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I've got a most probably silly question, but I can't find the answer to it:

If $\alpha$ is a countable limit ordinal, how can we be sure that there exist ordinals $\alpha_n$ such that $\alpha = \bigcup \{\alpha_n\mid n\in\omega\}$ (rather than the usual $\alpha = \bigcup \{\beta\mid \beta<\alpha\}$)?

Cheers!

Mathman
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Fedra
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1 Answers1

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Well, first of all $\{\beta\mid\beta<\alpha\}=\alpha$ is countable, therefore it can be enumerated as $\alpha_n$'s. But I suppose that you want to ask about a strictly increasing sequence.

First enumerate $\alpha$ as $\{\beta_n\mid n<\omega\}$. Then by induction define $\alpha_n$, where $\alpha_{n+1}=\beta_k$ where $k=\min\{m\mid\alpha_n<\beta_m\}$. You can now show that this is a strictly increasing and cofinal sequence.

Asaf Karagila
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  • Thanks for the answer. I'm new to ordinals and while I know (knew) that countable things can be enumerated as $\alpha_n$'s, I am now unsure why we can do so using only $\omega$, and not, say, $\omega^2$ indices to do so. Is that the definition of countability? Sorry for all the stupid questions. – Fedra May 05 '14 at 19:19
  • Well, being countable is by definition to have an injection into $\omega$, or in other words to be enumerated. – Asaf Karagila May 05 '14 at 19:21
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    Okay, in this case I'm fine. It's just weird how I am now unsure about so many things in set theory I didn't even think I could be unsure about a few weeks ago... :-) – Fedra May 05 '14 at 19:23
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    It's the graph of learning. Wait, you'll be sure again and then unsure again and then sure again... and so on and so forth. – Asaf Karagila May 05 '14 at 19:27
  • He he, so true. Anyway, thanks heaps again for helping me now for the second time! – Fedra May 05 '14 at 19:35
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    You're welcome. I'm probably gonna help you again in the future. – Asaf Karagila May 05 '14 at 19:39