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

How statement and proof provenance work

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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A uniformly continuous map from a dense subspace into a complete metric space extends uniquely to a uniformly continuous map on the whole space

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $(X, d_{X})$ be a metric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric), let $A \subseteq X$ be dense in $X$ (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space) and carry the subspace metric (Isometry, isometric embedding, and the subspace metric on a subset), let $(Y, d_{Y})$ be a complete metric space (Complete metric space: every Cauchy sequence converges in the space), and let $f : A \to Y$ be uniformly continuous (Uniform continuity of a map of metric spaces: one $δ$ serving every point). Then:

There is a uniformly continuous $g : X \to Y$ with $g (a) = f (a)$ for every $a \in A$ .
$g$ is the only continuous map $X \to Y$ extending $f$ (Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form).

The map $g$ is constructed explicitly below, as the unique point common to the closures of the images of the shrinking balls around $x$ ; no value of $g$ is selected, each is determined.

Facts & Assumptions

Given: The Axiom of Countable Choice; a metric space $(X, d_{X})$ , a dense $A \subseteq X$ , a complete metric space $(Y, d_{Y})$ , a uniformly continuous $f : A \to Y$ , and a real $ε > 0$ . For $x \in X$ and $n \in N$ write $U_{n} (x) : = B_{X} (x, 1 / (n + 1)) \cap A$ , $S_{n} (x) : = f [U_{n} (x)]$ and $T_{n} (x) : = \overline{S_{n} (x)}$ , the closure taken in $Y$ .

[A1]

Density: $\overline{A} = X$ , so $B_{X} (x, r) \cap A \neq \emptyset$ for every $x \in X$ and every real $r > 0$ (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space, Open ball, closed ball and sphere in a metric space).

[A2]

Uniform continuity of $f$ : for every real $ε > 0$ there is a real $δ > 0$ with $d_{Y} (f (a), f (a^{'})) < ε$ for all $a, a^{'} \in A$ with $d_{X} (a, a^{'}) < δ$ ; distances inside $A$ are those of $X$ (Uniform continuity of a map of metric spaces: one $δ$ serving every point, Isometry, isometric embedding, and the subspace metric on a subset).

[A3]

Completeness of $(Y, d_{Y})$ (Complete metric space: every Cauchy sequence converges in the space).

[L1]

Cantor's intersection theorem in a complete space: a sequence of nonempty closed bounded sets, nested and with diameters tending to $0$ , has exactly one common point (In a complete metric space nested nonempty closed sets whose diameters tend to $0$ meet in exactly one point, and this property characterises completeness).

[L2]

Closure by adherent points: $u \in \overline{S}$ means every ball around $u$ meets $S$ ; $S \subseteq \overline{S}$ ; $\overline{S}$ is closed and is the smallest closed superset of $S$ (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space, The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

[L3]

Diameter: for nonempty bounded $S$ , $diam (S) = \sup {d (u, v) : u, v \in S}$ , so any upper bound of those distances dominates the diameter; a nonempty set all of whose pairwise distances are below a real $β$ lies in a ball of radius $β + 1$ around any of its points, hence is bounded (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Complete ordered field (least-upper-bound property), Open ball, closed ball and sphere in a metric space).

[L4]

Reciprocals of naturals: $1 / (n + 1)$ is a positive real, decreasing in $n$ , and below every positive real from some index on (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Inverses of positives are positive, and reciprocation reverses order).

[L5]

A point lies in the closure of $A$ exactly when some sequence in $A$ converges to it; 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), limits in a metric space are unique (A sequence in a metric space has at most one limit), and a uniformly continuous map is 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).

[L7]

Triangle inequality (M3) and symmetry (M2) of a metric (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

Proof

technique · constructive

1.1

For every $x \in X$ and $n \in N$ the set $U_{n} (x)$ is nonempty by [A1], so $S_{n} (x)$ is nonempty and $T_{n} (x)$ is a nonempty closed subset of $Y$ .

A1L2construct

1.2

The radii decrease, so $U_{n + 1} (x) \subseteq U_{n} (x)$ and $S_{n + 1} (x) \subseteq S_{n} (x)$ ; since $T_{n} (x)$ is a closed superset of $S_{n + 1} (x)$ , minimality of the closure gives $T_{n + 1} (x) \subseteq T_{n} (x)$ .

L2L4

1.3

Fix a real $ε > 0$ , let $δ > 0$ be as in [A2] for $ε / 3$ , and let $N$ be a natural with $2 / (N + 1) < δ$ ; note that $N$ depends on $ε$ alone and not on $x$ .

A2L4choose

1.4

Towards uniform continuity, let $ε > 0$ be real, let $δ > 0$ be as in [A2] for $ε / 3$ , and put $δ^{'} : = δ / 3 > 0$ . Fix a natural $m$ with $1 / (m + 1) < δ / 3$ .

A2L4choose

1.5

For claim 2, let $h : X \to Y$ be continuous with $h (a) = f (a)$ for all $a \in A$ , and let $x \in X$ . Since $\overline{A} = X$ there is a sequence $(a_{k})$ in $A$ with $a_{k} \to x$ in $X$ .

A1L5

2.1

Let $n \geq N$ and $a, a^{'} \in U_{n} (x)$ . Then $d_{X} (a, a^{'}) \leq d_{X} (a, x) + d_{X} (x, a^{'}) < 2 / (n + 1) \leq 2 / (N + 1) < δ$ , so $d_{Y} (f (a), f (a^{'})) < ε / 3$ . Hence all pairwise distances in $S_{n} (x)$ are below $ε / 3$ , so $S_{n} (x)$ is bounded and $diam (S_{n} (x)) \leq ε / 3$ .

step 1.3A2L3L4L7

3.1

Let $n \geq N$ , let $u, v \in T_{n} (x)$ and let $η > 0$ be real. The balls $B_{Y} (u, η)$ and $B_{Y} (v, η)$ meet $S_{n} (x)$ , so there are $s, s^{'} \in S_{n} (x)$ with $d_{Y} (u, s) < η$ and $d_{Y} (v, s^{'}) < η$ , whence $d_{Y} (u, v) \leq d_{Y} (u, s) + d_{Y} (s, s^{'}) + d_{Y} (s^{'}, v) < ε / 3 + 2 η$ . As $η > 0$ was arbitrary, $d_{Y} (u, v) \leq ε / 3$ : were $d_{Y} (u, v) > ε / 3$ , the value $η : = (d_{Y} (u, v) - ε / 3) / 3$ would be positive and would give $d_{Y} (u, v) < d_{Y} (u, v)$ .

step 2.1L2L3L7

4.1

So for $n \geq N$ the set $T_{n} (x)$ is nonempty, closed and bounded with $diam (T_{n} (x)) \leq ε / 3 < ε$ .

step 1.1step 3.1L3

5.1

Apply steps 1.3 to 4.1 with $ε = 1$ to get a natural $N_{1}$ such that $T_{n} (x)$ is nonempty, closed and bounded for every $n \geq N_{1}$ and every $x \in X$ . Then $(T_{N_{1} + j} (x))_{j \in N}$ is nested by step 1.2, and its diameters tend to $0$ : given a real $ε > 0$ , the $N$ of step 1.3 satisfies $diam (T_{N_{1} + j} (x)) < ε$ for every $j \geq N$ , since then $N_{1} + j \geq N$ .

step 1.2step 4.1

6.1

By [L1] and [A3] the intersection $⋂_{j \in N} T_{N_{1} + j} (x)$ has exactly one element; and because the family $(T_{n} (x))_{n}$ is nested this intersection equals $⋂_{n \in N} T_{n} (x)$ , a set defined without reference to $N_{1}$ . Define $g (x)$ to be its unique element; this determines a function $g : X \to Y$ , and no choice is made, since the value is unique.

step 1.2step 5.1A3L1construct

7.1

$g$ extends $f$ : for $a \in A$ and every $n$ we have $a \in U_{n} (a)$ , so $f (a) \in S_{n} (a) \subseteq T_{n} (a)$ ; hence $f (a) \in ⋂_{n} T_{n} (a)$ , and by uniqueness $g (a) = f (a)$ .

step 6.1L2

7.2

Let $x, x^{'} \in X$ with $d_{X} (x, x^{'}) < δ^{'}$ . Since $g (x) \in T_{m} (x) = \overline{S_{m} (x)}$ , the ball $B_{Y} (g (x), ε / 3)$ meets $S_{m} (x)$ , so there is $a \in U_{m} (x)$ with $d_{Y} (g (x), f (a)) < ε / 3$ ; likewise there is $a^{'} \in U_{m} (x^{'})$ with $d_{Y} (g (x^{'}), f (a^{'})) < ε / 3$ .

step 6.1step 1.4L2

8.1

Then $d_{X} (a, a^{'}) \leq d_{X} (a, x) + d_{X} (x, x^{'}) + d_{X} (x^{'}, a^{'}) < δ / 3 + δ / 3 + δ / 3 = δ$ , so $d_{Y} (f (a), f (a^{'})) < ε / 3$ , and therefore $d_{Y} (g (x), g (x^{'})) \leq d_{Y} (g (x), f (a)) + d_{Y} (f (a), f (a^{'})) + d_{Y} (f (a^{'}), g (x^{'})) < ε$ .

step 1.4step 7.2A2L7

9.1

The real $δ^{'}$ depended on $ε$ alone, so $g$ is uniformly continuous; together with step 7.1 this establishes claim 1.

step 7.1step 1.4step 8.1

10.1

The map $g$ is continuous, being uniformly continuous, so $g (a_{k}) \to g (x)$ and $h (a_{k}) \to h (x)$ ; but $g (a_{k}) = f (a_{k}) = h (a_{k})$ for every $k$ , so one sequence in $Y$ converges to both $g (x)$ and $h (x)$ , whence $g (x) = h (x)$ by uniqueness of limits. As $x$ was arbitrary, $h = g$ .

step 7.1step 9.1step 1.5L6

11.1

The map $g$ of step 6.1 is a uniformly continuous extension of $f$ and is the only continuous one, which is claims 1 and 2.

step 9.1step 10.1discharge-construct∎

Remarks

Why the construction avoids the Axiom of Choice, and where choice reappears. The obvious construction sets $g (x) : = \lim_{k} f (a_{k})$ for a sequence $(a_{k})$ in $A$ converging to $x$ . That defines $g$ only after a sequence has been selected at every point of $X$ at once, which is a choice over a set that need not be countable. The construction above never selects: $g (x)$ is defined as the unique element of a set built from $x$ by a formula. Choice does appear, twice, and both times only inside a proof: ${A C}_{ω}$ is spent by In a complete metric space nested nonempty closed sets whose diameters tend to $0$ meet in exactly one point, and this property characterises completeness in step 6.1, and again by 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 in step 1.5, which produces one sequence for one point at a time.
Completeness of the target is what makes the intersection nonempty, and it cannot be weakened. Without it the shrinking closed sets $T_{n} (x)$ can have empty intersection, and there is then nothing to define $g (x)$ to be; the inclusion $Q ↪ Q$ read as a uniformly continuous map from the dense subspace $Q$ of $R$ into $Q$ has no continuous extension to $R$ for exactly that reason.
Uniform continuity of $f$ is what makes the diameters shrink, and ordinary continuity does not suffice: step 1.3 chooses one $δ$ before any point $x$ is fixed, and step 5.1 needs that same $δ$ at every $x$ simultaneously. This is the same pressure point as in A uniformly continuous map sends Cauchy sequences to Cauchy sequences.
The extension inherits the modulus, not the constants. The proof produces $δ^{'} = δ / 3$ from the $δ$ that $f$ supplies for $ε / 3$ , so a Lipschitz $f$ extends to a uniformly continuous $g$ ; that $g$ is in fact Lipschitz with the same constant is a separate argument, carried out for a concrete case in A Lipschitz function on $Q$ extends uniquely to a Lipschitz function on $R$ with the same constant ↗.

Depends on

Used by

Dependency tree · two levels

63 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

Uniform continuity (Wikipedia) (standard reference, not scraped)
Continuous linear extension (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 4 (standard reference, not scraped)