CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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An infinite particular-point space is pseudocompact and not compact

Statement refuted

False claim: every pseudocompact topological space is compact.

Let $X$ be an infinite set with a distinguished point $p$ , carrying the particular-point topology. Then $X$ is pseudocompact and not compact.

Facts & Assumptions

Given: An infinite set $X$ , a point $p \in X$ , and the particular-point topology, whose nonempty open sets are exactly the subsets containing $p$ .

[A1]

Every pseudocompact topological space is compact.

[L1]

The particular-point topology is a topology and has exactly the stated open sets (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

[L2]

The usual topology on $R$ is Hausdorff, so distinct real numbers have disjoint open neighbourhoods (The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded, Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not).

[L3]

A continuous map pulls back open sets to open sets, and compactness means that every open cover has a finite subcover (Continuity of a map of topological spaces at a point and globally, Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right).

[L4]

A space is pseudocompact exactly when every continuous real-valued map has bounded image (Pseudocompact space: every continuous real-valued function has bounded image).

Counterexample

technique · contradiction

1.1

Let $f : X \to R$ be continuous. If $f (x) \neq f (p)$ for some $x$ , choose disjoint open neighbourhoods $U$ of $f (x)$ and $V$ of $f (p)$ by [L2]. Then $f^{- 1} [U]$ is open, contains $x$ , and does not contain $p$ , contradicting [L1].

L1L2L3assume-contra

1.2

The family $U : = {{p, x} : x \in X ∖ {p}}$ consists of open sets and covers $X$ .

2.1

Hence every continuous $f : X \to R$ is constant, so its image is bounded. Thus $X$ is pseudocompact by [L4].

step 1.1L4

2.2

No finite subfamily covers $X$ , because its union contains $p$ and only finitely many other points, whereas $X ∖ {p}$ is infinite. Thus $X$ is not compact.

L3step 1.2

3.1

The pseudocompact noncompact space $X$ contradicts [A1], refuting the claim.

A1step 2.1step 2.2discharge-contradiction∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

36 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

Particular point topology (standard reference, not scraped)
Pseudocompact space (standard reference, not scraped)