False statementConstruction: Literature-sourcedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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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.

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FALSE: every Hausdorff space is regular

Statement

False claim: every Hausdorff space (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not) is regular (Regular spaces and $T_3$ spaces, with the source disagreement over whether regularity includes $T_1$ stated explicitly).

The refutation is the $K$ -topology on $\mathbb{R}$ , generated by the bounded open intervals together with the sets $(a,b) \setminus K$ for $K = \{\, 1/(n+1) : n \in \mathbb{N} \,\}$ : it is Hausdorff and $T_1$ , the set $K$ is closed in it, and the point $0$ and the closed set $K$ have no disjoint open neighbourhoods (The $K$ -topology on $\mathbb{R}$ , generated by the open intervals together with their complements of $K = \{1/(n+1) : n \in \mathbb{N}\}$ , is $T_1$ and Hausdorff but not regular).

Since regularity together with $T_1$ gives back the Hausdorff condition (Every Urysohn space is Hausdorff, every Hausdorff space is $T_1$ and hence $T_0$ , and every regular $T_1$ space is Urysohn), the arrow $T_2 \Rightarrow T_3$ is refuted as well: the witness is $T_1$ , so it is a Hausdorff space that is not $T_3$ .

Facts & Assumptions

Given: $\mathbb{R}$ with the $K$ -topology $\mathcal{T}_K$ of The $K$ -topology on $\mathbb{R}$ , generated by the open intervals together with their complements of $K = \{1/(n+1) : n \in \mathbb{N}\}$ , is $T_1$ and Hausdorff but not regular.

[A1]

A space is regular when for every closed $C$ and every point $x \notin C$ there are disjoint open $U \ni x$ and $V \supseteq C$ (Regular spaces and $T_3$ spaces, with the source disagreement over whether regularity includes $T_1$ stated explicitly, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[L1]

$(\mathbb{R}, \mathcal{T}_K)$ is Hausdorff and $T_1$ ; $K$ is closed in it; and there are no disjoint open $U \ni 0$ and $V \supseteq K$ (The $K$ -topology on $\mathbb{R}$ , generated by the open intervals together with their complements of $K = \{1/(n+1) : n \in \mathbb{N}\}$ , is $T_1$ and Hausdorff but not regular, claims 2, 3 and 4, Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not, $T_0$ (Kolmogorov) and $T_1$ (Frechet) spaces).

Refutation

technique · direct

1.1

$(\mathbb{R},\mathcal{T}_K)$ is Hausdorff.

1.2

$K$ is closed in $\mathcal{T}_K$ and $0 \notin K$ , every element of $K$ being positive.

2.1

By [L1] there are no disjoint open $U \ni 0$ and $V \supseteq K$ , so the pair consisting of the closed set $K$ and the point $0$ violates the condition of [A1] and $(\mathbb{R},\mathcal{T}_K)$ is not regular.

step 1.2A1L1

3.1

By steps 1.1 and 2.1 the space $(\mathbb{R},\mathcal{T}_K)$ is Hausdorff and not regular, so the claim is false.

step 1.1step 2.1∎

Remarks

The witness is $T_1$ , which makes the refutation as strong as possible. A Hausdorff space that failed regularity only for want of closed points would be an artefact of the convention fork of Regular spaces and $T_3$ spaces, with the source disagreement over whether regularity includes $T_1$ stated explicitly; this one is not, since it satisfies every axiom below $T_3$ and fails $T_3$ itself.
What the converse direction says. Regularity does not imply Hausdorff either, and the indiscrete two-point space shows it; so neither of $T_2$ and regular implies the other, and only the conjunction $T_3$ sits above $T_2$ in the chain (Every Urysohn space is Hausdorff, every Hausdorff space is $T_1$ and hence $T_0$ , and every regular $T_1$ space is Urysohn).
This is the only Hausdorff-not-regular witness on these two pages, and it is minted on the main page rather than the companion page precisely so that this false statement can cite it.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 69 results over 18 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

K-topology (Wikipedia) (standard reference, not scraped)
Regular space (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §31 (standard reference, not scraped)
R. Gardner, Introduction to Topology, notes on Munkres Section 31: The Separation Axioms (East Tennessee State University) (standard reference, not scraped)