d_{\infty} (f_{k}, f_{l})$ , so $∣ f_{k} (s) - f_{l} (s) ∣ \leq d_{\infty} (f_{k}, f_{l})$ .

1.2

Apply [A1] with $ε = 1$ to get $K_{1}$ with $d_{\infty} (f_{k}, f_{l}) < 1$ for all $k, l \geq K_{1}$ , and let $M \geq 0$ satisfy $∣ f_{K_{1}} (s) ∣ \leq M$ for every $s$ , which exists because $f_{K_{1}}$ is bounded.

A1L2

2.1

Hence for each fixed $s$ the real sequence $(f_{k} (s))_{k}$ is Cauchy, by [A1] and step 1.1; so it converges, and its limit is unique, so $f (s) : = \lim_{k} f_{k} (s)$ defines a function $f : S \to R$ . No choice is used, each value being a unique limit.

step 1.1A1L3

2.2

Let $ε > 0$ be real and take $K$ from [A1] for $ε / 2$ , so $d_{\infty} (f_{k}, f_{l}) < ε / 2$ for all $k, l \geq K$ . For a fixed $s$ and a fixed $k \geq K$ we get $∣ f_{k} (s) - f_{l} (s) ∣ \leq d_{\infty} (f_{k}, f_{l}) < ε / 2$ for every $l \geq K$ .

step 1.1A1

3.1

For every $s$ and every $l \geq K_{1}$ : $∣ f_{l} (s) ∣ \leq ∣ f_{K_{1}} (s) ∣ + ∣ f_{l} (s) - f_{K_{1}} (s) ∣ \leq M + 1$ by step 1.1; letting $l$ grow and using $∣ f_{l} (s) ∣ \to ∣ f (s) ∣$ gives $∣ f (s) ∣ \leq M + 1$ . So $f$ is bounded and $f \in B (S)$ .

step 1.1step 2.1step 1.2L2L4L5

3.2

Letting $l$ grow in step 2.2 and using $f_{l} (s) \to f (s)$ , hence $∣ f_{k} (s) - f_{l} (s) ∣ \to ∣ f_{k} (s) - f (s) ∣$ , gives $∣ f_{k} (s) - f (s) ∣ \leq ε / 2$ for every $s \in S$ and every $k \geq K$ .

step 2.1step 2.2L4L5

4.1

So $ε / 2$ is an upper bound of ${∣ f_{k} (s) - f (s) ∣ : s \in S}$ for every $k \geq K$ , whence $d_{\infty} (f_{k}, f) \leq ε / 2 < ε$ ; note that $d_{\infty} (f_{k}, f)$ is defined, both functions being bounded.

step 3.1step 3.2L1

5.1

Since $ε > 0$ was an arbitrary real, $f_{k} \to f$ in $(B (S), d_{\infty})$ with $f \in B (S)$ ; every Cauchy sequence therefore converges, and $(B (S), d_{\infty})$ is complete.

step 3.1step 4.1L6∎

Remarks

The two limits are taken in different orders, and that is the point. Step 2.1 fixes $s$ and lets $k$ grow, producing a candidate limit function; steps 2.2 to 4.1 fix $k$ and let the other index grow inside an estimate that is uniform in $s$ . It is the uniformity of the bound $ε / 2$ in $s$ , and nothing else, that converts pointwise convergence into convergence in $d_{\infty}$ .
Boundedness of the limit is a separate step and is genuinely needed. The metric $d_{\infty} (f_{k}, f)$ is only defined once $f$ is known to be bounded (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), so step 3.1 has to come before step 4.1. The extended real line is introduced later, but it is not used as a metric codomain or as a placeholder here; $d_{\infty} (f_{k}, f)$ is real-valued only after boundedness of $f$ is proved.
Nothing is assumed about $S$ beyond nonemptiness, which is what 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 needs so that the supremum is taken over a nonempty set. In particular $S$ carries no topology and no metric here; the functions are arbitrary bounded functions, not continuous ones.
Where the least-upper-bound property is spent. Twice: inside The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges at step 2.1, and in the very definition of $d_{\infty}$ (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). Completeness of $R$ is the whole engine of this example.

Depends on

Used by

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Sources

Uniform norm (Wikipedia) (standard reference, not scraped)
Complete metric space (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 7 (standard reference, not scraped)