CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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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.
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In the indiscrete topology every sequence converges to every point, and in the cofinite topology on an infinite set an injective sequence converges to every point

Statement refuted

Refuted: that a convergent sequence in a topological space has exactly one limit, and hence that the notation $\lim_{k} x_{k}$ denotes at that generality (Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure).

Witnesses.

Let $X$ carry the indiscrete topology and have at least two points (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies). Then every sequence in $X$ converges to every point of $X$ .
Let $X$ be infinite with the cofinite topology and let $(x_{k})$ be an injective sequence in $X$ (Injection, surjection, bijection). Then $(x_{k})$ converges to every point of $X$ .
Claim 2 is instantiated without any choice principle by $X : = N$ with the cofinite topology and the sequence $x_{k} : = k$ , which is injective outright and whose index set is infinite (The natural numbers $N$ (von Neumann), The pigeonhole principle on $N$ ).

No appeal is made to "every infinite set has a countably infinite subset". That statement is not a theorem of ZF, and claim 2 is a conditional statement about a sequence that is given; claim 3 supplies such a sequence explicitly on $N$ rather than extracting one from an arbitrary infinite set.

Facts & Assumptions

Given: A set $X$ with at least two points carrying the indiscrete topology; an infinite set $X$ carrying the cofinite topology, a point $p \in X$ and an injective sequence $(x_{k})$ in $X$ ; and $N$ with the cofinite topology and the sequence $x_{k} = k$ .

[A1]

$x_{k} \to p$ means that for every neighbourhood $N$ of $p$ there is $K \in N$ with $x_{k} \in N$ for all $k \geq K$ ; a neighbourhood of $p$ contains an open set containing $p$ (Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure, Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open).

[A2]

In the indiscrete topology the only open sets are $\emptyset$ and $X$ ; in the cofinite topology the open sets are $\emptyset$ together with the sets of finite complement (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).

[L1]

In the indiscrete topology every sequence converges to every point (The sequential closure is contained in the closure, continuity implies sequential continuity, and sequential limits need not be unique, claim 3).

[L2]

A subset of a finite set is finite (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies); $\approx$ is symmetric and transitive, and an injection restricts to a bijection onto its image (Equinumerous sets, $A \approx B$ and $A ⪯ B$ , Injection, surjection, bijection).

[L3]

A subset of $N$ that is not bounded above is countably infinite (Every subset of an at most countable set is at most countable); no finite set is countably infinite, since $N \approx̸ n$ for every natural $n$ (The pigeonhole principle on $N$ , claim 4, Finite, countably infinite, countable, uncountable).

[L4]

$N$ is infinite, that is not finite (The pigeonhole principle on $N$ , claim 4, Finite, countably infinite, countable, uncountable, The natural numbers $N$ (von Neumann)).

[L5]

In the cofinite topology on an infinite set no two nonempty open sets are disjoint (On an infinite set the cofinite topology has every infinite subset dense and no two nonempty open sets disjoint, claim 3).

Counterexample

technique · direct

1.1

Claim 1 is [L1]: in the indiscrete topology the only neighbourhood of any point is $X$ itself, so every sequence is eventually in every neighbourhood of every point. With at least two points, some sequence therefore has two distinct limits.

A1A2L1

1.2

Let $X$ be infinite with the cofinite topology, let $p \in X$ , let $(x_{k})$ be injective, and let $N$ be a neighbourhood of $p$ ; fix an open $U$ with $p \in U \subseteq N$ , so $U \neq \emptyset$ and $F : = X ∖ U$ is finite.

A1A2choose

1.3

The index set $S : = {k \in N : x_{k} \in F}$ is finite: the injectivity of $(x_{k})$ makes $k \mapsto x_{k}$ a bijection of $S$ onto its image, which is a subset of the finite set $F$ and hence finite, so $S$ is finite as well.

givenL2

2.1

A finite subset of $N$ is bounded above: if it were not, it would be countably infinite by [L3], and no finite set is countably infinite. So $S$ is bounded above, say by $K \in N$ .

step 1.3L3

3.1

For every $k$ with $k > K$ one has $k \notin S$ , that is $x_{k} \notin F$ , that is $x_{k} \in U \subseteq N$ ; so $(x_{k})$ is eventually in $N$ . As $N$ was an arbitrary neighbourhood of $p$ , $x_{k} \to p$ , and as $p$ was arbitrary this proves claim 2.

step 1.2step 2.1A1

4.1

Claim 3: $N$ is infinite by [L4], the sequence $x_{k} = k$ is injective, being the identity function of $N$ , and $N$ has at least two points; so claim 2 applies and $(k)_{k \in N}$ converges in the cofinite topology on $N$ to every natural number at once.

step 3.1L4

5.1

By steps 1.1 and 4.1 there are topological spaces in which a sequence has more than one limit; the notation $\lim_{k} x_{k}$ therefore does not denote in a general topological space, and the uniqueness available for sequences of reals and in metric spaces is a property of those settings and not of convergence as such. Both spaces also fail to separate their points by disjoint open sets, in the second case by [L5].

step 1.1step 4.1L5∎

Remarks

Uniqueness of limits is a separation property, not a fact about sequences. In both witnesses distinct points fail to have disjoint neighbourhoods: in the indiscrete topology the only neighbourhood of any point is the whole space, and in the cofinite topology on an infinite set any two nonempty open sets meet (On an infinite set the cofinite topology has every infinite subset dense and no two nonempty open sets disjoint). Where distinct points are separated by disjoint open sets — in particular in every metric space (Distinct points of a metric space have disjoint balls around them) — the argument that a sequence cannot be eventually inside two disjoint sets restores uniqueness (A sequence in a metric space has at most one limit).
Why claim 2 is stated for a given injective sequence. The proof uses the sequence itself; infinitude of the underlying set supplies no canonical enumeration. Claim 3 avoids any selection issue by naming $N$ and the identity sequence, for which injectivity is immediate.
A sequence in the cofinite topology need not be injective to have many limits, and need not have many limits if it is not: a constant sequence converges only to its value there, since singletons are closed. Injectivity is used in exactly one place, step 1.3, to make each finite set catch only finitely many indices.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

42 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

Limit of a sequence (Wikipedia) (standard reference, not scraped)
Trivial topology (Wikipedia) (standard reference, not scraped)
Cofiniteness (Wikipedia) (standard reference, not scraped)