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Let $\gamma=\sup\{\xi_n\}=\{\xi_n<\omega_1: n\in \omega\}$. How to prove $\gamma < \omega_1$?

Thanks for your help.

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up vote 3 down vote accepted

Since $\gamma=\bigcup_{n\in\omega}\xi_n$, it’s the union of countably many countable sets and is therefore countable; thus, it must be less than $\omega_1$.

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@Brain, I might be wrong, but I think $\gamma$ is a number... – Easy Apr 17 '13 at 2:46
@Easy: $\gamma$ is an ordinal number, which means that it is the set of smaller ordinals. – Brian M. Scott Apr 17 '13 at 2:47
IMO, it is important in these foundational set-theoretic settings to indicate use of choice principles; you used some form of countable choice. – Lord_Farin Apr 17 '13 at 10:16
@Lord_Farin: This isn’t particularly foundational, and in any case everything in sight is well-ordered, so no choice is required. (Besides, my default set theory is ZFC unless AC is explicitly an issue.) – Brian M. Scott Apr 17 '13 at 10:21
It does not matter that everything is well-ordered. One still has to choose bijections $\xi_n \to \Bbb N$ for each $n$. Both Wikipedia and T.Jech (in Set Theory) agree that some choice principle is necessary to show $\aleph_1$ is regular. – Lord_Farin Apr 17 '13 at 10:28

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