LemmaStatement: AI-adaptedProof: AI-adaptedprecheck 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 ${int}^{X} (A) \cap S$ , 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 ${cl}_{S} (V \cap S) \subseteq \overline{V} \cap S \subseteq U \cap S = W$ .

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 · two levels

12 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)

Regularity is hereditary, without a hidden T1T1​ hypothesis