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Specifically I would like to show:

Let $S_\Omega$ be the minimal uncountable well-ordered set. For any $a\in S_\Omega$, either $a$ has an immediate predecessor in $S_\Omega$, or there exists an increasing sequence $a_i$ in $S_\Omega$ with $a=\sup\{a_i\}$.

The part that I don't understand is how to find such a sequence $\{a_i\}$. I understand that if $a$ does not have an immediate predecessor, then for any $x\in S_\Omega$ and $x<a$, there exists $y\in S_\Omega$ s.t. $x<y<a$. By this one can inductively find a sequence $\{a_i\}$ s.t. $a_0<a_1<\cdots$ and $a_i<a$ for all $i$. But how can one guarantee that $\sup\{a_i\}=a$ in this case?

I encountered this question while proving something else (Munkres Topology, 2/e p.159 12(c)).

tvk
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  • While the phrasing might be a bit different from Munkres, this is essentially the same question. – Asaf Karagila Nov 11 '14 at 22:00
  • @AsafKaragila To be honest, I do not quite understand the question of which you marked this as a duplicate. I have never learned concepts like "limit ordinal" or "cofinality". I guess even if these two questions have this same essence (which I am not in a position to judge), this one is worded in a more elementary way to make them somewhat different (if you consider their respective readerships)? – tvk Nov 11 '14 at 22:17
  • You might be right. I am heading to dreamland, but I requested Brian to weigh in, and I trust his judgement on the matter. – Asaf Karagila Nov 11 '14 at 22:21
  • @FangJing: I’m going to leave this closed, since it looks as if your question is now answered to your satisfaction. This really is a duplicate in simpler language: in this context a limit ordinal is a member of $S_\Omega$ that has no immediate predecessor, and that a member of $S_\Omega$ has countable cofinality if and only if it is the limit of an increasing sequence. Camilo’s hint in the other question is a minor variant of my sketch here. – Brian M. Scott Nov 11 '14 at 22:32
  • @BrianM.Scott Yes, you have already answered my question. That's fine. Thank you! – tvk Nov 12 '14 at 00:49
  • @FangJing: You're welcome! – Brian M. Scott Nov 12 '14 at 02:01

1 Answers1

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SKETCH: Let $B=\{x\in S_\Omega:x<a\}$. $B$ is countable, so you can enumerate it as $\{x_n:n\in\Bbb N\}$. Now make sure to choose $a_k\ge x_k$ for each $k$.

Brian M. Scott
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