TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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 implications proved on this page: perfectly normal gives completely normal under countable choice, and completely normal gives normal; normal with $T_{1}$ gives $T_{3}$ ; completely regular gives regular; regular with $T_{1}$ gives Urysohn, hence Hausdorff, hence $T_{1}$ , hence $T_{0}$ ; and metrizable gives every one of them

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). The following implications hold, and each is proved by an earlier item of this page.

Perfectly normal implies completely normal, assuming the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )).
Completely normal implies normal, and perfectly normal implies normal.
Normal together with $T_{1}$ implies $T_{3}$ , that is regular together with $T_{1}$ .
Completely regular implies regular, and Tychonoff implies $T_{3}$ .
Regular together with $T_{1}$ implies Urysohn, which implies Hausdorff, which implies $T_{1}$ , which implies $T_{0}$ .
Metrizable implies every property named above: a metrizable space is perfectly normal, completely normal, normal, Tychonoff, completely regular, $T_{3}$ , regular, Urysohn, Hausdorff, $T_{1}$ and $T_{0}$ , with no choice principle used.

Reading the numbered axioms in order, clauses 1 to 5 give

$T_{6} \Rightarrow T_{5} \Rightarrow T_{4} \Rightarrow T_{3} \Rightarrow T_{2 \frac{1}{2}} \Rightarrow T_{2} \Rightarrow T_{1} \Rightarrow T_{0},$

the first arrow under ${A C}_{ω}$ , together with $T_{3 \frac{1}{2}} \Rightarrow T_{3}$ .

This is the whole of the classical chain that this page proves, and it is one arrow short of the classical chain. The implication $T_{4} \Rightarrow T_{3 \frac{1}{2}}$ — a normal $T_{1}$ space is completely regular — is Urysohn's lemma and is not available at this point in the reading order. Its absence is recorded, with what would license it, in this page's conventions remark; it is deliberately not asserted here, and no clause above may be read as giving it.

Facts & Assumptions

Given: A topological space $(X, T)$ , and the definitions of $T_{0}$ , $T_{1}$ , Hausdorff, Urysohn, regular, completely regular, normal, completely normal and perfectly normal ( $T_{0}$ (Kolmogorov) and $T_{1}$ (Frechet) spaces, Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not, Urysohn ( $T_{2 \frac{1}{2}}$ ) space: distinct points have neighbourhoods with disjoint closures, Regular spaces and $T_{3}$ spaces, with the source disagreement over whether regularity includes $T_{1}$ stated explicitly, Completely regular spaces and Tychonoff ( $T_{3 \frac{1}{2}}$ ) spaces, Normal spaces and $T_{4}$ spaces, with the source disagreement over whether normality includes $T_{1}$ stated explicitly, Completely normal ( $T_{5}$ ) and perfectly normal ( $T_{6}$ ) spaces).

[L1]

Assuming ${A C}_{ω}$ , every perfectly normal space is completely normal (Assuming countable choice, every perfectly normal space is completely normal: separated sets in a normal space whose open sets are all $F_{σ}$ can be separated by disjoint open sets, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L2]

Every completely normal space is normal, and every perfectly normal space is normal (Every completely normal space is normal, and every perfectly normal space is normal).

[L3]

A normal $T_{1}$ space is regular, hence $T_{3}$ (A normal $T_{1}$ space is regular, hence $T_{3}$ , hence Urysohn, Hausdorff, $T_{1}$ and $T_{0}$ ).

[L4]

Every completely regular space is regular, and every Tychonoff space is $T_{3}$ (Every completely regular space is regular, and every Tychonoff space is $T_{3}$ ).

[L5]

Every regular $T_{1}$ space is Urysohn, every Urysohn space is Hausdorff, and every Hausdorff space is $T_{1}$ and hence $T_{0}$ (Every Urysohn space is Hausdorff, every Hausdorff space is $T_{1}$ and hence $T_{0}$ , and every regular $T_{1}$ space is Urysohn).

[L6]

Every $T_{1}$ space has closed singletons, and conversely (A space is $T_{1}$ if and only if every singleton is closed, if and only if every finite subset is closed, if and only if its topology contains the cofinite topology).

[L7]

Every metric space is completely normal, hence normal, with no choice principle used (In a metric space any two separated sets have disjoint open neighbourhoods, so every metrizable space is completely normal).

[L8]

Every metrizable space is Tychonoff and perfectly normal, and hence satisfies every axiom named in clause 6 (In a metric space every closed set is a zero set and a $G_{δ}$ , and the distance function separates a point from a closed set, so every metrizable space is Tychonoff and perfectly normal, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not).

Proof

technique · direct

1.1

Clause 1 is [L1], whose hypothesis ${A C}_{ω}$ is carried into clause 1 unchanged.

1.2

Clause 2 is [L2].

1.3

Clause 3 is [L3].

1.4

Clause 4 is [L4].

1.5

Clause 5 is [L5], the first implication of which uses [L6] inside its own proof and needs nothing further here.

L5L6

1.6

Clause 6 is [L7] together with [L8].

L7L8

2.1

The displayed chain of numbered axioms is read off from steps 1.1 to 1.5, each numbered axiom being the corresponding unnumbered property together with $T_{1}$ , which is carried along every arrow: $T_{6}$ gives completely normal by step 1.1, hence $T_{5}$ ; $T_{5}$ gives normal by step 1.2, hence $T_{4}$ ; $T_{4}$ gives $T_{3}$ by step 1.3; and $T_{3}$ gives Urysohn, Hausdorff, $T_{1}$ and $T_{0}$ by step 1.5.

step 1.1step 1.2step 1.3step 1.5

2.2

The side arrow $T_{3 \frac{1}{2}} \Rightarrow T_{3}$ is the second half of step 1.4.

step 1.4

3.1

Steps 1.1 to 1.6, 2.1 and 2.2 are exactly clauses 1 to 6 and the two displayed chains, and no other implication is asserted.

step 1.6step 2.1step 2.2∎

Remarks

Every clause above is an implication and none is an equivalence. This page refutes four of the possible converses among its false statements — $T_{1}$ does not give Hausdorff, normal does not give Hausdorff, Hausdorff does not give regular, and unique sequential limits do not give Hausdorff — and asserts nothing about the others.
The $T_{1}$ hypothesis is where the numerals differ from the adjectives. Regular, completely regular, normal, completely normal and perfectly normal carry no $T_{1}$ in this library; $T_{3}$ , $T_{3 \frac{1}{2}}$ , $T_{4}$ , $T_{5}$ and $T_{6}$ are the conjunctions with $T_{1}$ . Clauses 3 and 5 are the two places the conjunction is genuinely needed for the next arrow, and they are what makes the numbered chain descend at all.
The countable choice in clause 1 is inherited, not introduced. It is spent in the proof of Assuming countable choice, every perfectly normal space is completely normal: separated sets in a normal space whose open sets are all $F_{σ}$ can be separated by disjoint open sets and nowhere else on this page; clause 6 in particular is choice free, since the metric proofs construct their open sets explicitly.

Depends on

Used by

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

Separation axiom (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §31-33 (standard reference, not scraped)
S. Willard, General Topology, §13-15 (standard reference, not scraped)
Normal space (Wikipedia) (standard reference, not scraped)
Tychonoff space (Wikipedia) (standard reference, not scraped)
Urysohn and completely Hausdorff spaces (Wikipedia) (standard reference, not scraped)