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 preserve $T_{0}$ , $T_{1}$ , and Hausdorffness

Statement

For any family $(X_{i})_{i \in I}$ , if every $X_{i}$ is $T_{0}$ , respectively $T_{1}$ , respectively Hausdorff, then $\prod_{i \in I} X_{i}$ is respectively $T_{0}$ , $T_{1}$ , respectively Hausdorff. The empty product is included.

Facts & Assumptions

Given: A family of spaces with the indicated separation property and two distinct points $x, y$ of its product $P$ .

[F1]

Distinct product points differ at a coordinate, and $π_{i}^{- 1} [U]$ is open whenever $U$ is open in $X_{i}$ (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]

The $T_{0}$ , $T_{1}$ , and Hausdorff conditions are respectively the stated one-sided, two-sided, and disjoint-open separations of distinct points ( $T_{0}$ (Kolmogorov) and $T_{1}$ (Frechet) spaces, Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not).

Proof

technique · direct

1.1F2

If $I = \emptyset$ , the product has one point and all three conditions hold vacuously.

1.2F1F2

Otherwise choose $i \in I$ with $x_{i} \neq y_{i}$ . For a $T_{0}$ factor, the inverse image under $π_{i}$ of an open set distinguishing $x_{i}, y_{i}$ distinguishes $x, y$ .

1.3F1F2

For a $T_{1}$ factor, pull back the two open sets separating $x_{i}$ from $y_{i}$ and $y_{i}$ from $x_{i}$ .

1.4F1F2

For a Hausdorff factor, pull back disjoint open neighbourhoods of $x_{i}, y_{i}$ ; their inverse images remain disjoint.

Arbitrary products preserve $T_{0}$ , $T_{1}$ , and Hausdorffness

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Arbitrary products preserve T0T0​, T1T1​, and Hausdorffness

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Arbitrary products preserve $T_{0}$ , $T_{1}$ , and Hausdorffness