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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Arbitrary products of completely regular spaces are completely regular

Statement

An arbitrary product of completely regular spaces is completely regular.

Facts & Assumptions

Given: A product $P = \prod_{i \in I} X_{i}$ of completely regular spaces, a closed $C \subseteq P$ , and $x \in P ∖ C$ .

[F1]

A basic product-open set has finite support, and the product universal property makes finite-coordinate maps continuous (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, A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice).

[F2]

Complete regularity gives $h_{i} : X_{i} \to [0, 1]$ with $h_{i} (x_{i}) = 1$ and $h_{i} [X_{i} ∖ U_{i}] = {0}$ when $x_{i} \in U_{i}$ is open (Completely regular spaces and Tychonoff ( $T_{3 \frac{1}{2}}$ ) spaces).

[L1]

A finite pointwise minimum of continuous $[0, 1]$ -valued maps is continuous (Finite pointwise minima of continuous maps to $[0, 1]$ are continuous).

[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 finite-support basic neighbourhood $B = ⋂_{i \in J} π_{i}^{- 1} [U_{i}]$ of $x$ contained in $P ∖ C$ .

1.2

The finite-choice result [L2] selects a map $h_{i}$ as in [F2] for every $i \in J$ ; put $h = \min_{i \in J} (h_{i} \circ π_{i})$ .

F2L1L2

2.1

The map $h$ is continuous and $h (x) = 1$ . If $y \in C$ , then $y \notin B$ , so $y_{i} \notin U_{i}$ for some $i \in J$ and $h (y) = 0$ .

F1L1step 1.1step 1.2

3.1

Thus $h$ separates $x$ from $C$ in the defining sense of complete regularity.

F2step 2.1∎

Depends on

Used by

Dependency tree · two levels

26 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)