my question is the following: I have a set $X$ $\lambda^+$-ordered (for $\lambda>\omega$) and I want to "extract" a subset $Y$ $\lambda$-ordered. Do I need AC to choose $\lambda$ different elements of $X$?
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No. By definition there is a bijection between $X$ and $\lambda^+$. Since $\lambda$ is a subset of $\lambda^+$ we can restrict this bijection.
If, however, you want to prove that the supremum of any $A\subseteq\lambda^+$ such that $|A|\leq|\lambda$ is below $\lambda^+$, then you will need to use the Axiom of Choice to choose an injection from each $\alpha\in A$ into $\lambda$. But this is a different question.
Asaf Karagila
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So, $\gamma$-ordered means ordered in type $\gamma$? – Andrés E. Caicedo Oct 22 '19 at 19:42
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Presumably, I think. In either case the title suggests that there is a bijection with $\lambda^+$ involved. – Asaf Karagila Oct 22 '19 at 19:43