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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.

8 results · all verified · 0 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 8 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Metrization: Urysohn, Nagata–Smirnov, Bing, Smirnov

1 · Prerequisites

2 · Summary

Metrizability is controlled here by families of open sets: locally finite and discrete decompositions of a basis, and normal sequences whose stars shrink around points. The development uses the library’s convention that regularity does not include T1, so the separation axiom is always stated separately. Choice is a sufficient hypothesis for the cover and well-ordering constructions used in the proofs.

The compatible-normal-sequence construction produces a metric, while metric spaces supply the two sigma-base forms. These implications yield the Nagata–Smirnov and Bing criteria, with the second-countable Urysohn theorem as a corollary. A locally finite merger of local bases then gives the paracompact Hausdorff local-metrization criterion of Smirnov.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-08-02Open item page →

Discrete families and σ-locally-finite and σ-discrete bases

Definition

Let X be a topological space. A family D of subsets of X is discrete if every xX has a neighbourhood meeting at most one member of D. It is therefore a locally finite family in the sense of Refinements, locally finite families, point-finite families, and star refinements.

An open basis B (Basis and subbasis for a topology, and the topology generated by a family of sets) is σ-locally finite if B=nNBn for locally finite families Bn, and σ-discrete if the families Bn can be taken discrete. Empty layers are permitted; the word σ records the countable indexing convention of Finite, countably infinite, countable, uncountable.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02Open item page →

Every discrete family is locally finite, so every σ-discrete basis is σ-locally finite

Statement

Every discrete family is locally finite. Consequently every σ-discrete basis is σ-locally finite.

Facts & Assumptions

Given: A discrete family D in a space X.

[L1]

A family is locally finite when every point has a neighbourhood meeting only finitely many of its members (Refinements, locally finite families, point-finite families, and star refinements).

Proof

technique · direct
1.1

For each xX, discreteness supplies a neighbourhood meeting at most one member of D. That is a finite number, so [L1] makes D locally finite.

L1
2.1

Applying step 1.1 separately to every discrete layer in a σ-discrete decomposition leaves the same countable decomposition and gives a σ-locally-finite basis.

step 1.1
DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-08-02Open item page →

A locally metrizable space: every point has a metrizable open neighbourhood

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-08-02Open item page →

Compatible normal sequences of open covers

Definition

For an open cover U and xX, write St(x,U)=St({x},U), using the star of Refinements, locally finite families, point-finite families, and star refinements. A sequence (Un)nN of open covers is normal if Un+1 star-refines Un for every n.

It is compatible with the topology if (i) for xy some n has ySt(x,Un), and (ii) for every open neighbourhood O of x (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open) some n has St(x,Un)O. The indexing starts at 0; an empty space has the legal empty cover at every index.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02Open item page →

A T1 space with a compatible normal sequence of open covers is metrizable

Statement

If a T1 space has a compatible normal sequence of open covers, then it is metrizable.

Facts & Assumptions

Given: A T1 space X and a compatible normal sequence (Un)nN.

Proof

technique · constructive
1.1

Put Vn={(x,y):ySt(x,Un)}. Each Vn is symmetric, contains the diagonal, and normality gives Vn+1Vn+1Vn: two successive Un+1-links lie in the star of one member, which is contained in a member of Un.

givenconstruct
2.1

Define d(x,y) as the smaller of 1 and the infimum of r=1k2nr over finite chains x=x0,,xk=y with (xr1,xr)Vnr, taking the infimum of an empty collection to be +. Reversing a chain gives symmetry. Concatenation gives the triangle inequality within a chain-connected component, while points in different components have distance 1; truncation at 1 preserves the triangle inequality. The diagonal chains give d(x,x)=0.

step 1.1construct
3.1

The containment in step 1.1 lets every chain of total weight below 2n1 be compressed, from its finest links upward, to a Vn-link. Hence d(x,y)<2n1 implies (x,y)Vn; conversely (x,y)Vn gives d(x,y)2n.

step 1.1step 2.1
4.1

Compatibility (i) and step 3.1 show d(x,y)>0 when xy. Compatibility (ii), the two bounds in step 3.1, and [L1] show that the d-balls and the original neighbourhoods contain one another at every point. Thus d is a metric inducing the given topology.

L1step 3.1discharge-construct
LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02Open item page →

Under choice, every metric space has a σ-locally-finite basis

Statement

Assume the Axiom of Choice. Every metric space has a σ-locally-finite open basis.

Facts & Assumptions

Given: A metric space (X,d) and the Axiom of Choice.

[L1]

Under choice every metric space is paracompact, so every open cover has a locally finite open refining cover (Stone's theorem, under choice: every metric space is paracompact).

Proof

technique · direct
1.1

For each nN, let Cn be the cover by balls of radius 2n3. By [L1], choose a locally finite open refining cover Vn of Cn.

L1choose
2.1

The family B=nVn is σ-locally finite. It is a basis: if xO with O open, [L2] gives ε>0 with Bd(x,ε)O; choose n with 2n2<ε, and a member VVn containing x. As V lies in some Bd(c,2n3) containing x, the triangle inequality gives VBd(x,2n2)O.

L2step 1.1
3.1

Thus B is the asserted σ-locally-finite basis.

step 2.1
RemarkRemark: AI-adaptedProof: Not applicable sources checked 2026-08-01 not proved hereOpen item page →
Recorded, not proved here. This statement is included so the library can refer to it honestly, with a citation to the literature. It is not proved anywhere in this library: the track that would prove it has not been developed here yet.

Under choice, a regular T1 space with a σ-locally-finite basis has a compatible normal sequence

Statement

Assume the Axiom of Choice. A regular T1 space with a σ-locally-finite basis has a compatible normal sequence of open covers.

Source convention. In Granath's source, regularity is defined only for a Fréchet space, i.e. a T1 space. The displayed library statement therefore names T1 separately rather than silently importing that convention.

Not proved in this library. This is a source-backed fallback rather than a local proof. The discarded local route chose a shrinking W(B,x) for every point of every basis member and claimed that its families stayed locally finite. That claim is false: with X=B=R and the one-member locally finite family {B}, the allowed shrinkings W(B,x)=(x1,x+1) have infinite local overlap at every point. The standard normal-cover construction needs additional machinery beyond that failed pointwise shrinking.

Why it remains visible. The Nagata--Smirnov comparison below depends on exactly this standard route. Its dependency marker therefore records that the result is externally sourced rather than pretending that the invalid local construction proves it.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02 rests on unproved materialOpen item page →
Rests on 1 statement not proved in this library. Every dependency marked below is recorded with a citation but is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Under choice, a space is metrizable if and only if it is regular, T1, and has a σ-locally-finite basis

Statement

Assume the Axiom of Choice. A space is metrizable if and only if it is regular, T1, and has a σ-locally-finite basis. Here regularity and T1 are separate hypotheses.

Facts & Assumptions

Given: The Axiom of Choice and a topological space X.

[L1]

Under choice every metric space has a σ-locally-finite basis (Under choice, every metric space has a σ-locally-finite basis).

[L2]

Recorded external fallback; not proved here. A regular T1 space with such a basis has a compatible normal sequence (Under choice, a regular T1 space with a σ-locally-finite basis has a compatible normal sequence ); a T1 space with such a sequence is metrizable (A T1 space with a compatible normal sequence of open covers is metrizable).

Proof

technique · cases
1.1

If X is metrizable, it is regular and T1, and [L1] gives the required basis.

assume-case forwardL1
1.2

If X is regular, T1, and has a σ-locally-finite basis, [L2] first gives a compatible normal sequence and then a compatible metric.

assume-case reverse L2
2.1

The two cases prove the two directions of the equivalence.

step 1.1step 1.2cases-exhaustive
RemarkRemark: AI-adaptedProof: Not applicable sources checked 2026-08-01 not proved hereOpen item page →
Recorded, not proved here. This statement is included so the library can refer to it honestly, with a citation to the literature. It is not proved anywhere in this library: the track that would prove it has not been developed here yet.

Under choice, every metric space has a σ-discrete basis

Statement

Assume the Axiom of Choice. Every metric space has a σ-discrete open basis.

Not proved in this library. This is a source-backed fallback rather than a local proof. The discarded construction assigns every point to a first centre and then asserts that the resulting cells are open. They need not be: in R, with integer centres at scale 1 ordered 0,1,1,2,, the cell assigned to 1 is [1,2). Intersecting it with a positive-distance condition does not repair that failure.

Why it remains visible. Bing's standard necessity direction needs a real discrete-open refinement argument. The external-dependency marker is retained on this result and its consequences so that none of them is mistaken for a locally established theorem.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02 rests on unproved materialOpen item page →
Rests on 1 statement not proved in this library. Every dependency marked below is recorded with a citation but is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Under choice, a space is metrizable if and only if it is regular, T1, and has a σ-discrete basis

Statement

Assume the Axiom of Choice. A space is metrizable if and only if it is regular, T1, and has a σ-discrete basis.

Facts & Assumptions

Given: The Axiom of Choice and a topological space X.

[L1]

Recorded external fallback; not proved here. Under choice every metric space has a σ-discrete basis (Under choice, every metric space has a σ-discrete basis ).

[L2]

Proof

technique · cases
1.1

If X is metrizable, [L1] gives a σ-discrete basis; metric spaces are regular and T1.

assume-case forward L1
1.2

If X is regular, T1, and has a σ-discrete basis, [L2] converts that basis to a σ-locally-finite one and then gives a metric.

assume-case reverseL2
2.1

The cases establish both directions.

step 1.1step 1.2cases-exhaustive
CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02 rests on unproved material (inherited)Open item page →
Rests on 1 statement not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are Under choice, a regular T₁ space with a σ-locally-finite basis has a compatible normal sequence. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Under choice, every regular T1 second-countable space is metrizable

Statement

Assume the Axiom of Choice. Every regular T1 second-countable space is metrizable.

Facts & Assumptions

Given: A regular T1 space with a countable basis (Bn)nN, and the Axiom of Choice.

[L1]

Nagata–Smirnov metrizes a regular T1 space with a σ-locally-finite basis (Under choice, a space is metrizable if and only if it is regular, T1, and has a σ-locally-finite basis).

Proof

technique · direct
1.1

Each singleton family {Bn} is locally finite, including when Bn=. Hence the displayed basis is σ-locally finite.

given
2.1

Apply [L1] to step 1.1 and the given regular and T1 hypotheses.

L1step 1.1
LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02Open item page →

A locally finite open cover by subspaces with σ-locally-finite bases yields a σ-locally-finite basis of the whole space

Statement

Let U be a locally finite open cover of X. If every UU, with its subspace topology, has a σ-locally-finite open basis nBU,n, then X has a σ-locally-finite open basis.

Facts & Assumptions

Given: A locally finite open cover U and the stated relative bases.

[L1]

A locally finite family has a neighbourhood at each point meeting only finitely many members (Refinements, locally finite families, point-finite families, and star refinements).

Proof

technique · direct
1.1

Since every UU is open in X, every member of a relative open basis BU,n is open in X by [L2]. Put Bn=UUBU,n.

L2construct
2.1

The family Bn is locally finite. At x, take from [L1] a neighbourhood meeting only finitely many U; within each of those finitely many U, local finiteness of BU,n supplies a neighbourhood meeting finitely many members, and their finite intersection meets only finitely many members of Bn.

L1step 1.1
2.2

If O is open and xO, choose UU containing x and then a member of the basis of U containing x and contained in OU. Thus nBn is a basis of X.

step 1.1
3.1

Steps 2.1 and 2.2 prove the result.

step 2.1step 2.2
TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02 rests on unproved material (inherited)Open item page →
Rests on 1 statement not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are Under choice, a regular T₁ space with a σ-locally-finite basis has a compatible normal sequence. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Under choice, a space is metrizable if and only if it is paracompact, Hausdorff, and locally metrizable

Statement

Assume the Axiom of Choice. A space is metrizable if and only if it is paracompact, Hausdorff, and locally metrizable.

Facts & Assumptions

Given: The Axiom of Choice and a topological space X.

[L1]

Under choice, every metric space is paracompact and has a σ-locally-finite basis (Stone's theorem, under choice: every metric space is paracompact, Under choice, every metric space has a σ-locally-finite basis).

[L2]

A paracompact Hausdorff space is regular, and Nagata–Smirnov applies to a regular T1 space with a σ-locally-finite basis (Every paracompact Hausdorff space is regular, Under choice, a space is metrizable if and only if it is regular, T1, and has a σ-locally-finite basis).

Proof

technique · cases
1.1

If X is metrizable, it is Hausdorff and locally metrizable by taking X itself as the open neighbourhood, and it is paracompact by [L1].

assume-case forwardL1
1.2

Conversely, local metrizability gives an open cover by metrizable subspaces. Paracompactness refines it by a locally finite open cover; every refining member is a metrizable subspace and has a σ-locally-finite basis by [L1]. The merger lemma A locally finite open cover by subspaces with σ-locally-finite bases yields a σ-locally-finite basis of the whole space gives such a basis for X.

assume-case reverseL1
2.1

Hausdorffness implies T1, and [L2] makes X regular; applying Nagata–Smirnov in [L2] to the basis from step 1.2 yields a metric.

L2step 1.2
3.1

The two cases prove the equivalence.

step 1.1step 2.1cases-exhaustive

5 · Examples, counterexamples and false statements

None yet.

Sources