Alphabeta Math
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.

Every closed subset of R\mathbb{R} is a zero set and a GδG_\delta, as the perfect-normality criterion predicts

Example

R\mathbb{R} with its usual topology is metrizable (The absolute value makes R\mathbb{R} a metric space: d(x,y)=xyd(x,y) = |x-y| is a metric, its open balls are the intervals (xr,x+r)(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 GδG_\delta, and the distance function separates a point from a closed set, so every metrizable space is Tychonoff and perfectly normal: every closed CRC \subseteq \mathbb{R} is a zero set (Zero sets and cozero sets of continuous real-valued functions) and a GδG_\delta (GδG_\delta and FσF_\sigma subsets of a topological space, agreeing with the real-line notion). Taking C:={0}C := \{0\} makes both witnesses explicit: C=Z(f)C = Z(f) for f(x):=xf(x) := |x|, and C=nN(1n+1, 1n+1)C = \bigcap_{n \in \mathbb{N}} \big(-\tfrac{1}{n+1},\ \tfrac{1}{n+1}\big).

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\mathbb{R} is already established directly from the metric.

Facts & Assumptions

Given: R\mathbb{R} with its usual topology, C={0}C = \{0\}, and f(x)=xf(x)=|x|.

[L1]

Every closed subset of a metric space is a zero set and a GδG_\delta: for CC \ne \varnothing closed, C=Z(xd(x,C))C = Z(x \mapsto d(x,C)) and C=n{x:d(x,C)<1/(n+1)}C = \bigcap_n \{x : d(x,C) < 1/(n+1)\} (In a metric space every closed set is a zero set and a GδG_\delta, 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}C=\{0\} is closed and nonempty; by [L1] with d(x,C)=xd(x,C)=|x| (step following [L2]), C=Z(f)C = Z(f) with f(x)=xf(x)=|x|, and C=n{x:x<1/(n+1)}C = \bigcap_n \{x : |x| < 1/(n+1)\}.

givenL1L2
1.2

{x:x<1/(n+1)}=(1n+1,1n+1)\{x : |x| < 1/(n+1)\} = \big(-\tfrac{1}{n+1}, \tfrac{1}{n+1}\big) for every nNn \in \mathbb{N}, directly unfolding the absolute-value inequality.

givenalgebra
2.1

By steps 1.1 and 1.2, {0}=Z(f)\{0\} = Z(f) with f(x)=xf(x)=|x|, and {0}=n(1n+1,1n+1)\{0\} = \bigcap_n \big(-\tfrac{1}{n+1},\tfrac{1}{n+1}\big), exhibiting {0}\{0\} as both a zero set and a GδG_\delta.

step 1.1step 1.2

Remarks

  • No general closed subset of R\mathbb{R} is exceptional here. The argument above uses nothing about {0}\{0\} beyond it being closed and nonempty in a metric space; the same two formulas, with d(x,C)d(x,C) in place of x|x|, exhibit any closed CRC \subseteq \mathbb{R} as a zero set and a GδG_\delta, 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 GδG_\delta 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

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 135 results over 18 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