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

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FALSE: a space in which every sequence has at most one limit is Hausdorff

Statement

False claim: if every sequence in a topological space has at most one limit (Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure), then the space 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 cocountable topology $T_{c o c}$ on $R$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), whose open sets are $\emptyset$ together with the complements of the at most countable subsets of $R$ . In it every convergent sequence is eventually constant, so limits are unique; and no two nonempty open sets are disjoint, so the space is not Hausdorff. It is nevertheless $T_{1}$ .

This is why Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure refuses the notation $\lim_{k} x_{k}$ in a general space and restores it only under a hypothesis. Uniqueness of sequential limits is strictly weaker than the Hausdorff condition, so it is uniqueness, and not the Hausdorff condition, that is the exact licensing condition for the symbol — and the two are not interchangeable.

Facts & Assumptions

Given: $R$ with the cocountable topology $T_{c o c}$ , a sequence $(x_{k})_{k \in N}$ in $R$ , and points $p, q \in R$ .

[A1]

$T_{c o c}$ consists of $\emptyset$ together with the sets whose complement in $R$ is at most countable; its closed sets are $R$ and the at most countable sets (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

[A2]

$x_{k} \to p$ means that for every neighbourhood $N$ of $p$ there is $K \in N$ with $x_{k} \in N$ for all $k \geq K$ ; 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).

[A3]

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

[L1]

The range ${x_{k} : k \in N}$ of a sequence is nonempty and at most countable, the sequence itself being a surjection of $N$ onto it; 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 $N$ , Every subset of an at most countable set is at most countable, Finite, countably infinite, countable, uncountable).

[L2]

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 ${A C}_{ω}$ , padded with copies of $\emptyset$ , and it needs no choice principle, exactly as The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies records for the cocountable topology itself.

[L3]

$R$ is uncountable ( $R$ is uncountable (Cantor's nested intervals, 1874)).

[L4]

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), $T_{0}$ (Kolmogorov) and $T_{1}$ (Frechet) spaces); a finite set is at most countable (Finite, countably infinite, countable, uncountable).

Refutation

technique · direct

1.1

Suppose $x_{k} \to p$ , and put $R : = {x_{k} : k \in N} ∖ {p}$ , which is at most countable by [L1].

A2L1assume-hyp

1.2

Let $U, V \in T_{c o c}$ be nonempty and suppose $U \cap V = \emptyset$ ; then $R = (R ∖ U) \cup (R ∖ V)$ is a union of two at most countable sets, hence at most countable by [L2], contradicting [L3].

A1L2L3assume-hyp

1.3

The cofinite topology on $R$ is contained in $T_{c o c}$ , a finite set being at most countable, so $(R, T_{c o c})$ is $T_{1}$ .

A1L4

2.1

Under step 1.1: $R ∖ R$ is open by [A1] and contains $p$ , so by [A2] there is $K$ with $x_{k} \in R ∖ R$ for all $k \geq K$ .

step 1.1A1A2

2.2

So no two nonempty open sets of $T_{c o c}$ are disjoint; taking $p = 0$ and $q = 1$ , any open $U ∋ p$ and $V ∋ q$ are nonempty and therefore meet, and $(R, T_{c o c})$ is not Hausdorff.

step 1.2A3

3.1

Under step 1.1: for $k \geq K$ the point $x_{k}$ lies in the range of the sequence and not in $R$ , hence $x_{k} = p$ ; so the sequence is eventually constant with value $p$ .

step 2.1

4.1

If also $x_{k} \to q$ with $q \neq p$ , then $R ∖ {p}$ is open by [A1], since ${p}$ is at most countable, and it contains $q$ ; so by [A2] there is $K^{'}$ with $x_{k} \in R ∖ {p}$ for all $k \geq K^{'}$ , contradicting step 3.1 at any index at least $\max {K, K^{'}}$ .

step 3.1A1A2

5.1

By step 4.1 every sequence in $(R, T_{c o c})$ has at most one limit.

step 4.1

6.1

By step 5.1 every sequence has at most one limit and by step 2.2 the space is not Hausdorff, so the claim is false; by step 1.3 the witness is moreover $T_{1}$ .

step 5.1step 2.2step 1.3∎

Remarks

The refutation is not about pathological sequences but about their scarcity. In the cocountable topology on an uncountable set a sequence can only reach at most countably many points, and every at most countable set is closed, so convergence degenerates to eventual constancy. Sequences are simply too small to detect this topology, which is also why nothing about it can be read off from sequential arguments.
What a countability hypothesis would change is not settled here. Whether adding first countability to the hypothesis rescues the claim is a question this library does not address, and nothing above asserts an answer. What is recorded is the metrizable case, where limits are unique and the space is Hausdorff for reasons independent of each other (Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not).
The converse is true and easy. In a Hausdorff space limits are unique: two distinct limits would have disjoint open neighbourhoods, each of which contains the sequence eventually, which is impossible. That direction is not what this item refutes.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 112 results over 22 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

Cocountable topology (Wikipedia) (standard reference, not scraped)
Hausdorff space (Wikipedia) (standard reference, not scraped)
S. Willard, General Topology, §13 (standard reference, not scraped)
Sequential space (Wikipedia) (standard reference, not scraped)