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Countable choice makes omega-one regular
Statement
In ZF plus , .
Facts & Assumptions
Countable unions of at most countable sets, assuming : Under countable choice, a countable union of at most countable sets is at most countable.
; and ; for a limit ordinal the value is an infinite cardinal with , so it is regular; and every cofinal subset of has cardinality at least , a value that is attained: The cofinality of a limit ordinal is an infinite cardinal at most that ordinal.
Proof
Given: The objects and hypotheses in the statement.
If , it must be omega: it is an infinite cardinal and omega is the only countable infinite initial ordinal. Thus there is a cofinal sequence in .
Each is a countable ordinal. Cofinality gives (or use without changing the argument). Countable choice makes that union countable, contradicting the definition of . The only remaining cofinality is .
Depends on
- Countable unions of at most countable sets, assuming $\mathrm{AC}_\omega$
- $\operatorname{cf}(\alpha) \le \alpha$; $\operatorname{cf}(0) = 0$ and $\operatorname{cf}(\alpha + 1) = 1$; for a limit ordinal $\lambda$ the value $\operatorname{cf}(\lambda)$ is an infinite cardinal with $\operatorname{cf}(\operatorname{cf}(\lambda)) = \operatorname{cf}(\lambda)$, so it is regular; and every cofinal subset of $\lambda$ has cardinality at least $\operatorname{cf}(\lambda)$, a value that is attained
Used by
Nothing in the library uses this result yet.
Dependency tree · two levels
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Sources
- Jech, The Axiom of Choice, §2.4.2, Corollary 2, p.20 (standard reference, not scraped)