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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FALSE: every $T_1$ space is Hausdorff

Statement

False claim: every $T_1$ space ( $T_0$ (Kolmogorov) and $T_1$ (Frechet) spaces) is Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not).

The refutation is the cofinite topology $\mathcal{T}_{\mathrm{cof}}$ on $\mathbb{R}$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), whose open sets are $\varnothing$ together with the complements of the finite subsets of $\mathbb{R}$ . It is $T_1$ , because its closed sets are exactly $\mathbb{R}$ and the finite sets; and it is not Hausdorff, because any two nonempty open sets meet, $\mathbb{R}$ being infinite. The witness is worked further on the companion page, where the same space is shown to fail regularity and normality as well.

Facts & Assumptions

Given: The set $\mathbb{R}$ with the cofinite topology $\mathcal{T}_{\mathrm{cof}}$ , and two points $x \ne y$ of $\mathbb{R}$ .

[A1]

A space is Hausdorff when any two 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, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[L1]

$\mathcal{T}_{\mathrm{cof}}$ consists of $\varnothing$ together with the sets whose complement is finite; a union of two finite sets is finite (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, facts (i) and (ii) of that item).

[L2]

A topology is $T_1$ exactly when it contains the cofinite topology on the same set (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 (d)).

[L3]

$\mathbb{R}$ is uncountable ( $\mathbb{R}$ is uncountable (Cantor's nested intervals, 1874)), and every finite set is at most countable (Finite, countably infinite, countable, uncountable); so $\mathbb{R}$ is not finite.

Refutation

technique · direct

1.1

$\mathcal{T}_{\mathrm{cof}} \subseteq \mathcal{T}_{\mathrm{cof}}$ , so the cofinite topology on $\mathbb{R}$ is $T_1$ .

1.2

$\mathbb{R}$ is not finite, since a finite set is at most countable and $\mathbb{R}$ is uncountable.

1.3

Let $U, V \in \mathcal{T}_{\mathrm{cof}}$ be nonempty and suppose $U \cap V = \varnothing$ ; then $\mathbb{R} = \mathbb{R} \setminus (U \cap V) = (\mathbb{R} \setminus U) \cup (\mathbb{R} \setminus V)$ , a union of two finite sets, hence finite by [L1].

L1assume-hyp

2.1

Step 1.3 contradicts step 1.2, so no two nonempty open sets of $\mathcal{T}_{\mathrm{cof}}$ are disjoint.

step 1.2step 1.3

3.1

Take $x \ne y$ in $\mathbb{R}$ , for instance $x = 0$ and $y = 1$ . Any open $U \ni x$ and open $V \ni y$ are nonempty, so $U \cap V \ne \varnothing$ by step 2.1, and $x$ and $y$ have no disjoint open neighbourhoods.

step 2.1

4.1

By step 1.1 the space $(\mathbb{R}, \mathcal{T}_{\mathrm{cof}})$ is $T_1$ , and by step 3.1 and [A1] it is not Hausdorff; so the claim is false.

step 1.1step 3.1A1∎

Remarks

The failure is as large as it can be, not a boundary case: in the cofinite topology on an infinite set no two nonempty open sets are disjoint, so the Hausdorff condition fails at every pair of distinct points at once.
What is true is the converse. Every Hausdorff space is $T_1$ (Every Urysohn space is Hausdorff, every Hausdorff space is $T_1$ and hence $T_0$ , and every regular $T_1$ space is Urysohn), so $T_1$ is strictly weaker, and this item is what makes "strictly" honest.
Any infinite set would do. $\mathbb{R}$ is chosen because its infinitude is already a theorem here ( $\mathbb{R}$ is uncountable (Cantor's nested intervals, 1874)); nothing in the argument uses the order or the arithmetic of $\mathbb{R}$ .

Depends on

Used by

The cofinite topology on an infinite set is T₁ but neither Hausdorff nor regular nor normal Example

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 86 results over 25 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

T1 space (Wikipedia) (standard reference, not scraped)
Cofiniteness (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §17 (standard reference, not scraped)

FALSE: every T1T_1T1​ space is Hausdorff