TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Every completely normal space is normal, and every perfectly normal space is normal

Statement

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

If $X$ is completely normal (Completely normal ( $T_5$ ) and perfectly normal ( $T_6$ ) spaces) then $X$ is normal (Normal spaces and $T_4$ spaces, with the source disagreement over whether normality includes $T_1$ stated explicitly).
If $X$ is perfectly normal then $X$ is normal.
Consequently $T_5$ implies $T_4$ and $T_6$ implies $T_4$ .

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 $(X,\mathcal{T})$ and closed sets $A, B \subseteq X$ with $A \cap B = \varnothing$ .

[A1]

$X$ completely normal: every pair of separated sets admits disjoint open supersets (Completely normal ( $T_5$ ) and perfectly normal ( $T_6$ ) spaces).

[A2]

$X$ perfectly normal: $X$ is normal and every closed subset of $X$ is a $G_\delta$ (Completely normal ( $T_5$ ) and perfectly normal ( $T_6$ ) spaces).

[L1]

$A$ and $B$ are separated when $\overline{A} \cap B = A \cap \overline{B} = \varnothing$ (Separated sets: $\overline{A} \cap B = A \cap \overline{B} = \varnothing$ ).

[L2]

A set is closed exactly when it equals its own closure (A point lies in the closure of $A$ iff every basic neighbourhood of it meets $A$ ; the closure is the smallest closed superset and equals $A$ together with its derived set, claim 2, Interior, closure, boundary, exterior, derived set and isolated point in a topological space).

[L3]

Normality is the assertion that disjoint closed sets admit disjoint open supersets (Normal spaces and $T_4$ spaces, with the source disagreement over whether normality includes $T_1$ stated explicitly).

Proof

technique · direct

1.1

$\overline{A} = A$ and $\overline{B} = B$ , both sets being closed.

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

$\overline{A} \cap B = A \cap B = \varnothing$ and $A \cap \overline{B} = A \cap B = \varnothing$ , 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 $U \supseteq A$ and $V \supseteq B$ ; since $A$ and $B$ were arbitrary disjoint closed sets, $X$ is normal, which is claim 1.

step 2.1A1L3

4.1

Adding the hypothesis $T_1$ to either of steps 3.1 and 1.2 turns $T_5$ , respectively $T_6$ , into $T_4$ , 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 $\mathbb{R}$ , 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

The implications proved on this page: perfectly normal gives completely normal under countable choice, and completely normal gives normal; normal with T₁ gives T₃; completely regular gives regular; regular with T₁ gives Urysohn, hence Hausdorff, hence T₁, hence T₀; and metrizable gives every one of them Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 37 results over 12 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

Normal space (Wikipedia) (standard reference, not scraped)
Separated sets (Wikipedia) (standard reference, not scraped)
S. Willard, General Topology, §15 (standard reference, not scraped)