ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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The $1$ -Lipschitz maps of a metric space into $R$ form a uniformly equicontinuous family, and the distance functions $x \mapsto d (x, A)$ all belong to it

Example

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 $R$ carry its usual metric (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). Put

$L : = {f : X \to R : f is Lipschitz with constant 1} \subseteq R^{X}$

(Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction, The topology of pointwise convergence on $Y^{X}$ , which is the product topology, and its restriction to $C (X, Y)$ ). Then:

$L$ is uniformly equicontinuous (Equicontinuity at a point, uniform equicontinuity, and pointwise boundedness of a family of maps between metric spaces), with $δ : = ε$ serving at every $ε$ ;
for every nonempty $A \subseteq X$ the distance function $φ_{A} (x) : = d (x, A)$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space) belongs to $L$ ;
if $X \neq \emptyset$ then $L$ is not pointwise bounded, since it contains every constant function.

So equicontinuity and pointwise boundedness are genuinely independent hypotheses: this family has the first and not the second, and the next counterexample on this page has the second and not the first.

Facts & Assumptions

Given: A metric space $(X, d)$ , the target $R$ with the metric $d_{R} (s, t) = ∣ s - t ∣$ , and the family $L$ displayed above.

[L1]

$f$ is Lipschitz with constant $1$ when $∣ f (x) - f (x^{'}) ∣ \leq d (x, x^{'})$ for all $x, x^{'} \in X$ (Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction, 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, Absolute value in an ordered field).

[L2]

A family $F$ is uniformly equicontinuous when for every real $ε > 0$ there is a real $δ > 0$ with $∣ f (x) - f (x^{'}) ∣ < ε$ for every $f \in F$ and all $x, x^{'}$ with $d (x, x^{'}) < δ$ ; and $F$ is pointwise bounded when each set ${f (x) : f \in F}$ is bounded (Equicontinuity at a point, uniform equicontinuity, and pointwise boundedness of a family of maps between metric spaces, Uniform continuity of a map of metric spaces: one $δ$ serving every point, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

[L3]

For nonempty $A \subseteq X$ the function $x \mapsto d (x, A)$ is defined and satisfies $∣ d (x, A) - d (x^{'}, A) ∣ \leq d (x, x^{'})$ ( $∣ d (x, A) - d (y, A) ∣ \leq d (x, y)$ , so the distance to a fixed nonempty set is $1$ -Lipschitz, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

[L4]

A subset $S \subseteq R$ is bounded exactly when it lies in some ball of $R$ , so an unbounded set of reals lies in no ball (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, 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).

[L5]

A Lipschitz map is uniformly continuous and hence continuous (Contraction implies Lipschitz implies uniformly continuous implies continuous; every Hölder map is uniformly continuous, and a Lipschitz map on a bounded space is Hölder for every exponent, claims 2 and 3, Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not).

Verification

technique · direct

1.1

Let $ε > 0$ be real and put $δ : = ε$ ; for every $f \in L$ and all $x, x^{'} \in X$ with $d (x, x^{'}) < δ$ we get $∣ f (x) - f (x^{'}) ∣ \leq d (x, x^{'}) < ε$ .

L1L2

1.2

For nonempty $A \subseteq X$ the function $φ_{A}$ is defined at every point and satisfies $∣ φ_{A} (x) - φ_{A} (x^{'}) ∣ \leq d (x, x^{'})$ , so it is Lipschitz with constant $1$ and lies in $L$ ; this is claim 2.

L1L3

2.1

As $ε$ was arbitrary, step 1.1 is exactly uniform equicontinuity of $L$ , which is claim 1; in particular every member of $L$ is uniformly continuous and continuous.

step 1.1L2L5

3.1

Every constant function $c : X \to R$ satisfies $∣ c (x) - c (x^{'}) ∣ = 0 \leq d (x, x^{'})$ , so lies in $L$ ; hence for $X \neq \emptyset$ and any $x \in X$ the set ${f (x) : f \in L}$ contains every real and so lies in no ball of $R$ , and $L$ is not pointwise bounded, which is claim 3.

L1L2L4∎

Remarks

A common constant is what makes the family equicontinuous, not Lipschitzness of each member. Every member of $L$ is Lipschitz, but so is every member of ${x \mapsto ι (k) x : k \in N}$ on $R$ , and that family is not equicontinuous at any point: the constants grow without bound. Fixing the constant at $1$ is the hypothesis doing the work, exactly as the last remark of Equicontinuity at a point, uniform equicontinuity, and pointwise boundedness of a family of maps between metric spaces records.
Claim 2 is why equicontinuity is worth defining at all here. The distance functions $φ_{A}$ are the standard supply of Lipschitz maps in a metric space, and they are what an Ascoli-type argument on a later page will use; that they all sit in one uniformly equicontinuous family is the reason such arguments do not need any hypothesis on $A$ beyond nonemptiness.
Claim 3 is a warning about reading the two hypotheses as one. Pointwise boundedness is a condition on the values and equicontinuity a condition on the variation; a family may satisfy either without the other, and the theorem that uses both needs both.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

67 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Equicontinuity (Wikipedia) (standard reference, not scraped)
Lipschitz continuity (Wikipedia) (standard reference, not scraped)

The 11-Lipschitz maps of a metric space into RR form a uniformly equicontinuous family, and the distance functions x↦d(x,A)x↦d(x,A) all belong to it