LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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.

Every Urysohn space is Hausdorff, every Hausdorff space is $T_{1}$ and hence $T_{0}$ , and every regular $T_{1}$ space is Urysohn

Statement

Let $(X, T)$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison). Then:

If $X$ is Urysohn (Urysohn ( $T_{2 \frac{1}{2}}$ ) space: distinct points have neighbourhoods with disjoint closures) then $X$ is Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not).
If $X$ is Hausdorff then $X$ is $T_{1}$ , and hence $T_{0}$ ( $T_{0}$ (Kolmogorov) and $T_{1}$ (Frechet) spaces).
If $X$ is regular (Regular spaces and $T_{3}$ spaces, with the source disagreement over whether regularity includes $T_{1}$ stated explicitly) and $T_{1}$ — that is, if $X$ is $T_{3}$ — then $X$ is Urysohn.

Consequently $T_{3}$ implies $T_{2 \frac{1}{2}}$ , which implies $T_{2}$ , which implies $T_{1}$ , which implies $T_{0}$ . Nothing here asserts that any of the four implications reverses; two of the failures are recorded among this page's false statements.

Facts & Assumptions

Given: A topological space $(X, T)$ and points $x, y \in X$ with $x \neq y$ .

[A1]

$X$ is Urysohn when distinct points have open neighbourhoods with disjoint closures (Urysohn ( $T_{2 \frac{1}{2}}$ ) space: distinct points have neighbourhoods with disjoint closures).

[A2]

$X$ is Hausdorff when distinct points have disjoint open neighbourhoods (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not).

[A3]

$X$ is $T_{1}$ when for distinct $x, y$ there are open $U ∋ x$ with $y \notin U$ and open $V ∋ y$ with $x \notin V$ ; every $T_{1}$ space is $T_{0}$ ( $T_{0}$ (Kolmogorov) and $T_{1}$ (Frechet) spaces).

[L1]

In a $T_{1}$ space every singleton is closed (A space is $T_{1}$ if and only if every singleton is closed, if and only if every finite subset is closed, if and only if its topology contains the cofinite topology, clause (b)).

[L2]

$X$ is regular exactly when for every $x$ and every open $U ∋ x$ there is 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 $x \in V \subseteq \overline{V} \subseteq U$ , clause (b)).

[L3]

$A \subseteq \overline{A}$ for every $A \subseteq X$ , and a set is closed exactly when its complement is open (Interior, closure, boundary, exterior, derived set and isolated point in a topological space, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

Proof

technique · direct

1.1A1assume-hyp

Assume $X$ is Urysohn and let $x \neq y$ ; [A1] gives open $U ∋ x$ and $V ∋ y$ with $\overline{U} \cap \overline{V} = \emptyset$ .

1.2A2assume-hyp

Assume $X$ is Hausdorff and let $x \neq y$ ; [A2] gives disjoint open $U ∋ x$ and $V ∋ y$ .

1.3L1L3assume-hyp

Assume $X$ is regular and $T_{1}$ and let $x \neq y$ ; by [L1] the set ${y}$ is closed, so $U_{0} : = X ∖ {y}$ is open by [L3] and contains $x$ .

2.1step 1.1A2L3

Under step 1.1: $U \cap V \subseteq \overline{U} \cap \overline{V} = \emptyset$ by [L3], so $U$ and $V$ are disjoint open neighbourhoods of $x$ and $y$ and $X$ is Hausdorff, which is claim 1.

2.2step 1.2A3

Under step 1.2: $y \notin U$ and $x \notin V$ , since $U \cap V = \emptyset$ with $y \in V$ and $x \in U$ ; so $U$ and $V$ witness the $T_{1}$ condition and $X$ is $T_{1}$ , hence $T_{0}$ , which is claim 2.

2.3step 1.3L2

Under step 1.3: [L2] applied to $x \in U_{0}$ gives an open $U$ with $x \in U \subseteq \overline{U} \subseteq U_{0} = X ∖ {y}$ , so $y \notin \overline{U}$ .

3.1step 2.3L2L3

Under step 2.3: $X ∖ \overline{U}$ is open by [L3] and contains $y$ , so [L2] applied to it gives an open $V$ with $y \in V \subseteq \overline{V} \subseteq X ∖ \overline{U}$ .

4.1step 3.1A1

Under step 3.1: $\overline{U} \cap \overline{V} = \emptyset$ , since $\overline{V} \subseteq X ∖ \overline{U}$ ; so $U$ and $V$ witness the Urysohn condition at the pair $x, y$ and $X$ is Urysohn, which is claim 3.

5.1step 2.1step 2.2step 4.1∎

Claims 1, 2 and 3 are steps 2.1, 2.2 and 4.1, and composing them gives the chain $T_{3} \Rightarrow T_{2 \frac{1}{2}} \Rightarrow T_{2} \Rightarrow T_{1} \Rightarrow T_{0}$ .

Remarks

Claim 3 is where the $T_{1}$ hypothesis earns its place. Regularity separates a point from a closed set, and the closed set used in the proof is the singleton ${y}$ ; without $T_{1}$ that singleton need not be closed and the argument has nothing to start from. The indiscrete topology on two points is regular and not Urysohn, which shows the hypothesis cannot simply be dropped.
Regularity is applied twice, and the second application is the whole point. The first shrink puts $y$ outside $\overline{U}$ ; the second separates $y$ from the closed set $\overline{U}$ , which is what upgrades disjointness of the sets to disjointness of their closures.
Claim 2 explains why the Hausdorff condition alone is often quoted as "points are closed". By A space is $T_{1}$ if and only if every singleton is closed, if and only if every finite subset is closed, if and only if its topology contains the cofinite topology the two are not the same condition; $T_{1}$ is strictly weaker, and the cofinite topology on an infinite set separates them.

Every Urysohn space is Hausdorff, every Hausdorff space is $T_{1}$ and hence $T_{0}$ , and every regular $T_{1}$ space is Urysohn

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Every Urysohn space is Hausdorff, every Hausdorff space is T1T1​ and hence T0T0​, and every regular T1T1​ space is Urysohn

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Every Urysohn space is Hausdorff, every Hausdorff space is $T_{1}$ and hence $T_{0}$ , and every regular $T_{1}$ space is Urysohn