Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29
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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 completely normal space is normal, and every perfectly normal space is normal

Statement

Let (X,T) be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

  1. If X is completely normal (Completely normal (T5) and perfectly normal (T6) spaces) then X is normal (Normal spaces and T4 spaces, with the source disagreement over whether normality includes T1 stated explicitly).
  2. If X is perfectly normal then X is normal.
  3. Consequently T5 implies T4 and T6 implies T4.

Claim 2 is immediate from the definition, normality being one of the two conjuncts of perfect normality; it is recorded here so that the chain assembled at the end of this page has a single item to cite for both implications. Claim 1 is the one with content, and its content is that disjoint closed sets are a special case of separated sets.

Facts & Assumptions

Given: A topological space (X,T) and closed sets A,B⊆X with A∩B=∅.

[A1]

X completely normal: every pair of separated sets admits disjoint open supersets (Completely normal (T5) and perfectly normal (T6) spaces).

[A2]

X perfectly normal: X is normal and every closed subset of X is a Gδ (Completely normal (T5) and perfectly normal (T6) spaces).

[L1]

A and B are separated when A‾∩B=A∩B‾=∅ (Separated sets: A‾∩B=A∩B‾=∅).

[L3]

Normality is the assertion that disjoint closed sets admit disjoint open supersets (Normal spaces and T4 spaces, with the source disagreement over whether normality includes T1 stated explicitly).

Proof

technique · direct
1.1

A‾=A and B‾=B, both sets being closed.

L2
1.2

If X is perfectly normal then X is normal, this being the first conjunct of [A2], which is claim 2.

A2L3
2.1

A‾∩B=A∩B=∅ and A∩B‾=A∩B=∅, so A and B are separated.

step 1.1L1
3.1

If X is completely normal, [A1] applied to the separated pair of step 2.1 gives disjoint open U⊇A and V⊇B; since A and B were arbitrary disjoint closed sets, X is normal, which is claim 1.

step 2.1A1L3
4.1

Adding the hypothesis T1 to either of steps 3.1 and 1.2 turns T5, respectively T6, into T4, which is claim 3.

step 3.1step 1.2∎

Remarks

  • Neither converse is proved here and neither is asserted. Whether a normal space must be completely normal, and whether a normal space must be perfectly normal, are left open on this page: any witness would need machinery this page does not have, and no false statement asserting a reversal is planted here.

  • Where the strength of complete normality actually shows. It is not in the closed case above but in pairs like (0,1) and (1,2) in R, which are separated and not closed. The metric theorem later on this page separates every such pair at once, which is why every metrizable space is completely normal and not merely normal.

Depends on

Used by

Dependency tree · two levels

15 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