LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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 $x\in S\setminus F$ .

[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 $x\in S\setminus F$ , 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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 60 results over 14 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

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

Complete regularity is hereditary, without a hidden T1T_1T1​ hypothesis