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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?2012-09-26
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    Pretty much every ordinal that you can "construct" is countable and hence has countable cofinality.2012-09-26
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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...2012-09-26
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    @StevenStadnicki I think ${\omega_1}$ has countable cofinality, despite being uncountable. http://en.wikipedia.org/wiki/Cofinality#Cofinality_of_ordinals_and_other_well-ordered_sets2012-09-26
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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 http://en.wikipedia.org/wiki/Regular_cardinal for the basics of regular cardinals, and why $\omega_1$ can't have countable cofinality assuming AC.2012-09-26
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    @StevenStadnicki yes, you're right, just realized. thanks2012-09-26

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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?2012-09-26
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    @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.2012-09-26
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    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.2012-09-26
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    @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.2012-09-26
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    @AsafKaragila, yes, you're right. But $\omega_1$ is the smallest ordinal with uncountable cofinality, right?2012-09-27
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    @tom4everitt: Yes, trivially too. Every ordinal below $\omega_1$ is countable, and of course that every countable ordinal has a countable cofinality! (This *does* require some choice though, as I remark in my answer.)2012-09-27
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    @AsafKaragila, cool, just wanted to make sure I had got it right. thanks again2012-09-27