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ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)verified 2026-07-29 (claude-fable-5)
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Every closed subset of R is a zero set and a Gδ, as the perfect-normality criterion predicts

Example

R with its usual topology is metrizable (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, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not), hence perfectly normal by In a metric space every closed set is a zero set and a Gδ, and the distance function separates a point from a closed set, so every metrizable space is Tychonoff and perfectly normal: every closed C⊆R is a zero set (Zero sets and cozero sets of continuous real-valued functions) and a Gδ (Gδ and Fσ subsets of a topological space, agreeing with the real-line notion). Taking C:={0} makes both witnesses explicit: C=Z(f) for f(x):=∣x∣, and C=⋂n∈N(−1n+1, 1n+1).

This is exactly what Under dependent choice a space is perfectly normal if and only if it is normal and every closed set is a zero set predicts of a perfectly normal space, illustrated by the metric case that theorem's own proof does not need to run through, since perfect normality of R is already established directly from the metric.

Facts & Assumptions

Given: R with its usual topology, C={0}, and f(x)=∣x∣.

[L1]

Every closed subset of a metric space is a zero set and a Gδ: for C≠∅ closed, C=Z(x↦d(x,C)) and C=⋂n{x:d(x,C)<1/(n+1)} (In a metric space every closed set is a zero set and a Gδ, and the distance function separates a point from a closed set, so every metrizable space is Tychonoff and perfectly normal, clauses 1–2).

Verification

technique · direct
1.1

C={0} is closed and nonempty; by [L1] with d(x,C)=∣x∣ (step following [L2]), C=Z(f) with f(x)=∣x∣, and C=⋂n{x:∣x∣<1/(n+1)}.

givenL1L2
1.2

{x:∣x∣<1/(n+1)}=(−1n+1,1n+1) for every n∈N, directly unfolding the absolute-value inequality.

givenalgebra
2.1

By steps 1.1 and 1.2, {0}=Z(f) with f(x)=∣x∣, and {0}=⋂n(−1n+1,1n+1), exhibiting {0} as both a zero set and a Gδ.

step 1.1step 1.2∎

Remarks

  • No general closed subset of R is exceptional here. The argument above uses nothing about {0} beyond it being closed and nonempty in a metric space; the same two formulas, with d(x,C) in place of ∣x∣, exhibit any closed C⊆R as a zero set and a Gδ, choice-free.

  • This does not exercise the harder half of Under dependent choice a space is perfectly normal if and only if it is normal and every closed set is a zero set. That theorem's forward direction builds a zero set from a Gδ presentation via a countable family of Urysohn functions; here the zero set is read off directly from the metric, with no such construction and no dependent choice.

Depends on

Used by

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Dependency tree · two levels

66 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