CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)verified 2026-07-29 (claude-fable-5)

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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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$\mathbb{N} \times \{a,b\}$ with the indiscrete topology on the second factor is limit point compact and not countably compact, so the hypothesis that singletons are closed is not decoration

Statement refuted

Refuted: that a limit point compact space is countably compact (Countably compact, Lindel"of, sequentially compact, limit point compact and $\sigma$ -compact spaces, and relatively compact subsets). The true statement carries a hypothesis: limit point compactness gives countable compactness when every singleton of the space is closed, and assuming countable choice (Compact implies countably compact, Lindel"of and limit point compact; countably compact together with Lindel"of implies compact; and, at the cost of countable or dependent choice, sequentially compact implies countably compact, countably compact implies limit point compact, and the converse holds when every singleton is closed, claim 4). The witness below satisfies every other part of that theorem's hypotheses and fails the singleton one, and it is not countably compact.

Witness. Let $\mathbb{N}$ carry the discrete topology and let $D = \{a,b\}$ with $a \ne b$ carry the indiscrete topology (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), and give

$X \;:=\; \mathbb{N} \times D$

the product topology (The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space). Then every nonempty subset of $X$ has a limit point in $X$ , so $X$ is limit point compact; the family $\{\, \{n\} \times D : n \in \mathbb{N} \,\}$ is an at most countable open cover with no finite subcover, so $X$ is not countably compact; and no singleton of $X$ is closed.

Facts & Assumptions

Given: $\mathbb{N}$ with the discrete topology, $D = \{a,b\}$ with the indiscrete topology, and $X = \mathbb{N} \times D$ with the product topology.

[L1]

The basic open sets of a binary product are the sets $U \times V$ with $U$ open in the first factor and $V$ open in the second; every open set is a union of them (The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[L2]

Every subset of $\mathbb{N}$ is open, and the open subsets of $D$ are $\varnothing$ and $D$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

[L3]

$p$ is a limit point of $A$ when every neighbourhood $N$ of $p$ satisfies $N \cap (A \setminus \{p\}) \ne \varnothing$ ; an open set containing $p$ is a neighbourhood of $p$ , and every neighbourhood of $p$ contains an open set containing $p$ (Interior, closure, boundary, exterior, derived set and isolated point in a topological space, Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open).

[L4]

A space is limit point compact when every infinite subset has a limit point in it, and countably compact when every at most countable open cover has a finite subcover; infinite means not finite (Countably compact, Lindel"of, sequentially compact, limit point compact and $\sigma$ -compact spaces, and relatively compact subsets, Finite, countably infinite, countable, uncountable, Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right, The natural numbers $\mathbb{N}$ (von Neumann)).

Counterexample

technique · direct

1.1

The nonempty open subsets of $X$ are exactly the sets $U \times D$ with $U \subseteq \mathbb{N}$ nonempty: by [L1] and [L2] every basic open set is $U \times \varnothing = \varnothing$ or $U \times D$ , and a union of sets of the second form is again of that form.

L1L2

2.1

Every nonempty $A \subseteq X$ has a limit point in $X$ . Take $(n,c) \in A$ and let $p := (n,c')$ be the point with the same first coordinate and $c' \ne c$ , which exists since $D$ has two elements. Every neighbourhood of $p$ contains an open set containing $p$ , hence by step 1.1 a set $U \times D$ with $n \in U$ , and that set contains $(n,c)$ , which lies in $A$ and differs from $p$ . So $p$ is a limit point of $A$ , and in particular every infinite subset of $X$ has one: $X$ is limit point compact.

L3L4step 1.1

2.2

The family $\{\, \{n\} \times D : n \in \mathbb{N} \,\}$ consists of open sets by step 1.1, is at most countable, and covers $X$ ; a finite subfamily is $\{n_0\} \times D, \dots, \{n_k\} \times D$ and its union misses $(m, a)$ for any $m$ different from all the $n_j$ , which exists because $\mathbb{N}$ is not finite. So $X$ is not countably compact.

L4step 1.1

3.1

No singleton of $X$ is closed: the complement of $\{(n,c)\}$ contains $(n,c')$ , and by step 1.1 every open set containing $(n,c')$ contains $\{n\} \times D$ and hence $(n,c)$ , so that complement is not open. This is the hypothesis of claim 4 of Compact implies countably compact, Lindel"of and limit point compact; countably compact together with Lindel"of implies compact; and, at the cost of countable or dependent choice, sequentially compact implies countably compact, countably compact implies limit point compact, and the converse holds when every singleton is closed, and steps 2.1 and 2.2 show that dropping it makes the implication fail.

L4step 1.1step 2.1step 2.2∎

Remarks

What the witness does and does not separate. It separates limit point compactness from countable compactness, and it does so for a reason that is entirely about separation of points: each point has a partner that no open set can distinguish it from, so every point of the space is a limit point of every set containing its partner. Limit point compactness is then satisfied for free.

The space is a product of two very simple spaces, and each factor contributes one half of the behaviour: the discrete factor supplies the countable open cover with no finite subcover, and the indiscrete factor supplies the partners that make every nonempty set have a limit point.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 90 results over 27 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

Limit point compact (Wikipedia) (standard reference, not scraped)
Countably compact space (Wikipedia) (standard reference, not scraped)

N×{a,b}\mathbb{N} \times \{a,b\}N×{a,b} with the indiscrete topology on the second factor is limit point compact and not countably compact, so the hypothesis that singletons are closed is not decoration