CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedaudited 2026-09-22

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Kelley's cofinite set is not closed

Statement refuted

The following two related claims both fail, but they are not equivalent instances of one claim:

in the cofinite space on an infinite set $A$ , every infinite subset with infinite complement is closed; and
the coordinate set $A$ is closed in the cofinite topology on $A \cup {\infty}$ when $A$ is infinite (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).

Counterexample

Take $A = N$ with the cofinite topology (The natural numbers $N$ (von Neumann), Finite, countably infinite, countable, uncountable) and let $E \subseteq N$ be the set of even naturals. For the second claim, give $Y = N \cup {\infty}$ its cofinite topology.

Facts & Assumptions

Given: The cofinite spaces on $A = N$ and $Y = N \cup {\infty}$ , and the set $E$ of even naturals.

[F1]

In the cofinite topology on a set $S$ , the open sets are $\emptyset$ and the sets with finite complement, and the closed sets are $S$ and the finite subsets; hence every finite set, in particular every singleton, is closed (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, 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, $T_{0}$ (Kolmogorov) and $T_{1}$ (Frechet) spaces).

[F2]

The repaired coordinate of The isolated-point repair of Kelley's choice space is a different space: there $A$ is closed because the added point is isolated, which is why the cofinite presentation on $A \cup {\infty}$ is not the coordinate used in the product argument.

Verification

1.1

$E$ is infinite: the map $k \mapsto 2 k$ is injective from $N$ onto $E$ , so $E$ is countably infinite.

givenF1

1.2

$N ∖ E$ , the set of odd naturals, is infinite: $k \mapsto 2 k + 1$ is injective from $N$ into it, so it is not finite.

given

1.3

In the cofinite space $Y$ , the coordinate set $A = N$ is not closed. Indeed, its complement is the singleton ${\infty}$ ; this set is nonempty but is not open because its complement $A$ is infinite.

givenF1

2.1

$E$ is not closed: if $E$ were closed then its complement $N ∖ E$ would be open. It is nonempty because $1$ is odd, and it is not cofinite because its complement $E$ is infinite by step 1.1. Thus $N ∖ E$ is neither empty nor cofinite, contrary to [F1].

step 1.1step 1.2F1

3.1

Step 2.1 refutes the first claim using an infinite subset whose complement is infinite, whereas step 1.3 separately refutes the coordinate claim using a subset whose complement is finite. The isolated-point repair of [F2] meets the latter closedness obligation by making ${\infty}$ open.

step 2.1step 1.3F2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Kyriakos Keremedis and Eleftherios Tachtsis, Wallman Compactifications and Tychonoff's Compactness Theorem in ZF (standard reference, not scraped)