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How to construct an ordinal with uncountable cofinality? All the very "large" ordinals I can think of, such as $\omega_\omega^{\omega_\omega}$, still seem to have countable cofinality. I need a better intuitive sense of what such a large ordinal can be.

Relevant links: http://en.wikipedia.org/wiki/Cofinality#Cofinality_of_ordinals_and_other_well-ordered_sets http://en.wikipedia.org/wiki/Ordinal_number

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It's difficult to really get a sense for uncountable cofinalities, because by definition there's no (integer-)indexed sequence 'leading up to' that ordinal, but why not simply $\omega_1$, the ordinal of all countable ordinals ordered by set inclusion? –  Steven Stadnicki Sep 26 '12 at 21:44
    
Pretty much every ordinal that you can "construct" is countable and hence has countable cofinality. –  Levon Haykazyan Sep 26 '12 at 21:47
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@LevonHaykazyan: It's hard for me to conceive a sensible notion of constructibility in which $\omega_\omega^{\omega_\omega}$ is constructible, but $\omega_1$ isn't... –  tomasz Sep 26 '12 at 22:00
    
@StevenStadnicki I think ${\omega_1}$ has countable cofinality, despite being uncountable. en.wikipedia.org/wiki/… –  tom4everitt Sep 26 '12 at 23:06
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@tom4everitt Assuming the Axiom of Choice, $\omega_1$ has cofinality precisely $\omega_1$; its cardinal $\aleph_1$ is a so-called regular cardinal. See en.wikipedia.org/wiki/Regular_cardinal for the basics of regular cardinals, and why $\omega_1$ can't have countable cofinality assuming AC. –  Steven Stadnicki Sep 26 '12 at 23:10

1 Answer 1

For every ordinal $\alpha$ consider $\alpha+\omega_1$ (ordinal addition). Note that if $\alpha$ is countable (or finite) then the sum is equal to $\omega_1$ which has uncountable cofinality by the virtue of being a regular cardinal. In fact this trick works with any regular uncountable cardinal.

You can always use uncountable cardinals (with uncountable cofinality) as indices, e.g. $\omega_{\omega_{\omega_1}}$

By the way, if you feel that you can construct $\omega_\omega$ which is pretty uncountable, you already have many ordinals with uncountable cofinalities below it.


One note on constructive-ness of ordinals with uncountable cofinality, it requires some choice to prove there exists an ordinal with an uncountable cofinality, since it is consistent with ZF that there are none.

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Why do you restrict $\alpha$ to uncountable ordinals? –  tomasz Sep 26 '12 at 22:01
    
@tomasz: Nothing particular, really. Just with countable ordinals you get $\omega_1$ which seems a bit... trivial? I suppose I will edit that in the answer. –  Asaf Karagila Sep 26 '12 at 22:02
    
So $\omega_1$ has uncountable cofinality? Perfect, I think I assumed it didn't since $\omega_\omega$ doesn't (which is much larger), but I see it now, I think. Thanks a lot. –  tom4everitt Sep 26 '12 at 23:16
    
@tom4everitt: Cofinality is not monotone, every $\omega$-many ordinals it returns to $\omega$, and every $\omega_1$ is [necessarily] hits $\omega_1$ again, and so on. –  Asaf Karagila Sep 26 '12 at 23:17
    
@AsafKaragila, yes, you're right. But $\omega_1$ is the smallest ordinal with uncountable cofinality, right? –  tom4everitt Sep 27 '12 at 0:00

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