## V=L → ω_1=ω_1^CK ?

For the discussion of math. Duh.

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arbiteroftruth
Posts: 481
Joined: Wed Sep 21, 2011 3:44 am UTC

### V=L → ω_1=ω_1^CK ?

Does the axiom of constructibility imply that ω1 is the Church-Kleene ordinal? Intuitively, it seems so. If I understand correctly, ω1CK is the first ordinal without a *constructible* bijection to the natural numbers, and the axiom of constructibility implies that if there is no constructible bijection, then there's no bijection at all, and thus ω1CK is the first uncountable ordinal in such a set theory, which is the definition of ω1.

Is this right, or am I missing something? And if I'm missing something, would there be some similar situation that would imply this result? An "axiom of computability" for example?

Posts: 90
Joined: Fri May 14, 2010 12:49 pm UTC

### Re: V=L → ω_1=ω_1^CK ?

No. [;\omega_1^{CK};] is the limit of a countable sequence of countable ordinals, so it is countable (given the Axiom of Choice, which is implied by V=L).

[;\omega_1^{CK};] is the first ordinal without a recursive bijection to the natural numbers. But "constructible" in the axiom of constructibility does not mean "recursive" - it means "defined by a formula of ZF".

SAI Peregrinus
Posts: 16
Joined: Mon Mar 30, 2015 8:02 pm UTC

### Re: V=L → ω_1=ω_1^CK ?

ω_1 is the first uncountable ordinal.
ω_1^CK is a countable ordinal, and thus has a smaller cardinality than ω_1. It's just got somewhat confusing notation.