CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-09 (gpt-5.6-terra-codex-subscription)

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Assuming countable choice, perfect normality, and hence $T_{6}$ , is hereditary

Statement

Assuming the Axiom of Countable Choice, perfect normality is hereditary. Consequently $T_{6}$ is hereditary.

Facts & Assumptions

Given: The Axiom of Countable Choice and a subspace $S$ of a perfectly normal space $X$ .

[A1]

The Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L1]

Under [A1], 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).

[L2]

A space is completely normal exactly when every one of its subspaces is normal; $T_{1}$ is hereditary (A space is completely normal if and only if every subspace is normal, $T_{0}$ , $T_{1}$ , and Hausdorffness are hereditary).

[F1]

A closed set of $S$ is $C \cap S$ for ambient closed $C$ ; a $G_{δ}$ is a countable intersection of open sets (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace, $G_{δ}$ and $F_{σ}$ subsets of a topological space, agreeing with the real-line notion).

Proof

technique · direct

1.1

By [L1], $X$ is completely normal. Every subspace of $S$ is then a subspace of $X$ , hence normal by [L2]; applying [L2] to $S$ shows that $S$ is completely normal. The $T_{1}$ clause of [L2] also shows that $S$ is $T_{1}$ when $X$ is $T_{6}$ .

A1L1L2

1.2

Let $F$ be closed in $S$ . Write $F = C \cap S$ with $C$ closed in $X$ ; perfect normality writes $C = ⋂_{n \in N} U_{n}$ with every $U_{n}$ open in $X$ .

2.1

Then $F = ⋂_{n \in N} (U_{n} \cap S)$ , a $G_{δ}$ of $S$ . Thus $S$ is perfectly normal, and with its inherited $T_{1}$ property it is $T_{6}$ when $X$ is $T_{6}$ .

F1step 1.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

35 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

R. Engelking, General Topology, §1.5 (standard reference, not scraped)
Normal space (Wikipedia) (standard reference, not scraped)

Assuming countable choice, perfect normality, and hence T6T6​, is hereditary