False statementConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14

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ZF proves that omega one is regular

Statement refuted

ZF proves $cf (ω_{1}) = ω_{1}$ ; equivalently, ZF proves that $ω_{1}$ is regular.

Assuming $Con (Z F)$ , this statement is false: it is not a theorem of ZF.

Facts & Assumptions

Given: $Con (Z F)$ . The conclusion is conditional syntactic nonprovability, not the assertion of a transitive model from bare consistency.

[F1]

Relative consistency of the Feferman–Levy choice failures over ZF proves consistency of ZF with $cf (ω_{1}) = ω$ .

[F2]

Cofinality $cf (α)$ , and regular and singular cardinals defines an infinite cardinal $κ$ to be regular exactly when $cf (κ) = κ$ .

[F3]

$ω_{1}$ is uncountable, every ordinal below it is at most countable, it is a cardinal and a limit ordinal, and its existence is a theorem of ZF proves in ZF that $ω_{1}$ is uncountable while $ω$ is countable, so $ω \neq ω_{1}$ .

Proof

technique · contradiction with the consistent Feferman--Levy target theory

1.1

Let $T$ be the consistent theory supplied by F1. It contains ZF and the exact equality $cf (ω_{1}) = ω$ . By F3, the ZF part of $T$ proves $ω \neq ω_{1}$ . The ordinals here are the target model's own $ω$ and $ω_{1}$ ; no ground-model ordinal is being substituted.

F1F3

Boundary check. The displayed cofinality is neither the empty nor a finite cofinality: its value is the infinite ordinal $ω$ . The possible degenerate equality $ω = ω_{1}$ is ruled out inside ZF by F3. Thus the contradiction below compares exact ordinal endpoints and does not use a Choice-based cardinal comparison.

2.1

Suppose for contradiction that ZF proved the statement refuted. By F2, $T$ would then prove $cf (ω_{1}) = ω_{1}$ . Together with step 1.1 it would prove $ω = ω_{1}$ , contradicting the ZF theorem recorded there. This would make $T$ inconsistent, contrary to F1. Hence, under $Con (Z F)$ , regularity of $ω_{1}$ is not provable in ZF.

F1F2step 1.1assume-contradischarge-contradiction∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

23 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

Thomas Jech, The Axiom of Choice, discussion after Theorem 10.6 and Problems 2–3, printed pp. 144, 148 (standard reference, not scraped)