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.

Arbitrary products of regular spaces are regular

Statement

An arbitrary product of regular spaces is regular.

Facts & Assumptions

Given: A point $x$ in a product $P = \prod_{i \in I} X_{i}$ of regular spaces and an open set $W \subseteq P$ containing $x$ .

[F1]

A basic product-open neighbourhood of $x$ restricts only finitely many coordinates and lies inside $W$ (The product set $\prod_{i \in I} X_{i}$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space).

[L1]

In a regular factor, $x_{i} \in U_{i}$ open gives open $V_{i}$ with $x_{i} \in V_{i} \subseteq \overline{V_{i}} \subseteq U_{i}$ (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]

A family indexed by a natural number whose members are nonempty has a choice function, without any choice axiom (Every natural-number-indexed list of nonempty sets has a choice function on its family of values).

Proof

technique · direct

1.1

Choose a basic neighbourhood $B = ⋂_{i \in J} π_{i}^{- 1} [U_{i}]$ of $x$ inside $W$ , where $J$ is finite.

1.2

Since $J$ is finite, [L2] makes these factorwise choices simultaneously: take open $V_{i}$ with $x_{i} \in V_{i} \subseteq \overline{V_{i}} \subseteq U_{i}$ for every $i \in J$ .

L1L2

2.1

Put $V = ⋂_{i \in J} π_{i}^{- 1} [V_{i}]$ . It is open, contains $x$ , and its closure lies in $⋂_{i \in J} π_{i}^{- 1} [\overline{V_{i}}] \subseteq B \subseteq W$ .

F1step 1.2

3.1

The closed-neighbourhood characterization proves $P$ regular.

L1step 2.1∎

Depends on

Used by

Dependency tree · two levels

17 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)