ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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The cocountable topology on $\mathbb{R}$ is $T_1$ , has unique sequential limits, and is neither Hausdorff nor regular nor normal

Example

Give $\mathbb{R}$ the cocountable topology $\mathcal{T}_{\mathrm{coc}} = \{\varnothing\} \cup \{\, U \subseteq \mathbb{R} : \mathbb{R} \setminus U \text{ is at most countable} \,\}$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), whose closed sets are $\mathbb{R}$ together with the at most countable subsets of $\mathbb{R}$ (Finite, countably infinite, countable, uncountable). Then:

$(\mathbb{R}, \mathcal{T}_{\mathrm{coc}})$ is $T_1$ ( $T_0$ (Kolmogorov) and $T_1$ (Frechet) spaces).
Every convergent sequence is eventually constant, so every sequence has at most one limit (Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure).
No two nonempty open sets are disjoint. Consequently the space is not Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not), not regular (Regular spaces and $T_3$ spaces, with the source disagreement over whether regularity includes $T_1$ stated explicitly) and not normal (Normal spaces and $T_4$ spaces, with the source disagreement over whether normality includes $T_1$ stated explicitly).

Clauses 1, 2 and the Hausdorff half of clause 3 are what refute the claim that unique sequential limits force the Hausdorff condition (FALSE: a space in which every sequence has at most one limit is Hausdorff); clause 3's other two halves place the space in the hierarchy exactly where the cofinite topology sits, at $T_1$ and no higher.

Facts & Assumptions

Given: $\mathbb{R}$ with the cocountable topology $\mathcal{T}_{\mathrm{coc}}$ , a sequence $(x_k)_{k \in \mathbb{N}}$ in $\mathbb{R}$ , and points $p, q, u, v, w \in \mathbb{R}$ .

[A1]

$U \in \mathcal{T}_{\mathrm{coc}}$ exactly when $U = \varnothing$ or $\mathbb{R} \setminus U$ is at most countable; the closed sets are $\mathbb{R}$ and the at most countable subsets (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison, Interior, closure, boundary, exterior, derived set and isolated point in a topological space).

[A2]

$x_k \to p$ means: for every neighbourhood $N$ of $p$ there is $K \in \mathbb{N}$ with $x_k \in N$ for all $k \ge K$ ; an open set containing $p$ is such a neighbourhood (Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure).

[L1]

A topology is $T_1$ exactly when it contains the cofinite topology on the same set; a finite set is at most countable (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), $T_0$ (Kolmogorov) and $T_1$ (Frechet) spaces, Finite, countably infinite, countable, uncountable).

[L2]

The range of a sequence is nonempty and at most countable, and a subset of an at most countable set is at most countable (A nonempty set is at most countable iff it is a surjective image of $\mathbb{N}$ , Every subset of an at most countable set is at most countable).

[L3]

A union of two at most countable sets is at most countable: this is the two-set instance of Countable unions of at most countable sets, assuming $\mathrm{AC}_\omega$ padded with copies of $\varnothing$ , and it needs no choice principle, as The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies records.

[L4]

$\mathbb{R}$ is uncountable ( $\mathbb{R}$ is uncountable (Cantor's nested intervals, 1874)), so in particular it has at least three distinct points.

[L5]

Hausdorff, regular and normal are as in Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not, Regular spaces and $T_3$ spaces, with the source disagreement over whether regularity includes $T_1$ stated explicitly and Normal spaces and $T_4$ spaces, with the source disagreement over whether normality includes $T_1$ stated explicitly.

Verification

technique · direct

1.1

The cofinite topology on $\mathbb{R}$ is contained in $\mathcal{T}_{\mathrm{coc}}$ , a finite complement being at most countable, so the space is $T_1$ , which is claim 1.

A1L1

1.2

Fix three distinct points $u, v, w$ of $\mathbb{R}$ , for instance $0$ , $1$ and $2$ .

1.3

Suppose $x_k \to p$ and put $R := \{\, x_k : k \in \mathbb{N} \,\} \setminus \{p\}$ , at most countable by [L2]; then $\mathbb{R} \setminus R$ is open by [A1] and contains $p$ , so [A2] gives $K$ with $x_k \notin R$ for all $k \ge K$ .

A1A2L2assume-hyp

1.4

Let $U, V$ be nonempty open sets and suppose $U \cap V = \varnothing$ ; then $\mathbb{R} = (\mathbb{R} \setminus U) \cup (\mathbb{R} \setminus V)$ is at most countable by [A1] and [L3], contradicting [L4]. So no two nonempty open sets are disjoint.

A1L3L4assume-hyp

2.1

Under step 1.3: for $k \ge K$ the point $x_k$ lies in the range of the sequence and outside $R$ , hence equals $p$ ; so the sequence is eventually constant with value $p$ .

step 1.3

2.2

Any open $U \ni u$ and open $V \ni v$ are nonempty, hence meet by step 1.4, so the space is not Hausdorff.

step 1.2step 1.4L5

2.3

$\{v\}$ is at most countable, hence closed by [A1], and $u \notin \{v\}$ ; any open $U \ni u$ and open $V \supseteq \{v\}$ are nonempty, hence meet by step 1.4, so the space is not regular.

step 1.2step 1.4A1L5

2.4

$\{v\}$ and $\{w\}$ are disjoint nonempty closed sets by [A1] and step 1.2; any open sets containing them are nonempty, hence meet by step 1.4, so the space is not normal.

step 1.2step 1.4A1L5

3.1

Under step 1.3: if also $x_k \to q$ with $q \ne p$ , then $\mathbb{R} \setminus \{p\}$ is open by [A1] and contains $q$ , so [A2] gives $K'$ with $x_k \ne p$ for all $k \ge K'$ , contradicting step 2.1 at any index at least both $K$ and $K'$ . So a sequence has at most one limit, which with step 2.1 is claim 2.

step 2.1A1A2

4.1

Steps 2.2, 2.3 and 2.4 complete claim 3, step 3.1 is claim 2 and step 1.1 is claim 1.

step 1.1step 3.1step 2.2step 2.3step 2.4∎

Remarks

Sequences cannot see this topology. A sequence reaches at most countably many points, and every at most countable set is closed, so the complement of the values other than the limit is an open set that forces the sequence to be eventually constant. Uniqueness of limits is therefore free, and it carries no separation information at all — which is the point of FALSE: a space in which every sequence has at most one limit is Hausdorff.
The failure of $T_2$ , $T_3$ and $T_4$ has the same one-line cause as in the cofinite case: two at most countable sets cannot cover an uncountable one, so two nonempty open sets always meet. What changes between the two examples is only how large a set has to be for the topology to be interesting: infinite for cofinite, uncountable for cocountable.
On an at most countable set the cocountable topology is discrete (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), so the uncountability of $\mathbb{R}$ is doing real work here and not merely supplying a familiar underlying set.

Depends on

Used by

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Sources

Cocountable topology (Wikipedia) (standard reference, not scraped)
Separation axiom (Wikipedia) (standard reference, not scraped)
L. Steen and J. Seebach, Counterexamples in Topology, §20 (standard reference, not scraped)

The cocountable topology on R\mathbb{R}R is T1T_1T1​, has unique sequential limits, and is neither Hausdorff nor regular nor normal