ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-09 (gpt-5.6-terra-codex-subscription)

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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In a metric space the function $d (x, A) / (d (x, A) + d (x, B))$ separates two disjoint closed sets outright, so the metric case spends no choice principle

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) with its metric topology (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), metrizable (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not) and hence normal (In a metric space any two separated sets have disjoint open neighbourhoods, so every metrizable space is completely normal), and let $A, B \subseteq X$ be disjoint, nonempty and closed. Define $h : X \to [0, 1]$ by

$h (x) : = \frac{d (x, A)}{d (x, A) + d (x, B)} .$

Then $h$ is a witness for Urysohn's lemma, under the axiom of dependent choice: in a normal space two disjoint closed sets are separated by a continuous function into $[0, 1]$ , and conversely such a space is normal applied to $A$ and $B$ , and it is written down by a single formula: no choice principle, dependent or otherwise, is spent in producing it, in contrast with the general construction inside that theorem.

Facts & Assumptions

Given: A metric space $(X, d)$ and disjoint, nonempty, closed $A, B \subseteq X$ .

[L1]

For nonempty $S \subseteq X$ , $d (\cdot, S)$ is $1$ -Lipschitz, hence continuous; $d (x, S) \geq 0$ ; and $\overline{S} = {x : d (x, S) = 0}$ ( $∣ d (x, A) - d (y, A) ∣ \leq d (x, y)$ , 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; 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, claim 1). In particular, when $S$ is closed, $d (x, S) = 0$ exactly when $x \in S$ .

Verification

technique · direct

1.1

For every $x \in X$ : $d (x, A) \geq 0$ and $d (x, B) \geq 0$ by [L1], and they are not both $0$ , since $d (x, A) = d (x, B) = 0$ would give $x \in A \cap B = \emptyset$ by [L1] (A, B closed); so $d (x, A) + d (x, B) > 0$ and $h (x)$ is a well-defined real number.

givenL1algebra

2.1

$0 \leq h (x) \leq 1$ for every $x$ , since $0 \leq d (x, A) \leq d (x, A) + d (x, B)$ by step 1.1.

step 1.1algebra

2.2

$h$ is continuous: it is the quotient of the continuous functions $d (\cdot, A)$ and $d (\cdot, A) + d (\cdot, B)$ (both continuous by [L1], the second a sum of continuous functions), and the denominator is nowhere $0$ by step 1.1.

step 1.1L1

2.3

For $x \in A$ : $d (x, A) = 0$ by [L1], so $h (x) = 0 / (0 + d (x, B)) = 0$ . For $x \in B$ : $d (x, B) = 0$ , and $d (x, A) \neq 0$ by step 1.1, so $h (x) = d (x, A) / (d (x, A) + 0) = 1$ .

step 1.1L1algebra

3.1

By steps 2.1, 2.2 and 2.3, $h : X \to [0, 1]$ is continuous with $A \subseteq h^{- 1} ({0})$ and $B \subseteq h^{- 1} ({1})$ , exactly the conclusion of Urysohn's lemma, under the axiom of dependent choice: in a normal space two disjoint closed sets are separated by a continuous function into $[0, 1]$ , and conversely such a space is normal for the pair $A, B$ .

step 2.1step 2.2step 2.3∎

Remarks

Nothing is selected. Every value $h (x)$ is computed from $d (x, A)$ and $d (x, B)$ , both determined by $x$ , $A$ , $B$ and $d$ alone; no step above fixes a witness from a nonempty set of alternatives. This is the same observation In a metric space any two separated sets have disjoint open neighbourhoods, so every metrizable space is completely normal and In a metric space every closed set is a zero set and a $G_{δ}$ , and the distance function separates a point from a closed set, so every metrizable space is Tychonoff and perfectly normal make about their own constructions, and it is why the metric case of every separation theorem on this page needs no choice hypothesis at all.
The nonemptiness of $A$ and $B$ is not decoration. If $A = \emptyset$ then $d (x, A)$ is undefined for every $x$ ( $∣ d (x, A) - d (y, A) ∣ \leq d (x, y)$ , so the distance to a fixed nonempty set is $1$ -Lipschitz presupposes a nonempty set), and the constant function $1$ serves instead; the formula above is written for the case that matters, where both sets carry a point to measure distance from.

Depends on

Used by

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Sources

Urysohn's lemma (Wikipedia) (standard reference, not scraped)
Gabriel Nagy, Real Analysis, Lecture 6: Metric spaces (standard reference, not scraped)

In a metric space the function d(x,A)/(d(x,A)+d(x,B))d(x,A)/(d(x,A)+d(x,B)) separates two disjoint closed sets outright, so the metric case spends no choice principle