ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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The $1$ -Lipschitz maps of a metric space into $\mathbb{R}$ 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 $\mathbb{R}$ carry its usual metric (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). Put

$\mathcal{L} \;:=\; \{\, f : X \to \mathbb{R} \;:\; f \text{ is Lipschitz with constant } 1 \,\} \;\subseteq\; \mathbb{R}^{X}$

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

$\mathcal{L}$ is uniformly equicontinuous (Equicontinuity at a point, uniform equicontinuity, and pointwise boundedness of a family of maps between metric spaces), with $\delta := \varepsilon$ serving at every $\varepsilon$ ;
for every nonempty $A \subseteq X$ the distance function $\varphi_A(x) := d(x,A)$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space) belongs to $\mathcal{L}$ ;
if $X \ne \varnothing$ then $\mathcal{L}$ 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 $\mathbb{R}$ with the metric $d_{\mathbb{R}}(s,t) = |s-t|$ , and the family $\mathcal{L}$ displayed above.

[L1]

$f$ is Lipschitz with constant $1$ when $|f(x) - f(x')| \le d(x,x')$ for all $x, x' \in X$ (Lipschitz map, $\alpha$ -Hölder map for rational $0 < \alpha \le 1$ , and contraction, 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, Absolute value in an ordered field).

[L2]

A family $\mathcal{F}$ is uniformly equicontinuous when for every real $\varepsilon > 0$ there is a real $\delta > 0$ with $|f(x) - f(x')| < \varepsilon$ for every $f \in \mathcal{F}$ and all $x,x'$ with $d(x,x') < \delta$ ; and $\mathcal{F}$ is pointwise bounded when each set $\{\, f(x) : f \in \mathcal{F} \,\}$ 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 $\delta$ 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 $|d(x,A) - d(x',A)| \le d(x,x')$ ( $|d(x,A) - d(y,A)| \le d(x,y)$ , 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 $S \subseteq \mathbb{R}$ is bounded exactly when it lies in some ball of $\mathbb{R}$ , 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 $\mathbb{R}$ 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).

[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 $\varepsilon$ - $\delta$ 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 $\varepsilon > 0$ be real and put $\delta := \varepsilon$ ; for every $f \in \mathcal{L}$ and all $x, x' \in X$ with $d(x,x') < \delta$ we get $|f(x) - f(x')| \le d(x,x') < \varepsilon$ .

L1L2

1.2

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

L1L3

2.1

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

step 1.1L2L5

3.1

Every constant function $c : X \to \mathbb{R}$ satisfies $|c(x)-c(x')| = 0 \le d(x,x')$ , so lies in $\mathcal{L}$ ; hence for $X \ne \varnothing$ and any $x \in X$ the set $\{\, f(x) : f \in \mathcal{L} \,\}$ contains every real and so lies in no ball of $\mathbb{R}$ , and $\mathcal{L}$ 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 $\mathcal{L}$ is Lipschitz, but so is every member of $\{\, x \mapsto \iota(k) x : k \in \mathbb{N} \,\}$ on $\mathbb{R}$ , 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 $\varphi_A$ 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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 123 results over 26 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

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

The 111-Lipschitz maps of a metric space into R\mathbb{R}R form a uniformly equicontinuous family, and the distance functions x↦d(x,A)x \mapsto d(x,A)x↦d(x,A) all belong to it