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False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31
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Assuming choice, refuted: paracompactness is hereditary

Statement

Assuming the Axiom of Choice, paracompactness is hereditary.

Facts & Assumptions

Given: The Axiom of Choice and the ordinal spaces ω1ω1+1\omega_1\subseteq\omega_1+1.

[A1]

Choice implies the countable choice used by the ordinal compactness theorem (The Axiom of Choice).

[L2]

Under choice, a countably compact paracompact Hausdorff space is compact (Assuming countable choice, every countably compact paracompact Hausdorff space is compact).

[L3]

A compact space is paracompact (Every compact space is paracompact).

[L4]

Every ordinal in its order topology is T1T_1 and Hausdorff, so each singleton is closed (Every ordinal with its order topology has a basis of clopen sets, and is T1T_1, Hausdorff and regular, clauses 2 and 3).

Refutation

technique · direct
1.1

By [A1] and [L1], ω1+1\omega_1+1 is compact, hence paracompact by [L3], and its initial segment ω1\omega_1 is countably compact but noncompact.

A1L1L3
1.2

The initial segment ω1\omega_1 is open in ω1+1\omega_1+1, since its complement is the closed singleton consisting of the top endpoint.

L4
2.1

If ω1\omega_1 were paracompact, its Hausdorffness from [L4] would let [L2] make it compact, contradicting step 1.1.

L2L4step 1.1
3.1

Thus a paracompact space has the nonparacompact subspace ω1\omega_1, which refutes the displayed hereditary assertion.

step 1.1step 1.2step 2.1

Depends on

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Sources