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

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A Lipschitz function on $Q$ extends uniquely to a Lipschitz function on $R$ with the same constant

Example

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Regard $Q$ as a subspace of $R$ with the metric inherited from the usual metric (The rationals embed densely in the reals, 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, Isometry, isometric embedding, and the subspace metric on a subset), let $L \in R$ with $L \geq 0$ , and let $f : Q \to R$ be Lipschitz with constant $L$ (Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction), that is

$∣ f (p) - f (q) ∣ \leq L ∣ p - q ∣ for all p, q \in Q .$

Then there is exactly one continuous $F : R \to R$ with $F (q) = f (q)$ for every $q \in Q$ , and that $F$ is again Lipschitz with the same constant $L$ .

Facts & Assumptions

Given: The Axiom of Countable Choice; $Q$ as a metric subspace of $R$ ; a real $L \geq 0$ ; a Lipschitz $f : Q \to R$ with constant $L$ ; reals $x, y$ .

[A1]

Lipschitz hypothesis: $∣ f (p) - f (q) ∣ \leq L ∣ p - q ∣$ for all $p, q \in Q$ (Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction).

[L1]

A Lipschitz map is uniformly 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, Uniform continuity of a map of metric spaces: one $δ$ serving every point).

[L2]

$Q$ is dense in $R$ : every ball of $R$ contains a rational (The rationals embed densely in the reals, Interior, closure, boundary, limit point, isolated point and dense subset of a metric space, Open ball, closed ball and sphere in a metric space).

[L3]

$R$ with the usual metric is complete ( $R$ and $R^{n}$ for $n \geq 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $R$ , Complete metric space: every Cauchy sequence converges in the 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, Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L4]

Extension from a dense subspace into a complete space: a uniformly continuous map extends to a uniformly continuous map, and that extension is the only continuous one (A uniformly continuous map from a dense subspace into a complete metric space extends uniquely to a uniformly continuous map on the whole space).

[L5]

A point of the closure is the limit of a sequence from the set; this direction spends ${A C}_{ω}$ (A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed, The Axiom of Countable Choice ( ${A C}_{ω}$ ), Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ).

[L6]

A continuous map is sequentially continuous (Metric continuity characterisations, with countable choice for the sequential converse, Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form).

[L7]

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

[L8]

$∣ ∣ a ∣ - ∣ b ∣ ∣ \leq ∣ a - b ∣$ for reals, which is the reverse triangle inequality of the usual metric of $R$ with third point $0$ (The reverse triangle inequality $∣ d (x, z) - d (y, z) ∣ \leq d (x, y)$ in any metric space, Basic properties of the absolute value).

Verification

technique · direct

1.1A1L1

By [A1] and [L1] the map $f$ is uniformly continuous on the subspace $Q$ of $R$ .

1.2L2L5

Let $x, y \in R$ . By density and [L5] there are sequences $(p_{k})$ and $(q_{k})$ of rationals with $p_{k} \to x$ and $q_{k} \to y$ .

2.1step 1.1L2L3L4

$Q$ is dense in $R$ and $R$ is complete, so [L4] supplies a uniformly continuous $F : R \to R$ with $F (q) = f (q)$ for every rational $q$ , and $F$ is the only continuous map $R \to R$ with that property.

2.2step 1.2L7L8

Likewise $p_{k} - q_{k} \to x - y$ , so $∣ p_{k} - q_{k} ∣ \to ∣ x - y ∣$ and $L ∣ p_{k} - q_{k} ∣ \to L ∣ x - y ∣$ .

3.1step 2.1step 1.2L6L7L8

$F$ is continuous, being uniformly continuous, so $F (p_{k}) \to F (x)$ and $F (q_{k}) \to F (y)$ ; hence $F (p_{k}) - F (q_{k}) \to F (x) - F (y)$ and, by [L8], $∣ F (p_{k}) - F (q_{k}) ∣ \to ∣ F (x) - F (y) ∣$ .

3.2step 2.1step 1.2A1

For every $k$ the terms $p_{k}, q_{k}$ are rational and $F$ agrees with $f$ on them, so $∣ F (p_{k}) - F (q_{k}) ∣ = ∣ f (p_{k}) - f (q_{k}) ∣ \leq L ∣ p_{k} - q_{k} ∣$ by [A1].

4.1step 3.1step 2.2step 3.2L7

Passing to the limit in step 3.2, using steps 3.1 and 2.2, gives $∣ F (x) - F (y) ∣ \leq L ∣ x - y ∣$ ; as $x, y$ were arbitrary reals, $F$ is Lipschitz with constant $L$ .

5.1step 2.1step 4.1∎

So $F$ exists, is the unique continuous extension of $f$ , and is Lipschitz with the same constant $L$ .

Remarks

The extension theorem gives uniform continuity; the constant is recovered separately. A uniformly continuous map from a dense subspace into a complete metric space extends uniquely to a uniformly continuous map on the whole space transports a modulus of continuity, not a Lipschitz constant, so step 4.1 is genuinely needed. The mechanism is generic: a non-strict inequality that holds on a dense set and whose two sides are continuous holds everywhere, by Limits preserve non-strict inequalities.
Nothing here is special to $Q$ and $R$ . The same argument extends a Lipschitz map from any dense subspace of any metric space into any complete metric space, with the same constant; $Q \subseteq R$ is used only because it is the density statement this library already has (The rationals embed densely in the reals).
Uniqueness is what makes "the" extension meaningful, and it needs only continuity, not the Lipschitz condition: two continuous maps agreeing on a dense set agree everywhere (A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed, Metric continuity characterisations, with countable choice for the sequential converse).
The hypothesis cannot be relaxed to continuity. A continuous $f : Q \to R$ need not extend continuously to $R$ at all, and the reason is the same one that defeats every argument of this kind: continuity does not control a function along a Cauchy sequence (A uniformly continuous map sends Cauchy sequences to Cauchy sequences, $x \mapsto 1 / x$ is continuous on $(0, 1)$ and sends the Cauchy sequence $(1 / (k + 2))_{k \geq 0}$ to an unbounded one).

A Lipschitz function on $Q$ extends uniquely to a Lipschitz function on $R$ with the same constant

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources

A Lipschitz function on QQ extends uniquely to a Lipschitz function on RR with the same constant

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources

A Lipschitz function on $Q$ extends uniquely to a Lipschitz function on $R$ with the same constant