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

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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

1.1

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

A1assume-hyp

1.2

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

A2assume-hyp

1.3

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 $x$ .

L1L3assume-hyp

2.1

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.

step 1.1A2L3

2.2

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.

step 1.2A3

2.3

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 $y \notin \overline{U}$ .

step 1.3L2

3.1

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

step 2.3L2L3

4.1

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.

step 3.1A1

5.1

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

step 2.1step 2.2step 4.1∎

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.

Depends on

Used by

Dependency tree · two levels

16 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

Urysohn and completely Hausdorff spaces (Wikipedia) (standard reference, not scraped)
Regular space (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §31 (standard reference, not scraped)
Separation axiom (Wikipedia) (standard reference, not scraped)

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