LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31

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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.

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Complete regularity is hereditary, without a hidden $T_{1}$ hypothesis

Statement

Complete regularity, with no $T_{1}$ condition built into its name, is hereditary.

Facts & Assumptions

Given: A completely regular space $X$ , a subspace $S \subseteq X$ , a closed set $F$ of $S$ , and $x \in S ∖ F$ .

[F1]

A closed subset of $S$ has the form $C \cap S$ for a closed $C \subseteq X$ (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).

[F2]

Complete regularity supplies a continuous $f : X \to [0, 1]$ with $f (x) = 1$ and $f [C] = {0}$ when $C$ is closed and misses $x$ (Completely regular spaces and Tychonoff ( $T_{3 \frac{1}{2}}$ ) spaces).

[L1]

A restriction of a continuous map to a subspace is continuous (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, For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f (\overline{A}) \subseteq \overline{f (A)}$ ).

Proof

technique · direct

1.1

Choose closed $C \subseteq X$ with $F = C \cap S$ ; since $x \in S ∖ F$ , one has $x \notin C$ .

1.2

Choose $f : X \to [0, 1]$ continuous with $f (x) = 1$ and $f [C] = {0}$ .

2.1

The restriction $f ∣_{S} : S \to [0, 1]$ is continuous, takes $x$ to $1$ , and vanishes on $F \subseteq C$ .

L1step 1.2

3.1

Thus $S$ is completely regular, and the arbitrariness of $S$ proves heredity.

F2step 2.1∎

Depends on

Used by

Dependency tree · two levels

18 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

J. P. May, An Outline Summary of Basic Point Set Topology, §6 (standard reference, not scraped)

Complete regularity is hereditary, without a hidden T1T1​ hypothesis