Let $\alpha$ be a limit ordinal which is not a cardinal, and $\kappa=|\alpha|$. Then there exists a bijection from $\kappa$ to $\alpha$, or equivalently, a one-to-one sequence $\langle \alpha_\xi \mid \xi< \kappa \rangle$ of ordinals such that $\{\alpha_\xi \mid \xi< \kappa\}=\alpha$.

My textbook states:

Now we can find (by transfinite recursion) a subsequence which is strictly increasing and has limit $\alpha$.

I have tried but to no avail in constructing such subsequence, please shed me some lights.

Thank you so much!

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    $\begingroup$ Pick one, then continue picking the least indexed ordinal which is strictly larger. $\endgroup$ – Asaf Karagila Dec 27 '18 at 7:37
  • $\begingroup$ @AsafKaragila. He means ordinal sequence. $\endgroup$ – William Elliot Dec 27 '18 at 8:39
  • $\begingroup$ Hi @AsafKaragila, is it correct that this subsequence is cofinal in $\alpha$? $\endgroup$ – Le Anh Dung Dec 27 '18 at 8:57
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    $\begingroup$ @William: Yes, and? Your answer is different how? $\endgroup$ – Asaf Karagila Dec 27 '18 at 9:19
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    $\begingroup$ @LeAnhDung: Why wouldn't it be? $\endgroup$ – Asaf Karagila Dec 27 '18 at 9:19

Let K = { a$_t$ : t < $\kappa$ }.

Define, using strong transfinite induction
s$_u$ = min{ a$_t$ in K : for all j < u, s$_j$ < a$_t$ }.


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