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Countable choice makes omega-one regular

Statement

In ZF plus ${A C}_{ω}$ , $cf (ω_{1}) = ω_{1}$ .

Facts & Assumptions

[F1]

Countable unions of at most countable sets, assuming ${A C}_{ω}$ : Under countable choice, a countable union of at most countable sets is at most countable.

[F2]

$cf (α) \leq α$ ; $cf (0) = 0$ and $cf (α + 1) = 1$ ; for a limit ordinal $λ$ the value $cf (λ)$ is an infinite cardinal with $cf (cf (λ)) = cf (λ)$ , so it is regular; and every cofinal subset of $λ$ has cardinality at least $cf (λ)$ , 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.

1.1F2

If $cf (ω_{1}) < ω_{1}$ , it must be omega: it is an infinite cardinal and omega is the only countable infinite initial ordinal. Thus there is a cofinal sequence $(α_{n})_{n < ω}$ in $ω_{1}$ .

2.1F1F2step 1.1∎

Each $α_{n}$ is a countable ordinal. Cofinality gives $ω_{1} = ⋃_{n} α_{n}$ (or use $α_{n} + 1$ without changing the argument). Countable choice makes that union countable, contradicting the definition of $ω_{1}$ . The only remaining cofinality is $ω_{1}$ .

Depends on

Dependency tree · 2 levels

Sources

Jech, The Axiom of Choice, §2.4.2, Corollary 2, p.20