LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31

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.

Regularity is hereditary, without a hidden $T_1$ hypothesis

Statement

Regularity, with no $T_1$ condition built into its name, is hereditary.

Facts & Assumptions

Given: A regular space $X$ , a subspace $S\subseteq X$ , a point $x\in S$ , and an open set $W$ of $S$ containing $x$ .

[L1]

In a regular space, $x\in U$ open gives an open $V$ with $x\in V\subseteq\overline V\subseteq U$ (A space is regular if and only if every point has a neighbourhood base of closed neighbourhoods, if and only if $x \in U$ open gives an open $V$ with $x \in V \subseteq \overline{V} \subseteq U$ ).

[L2]

Every open set of $S$ is a trace $U\cap S$ , and closure in $S$ is the ambient closure intersected with $S$ (For $A \subseteq S \subseteq X$ the closure of $A$ in $S$ is $\overline{A}^{X} \cap S$ , while the interior only contains $\operatorname{int}^{X}(A) \cap S$ , with equality when $S$ is open; and a dense subset of $X$ traces to a dense subset of every open $S$ ).

Proof

technique · direct

1.1

Write $W=U\cap S$ for an open $U\subseteq X$ containing $x$ .

1.2

Choose open $V\subseteq X$ with $x\in V\subseteq\overline V\subseteq U$ .

2.1

The trace $V\cap S$ is open in $S$ , contains $x$ , and has $\operatorname{cl}_S(V\cap S)\subseteq\overline V\cap S\subseteq U\cap S=W$ .

L2step 1.2

3.1

The closed-neighbourhood characterization now makes $S$ regular; as $S$ was arbitrary, regularity is hereditary.

L1step 2.1∎

Depends on

Used by

T₀, T₁, T₂, regularity, T₃, complete regularity, and Tychonoffness are hereditary Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 27 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

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

Regularity is hereditary, without a hidden T1T_1T1​ hypothesis