ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)verified 2026-07-29 (claude-fable-5)

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.

The sets $U_0, U_1, U_{1/2}, U_{1/4}, U_{3/4}$ of the Urysohn construction computed for two disjoint closed subsets of $\mathbb{R}$

Example

Take $A := (-\infty,0]$ and $B := [1,\infty)$ in $\mathbb{R}$ , disjoint closed sets of the normal space $\mathbb{R}$ (In a metric space any two separated sets have disjoint open neighbourhoods, so every metrizable space is completely normal, Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length). For $r$ a dyadic rational of $[0,1]$ with $r<1$ (The dyadic rationals of $[0,1]$ , their finite levels $D_n$ , and their density in $[0,1]$ ) put

$U_r \;:=\; \Big(-\infty,\ \tfrac{1+r}{2}\Big), \qquad U_1 := \mathbb{R}.$

These are open, and satisfy $\overline{U_r} \subseteq U_s$ for every $r<s$ in $D$ and $U_1 = \mathbb{R}$ , so $(U_r)_{r \in D}$ is a legitimate instance of the family hypothesised in If $(U_r)_{r \in D}$ are open with $\overline{U_r} \subseteq U_s$ whenever $r < s$ and $U_1 = X$ , then $x \mapsto \inf\{ r \in D : x \in U_r \}$ is a continuous map $X \to [0,1]$ , and no choice principle is used and could arise from the construction inside 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 $A,B$ . In particular:

$U_0 = \big(-\infty,\ \tfrac12\big), \quad U_{1/4} = \big(-\infty,\ \tfrac58\big), \quad U_{1/2} = \big(-\infty,\ \tfrac34\big), \quad U_{3/4} = \big(-\infty,\ \tfrac78\big), \quad U_1 = \mathbb{R}.$

Facts & Assumptions

Given: $A = (-\infty,0]$ , $B=[1,\infty)$ in $\mathbb{R}$ , and $U_r := (-\infty, (1+r)/2)$ for dyadic $r<1$ , $U_1 := \mathbb{R}$ .

[L1]

$\overline{(-\infty,c)} = (-\infty,c]$ for real $c$ , and $(-\infty,c) \subseteq (-\infty,c')$ exactly when $c \le c'$ . The closure identity is two lines from the cited items: $(-\infty,c]$ is closed, since its complement $(c,\infty)$ contains an interval $(x - r, x + r)$ around each of its points $x$ (take $r = x - c$ ), which is the open-set criterion of 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 claim 3; and $c$ lies in the closure of $(-\infty,c)$ , since every interval $(c - r, c + r)$ with $r > 0$ meets it, so by A point lies in the closure of $A$ iff every basic neighbourhood of it meets $A$ ; the closure is the smallest closed superset and equals $A$ together with its derived set claims 1 and 2 the closure of $(-\infty,c)$ is $(-\infty,c]$ exactly. The inclusion clause is immediate from Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length and the order. (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, A point lies in the closure of $A$ iff every basic neighbourhood of it meets $A$ ; the closure is the smallest closed superset and equals $A$ together with its derived set, Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length)

[L2]

$A \subseteq U_0$ : every $x \le 0$ satisfies $x < 1/2$ .

Verification

technique · direct

1.1

For dyadic $r<s<1$ : $\overline{U_r} = \big(-\infty,\ (1+r)/2\big]$ by [L1], and $(1+r)/2 < (1+s)/2$ since $r<s$ , so $\overline{U_r} \subseteq \big(-\infty,\ (1+s)/2\big) = U_s$ .

givenL1algebra

1.2

For dyadic $r<1$ : $\overline{U_r} = \big(-\infty,(1+r)/2\big]$ , and $(1+r)/2 < 1$ since $r<1$ , so $\overline{U_r} \subseteq (-\infty,1) \subseteq \mathbb{R} = U_1$ .

givenL1algebra

1.3

$A \subseteq U_0$ by [L2]; and $B \subseteq \mathbb{R} \setminus U_r$ for every dyadic $r<1$ , since $(1+r)/2 < 1 \le x$ for $x \in B$ , so $x \notin U_r$ .

givenL2algebra

2.1

By step 1.1 and step 1.2, $\overline{U_r} \subseteq U_s$ for every $r<s$ in $D$ , and $U_1 = \mathbb{R}$ ; so $(U_r)_{r\in D}$ satisfies the hypotheses of If $(U_r)_{r \in D}$ are open with $\overline{U_r} \subseteq U_s$ whenever $r < s$ and $U_1 = X$ , then $x \mapsto \inf\{ r \in D : x \in U_r \}$ is a continuous map $X \to [0,1]$ , and no choice principle is used, and $f(x) := \inf(\{r\in D : x \in U_r\}\cup\{1\})$ is continuous $\mathbb{R} \to [0,1]$ . By step 1.3, $A \subseteq f^{-1}(\{0\})$ (every $r \ge 0$ works for $x\in A$ , so $f(x) \le 0$ , and $f \ge 0$ always) and $B \subseteq f^{-1}(\{1\})$ (no dyadic $r<1$ works for $x \in B$ ).

step 1.1step 1.2step 1.3∎

Remarks

The five requested sets are read off the general formula. $U_0, U_{1/4}, U_{1/2}, U_{3/4}$ are nested, each strictly inside the next by step 1.1, and $U_1 = \mathbb{R}$ is the point at which the family widens all at once, in line with the discussion in 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's own Remarks of why the recursion tracks $X \setminus B$ rather than $X$ until the very last step.
This is one legitimate family among many. 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 never claims uniqueness, and the family here is not literally the output of the dependent-choice recursion in that item's proof — it is a hand-picked family satisfying the same two hypotheses, chosen because its members have closed forms.

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: 122 results over 21 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

Urysohn's lemma (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §33 (standard reference, not scraped)

The sets U0,U1,U1/2,U1/4,U3/4U_0, U_1, U_{1/2}, U_{1/4}, U_{3/4}U0​,U1​,U1/2​,U1/4​,U3/4​ of the Urysohn construction computed for two disjoint closed subsets of R\mathbb{R}R