Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Thanks for your help.

share|improve this question

1 Answer 1

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 $\omega_1$.

share|improve this answer
    
@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
1  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.