Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-generatedverified 2026-09-26 (gpt-6-sol)
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 particular-point topology is T0, it is not T1 and not regular once the set has at least two points, and it is not normal once the set has at least three

Example

Let X be a set, fix p∈X, and give X the particular-point topology Tp={∅}∪{ U⊆X:p∈U } (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), whose closed sets are X together with the subsets of X that do not contain p. Then:

  1. (X,Tp) is T0 (T0 (Kolmogorov) and T1 (Frechet) spaces), for every X and every p.
  2. If X has at least two points then (X,Tp) is not T1, and hence not Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not).
  3. If X has at least two points then (X,Tp) is not regular (Regular spaces and T3 spaces, with the source disagreement over whether regularity includes T1 stated explicitly).
  4. If X has at least three points then (X,Tp) is not normal (Normal spaces and T4 spaces, with the source disagreement over whether normality includes T1 stated explicitly).

Clause 4 needs one point more than clause 3, and the extra point is not slack: on a two-point set the particular-point topology is Sierpinski space, which is normal. So this family separates the two failures, and it shows that "not regular" and "not normal" begin at different sizes.

Facts & Assumptions

Verification

technique · direct
1.1A1

Every nonempty open set contains p, by [A1].

1.2A1L1

Let x≠y in X. If x=p then {p} is open by [A1], contains x and not y; if y=p the same argument applies with the roles exchanged; and if neither is p then {x,p} is open by [A1], contains x and not y. So (X,Tp) is T0, which is claim 1.

1.3A1L1assume-hyp

Suppose X has at least two points, so that X∖{p}≠∅. Then {p}≠X, and {p} contains p, so {p} is not closed by [A1]; hence (X,Tp) is not T1 by [L1], and not Hausdorff, which is claim 2.

1.4A1assume-hyp

Suppose X has at least three points and fix x,y∈X with x≠y, x≠p and y≠p. Then {x} and {y} are disjoint nonempty closed sets by [A1].

1.5A1L2

If X={p,x}, its closed sets are ∅, {x}, and X by [A1]. No two disjoint closed sets are both nonempty, so the normality condition in [L2] holds: for each disjoint pair, take the open neighbourhood of the empty set to be ∅ and that of the other set to be X.

2.1step 1.3A1

Under step 1.3 fix x∈X with x≠p; then {x} does not contain p, so {x} is closed by [A1], and p∉{x}.

2.2step 1.1step 1.4L2

Under step 1.4: any open U⊇{x} and open V⊇{y} are nonempty, hence both contain p by step 1.1, so U∩V≠∅ and (X,Tp) is not normal, which is claim 4.

3.1step 1.1step 2.1L2

Under step 1.3: any open U with p∈U and any open V with {x}⊆V are both nonempty, so both contain p by step 1.1 and U∩V≠∅. Hence the point p and the closed set {x} cannot be separated and (X,Tp) is not regular, which is claim 3.

4.1step 1.2step 1.3step 3.1step 2.2step 1.5∎

Steps 1.2, 1.3, 3.1 and 2.2 are claims 1 to 4, and step 1.5 verifies the two-point observation.

Remarks

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources