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TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-12
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Balogh continuum-sized ZFC Dowker space

Statement

Assume AC. There exists a Hausdorff hereditarily normal Dowker space of cardinality 20. It is a union of countably many relatively discrete subspaces, and its product with the usual closed interval [0,1] is not normal. No CH assumption is required.

Facts & Assumptions

Given: κ=20 in ZFC.

[F1]

The proved combinatorial map defines the Balogh space on X=κ×ω (Balogh continuum topology).

[F2]

Its topology is T1 and each level Ln is relatively discrete (Balogh neighborhood basis).

[F3]

The space is hereditarily normal and Hausdorff (Balogh hereditary normality).

[F4]

It is not countably paracompact (Balogh failure of countable shrinking).

[F5]

A normal T1 space failing countable paracompactness is a Dowker space (Countable paracompactness and Dowker spaces).

[F6]

Normality of the interval product of a T1 space implies countable paracompactness of that space (Dowker product characterization).

[A1]

AC is assumed for the construction and cardinal comparisons (The Axiom of Choice).

Proof

1.1

Take the space in F1, whose defining map exists by the proved combinatorial construction. F2 and F3 under A1 give its T1, Hausdorff and hereditary-normality properties. F4 gives failure of countable paracompactness. In particular it is normal, and F5 makes it a Dowker space.

F1F2F3F4F5A1
1.2

The map α(α,0) injects κ into X. In the reverse direction the underlying set is exactly κ×ω by F1. The cardinal κ is infinite: the binary functions with a single value one at coordinate n give an injection ωω2. Thus F7 gives κ×ω=κ, under the cardinal identifications allowed by A1. This proves X=20. Also X=n<ωLn directly from its underlying set, and F2 gives relative discreteness of each level.

F1F2F7A1
2.1

If X×[0,1] were normal, F6 would apply since X is T1 by step 1.1 and force countable paracompactness, contradicting F4. Hence the product is not normal. Steps 1.1–1.2 give the remaining assertions. Every assumption used was ZFC; no equation identifying the continuum with 1 was needed. QED.

step 1.1step 1.2F4F6

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Sources