# Let $(\alpha_\xi\mid\xi<\kappa)$ be a sequence such that $\{\alpha_\xi\mid\xi<\kappa\}=\alpha$. Find an increasing subsequence that has limit $\alpha$

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!

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

Let K = { a$$_t$$ : t < $$\kappa$$ }.
s$$_u$$ = min{ a$$_t$$ in K : for all j < u, s$$_j$$ < a$$_t$$ }.