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GCH for alephs means the statement that, for any aleph $\kappa$, there are no cardinals $\mathfrak{r}$ such that $\kappa<\mathfrak{r}<2^\kappa$.

Does GCH for alephs imply the axiom of choice?

Remark. Lindenbaum and Tarski assert in ``Communication sur les recherches de la théorie des ensembles'' without proof that GCH for alephs is equivalent to Cantor's aleph hypothesis that $2^{\aleph_\alpha}=\aleph_{\alpha+1}$ for all ordinals $\alpha$ (see page 188, No. 96). But as we know (although Lindenbaum and Tarski possibly do not kown) Cantor's aleph hypothesis implies AC. This means that Lindenbaum and Tarski in fact also assert that GCH for alephs implies the axiom of choice. But I do not see how to prove it.

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The answer is positive, yes.

Note that $2^\kappa\leq 2^{\kappa^+}$, and therefore $\kappa^+\leq\kappa^++2^\kappa\leq 2^{\kappa^+}$. So either $2^\kappa=2^{\kappa^+}$ or $2^\kappa=\kappa^+$.

In the first case $\kappa^+$, $\kappa<\kappa^+<2^\kappa$ is impossible. So the latter case holds.

Therefore the power set of an ordinal can be well-ordered, and the Axiom of Choice follows in $\sf ZF$.

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    $\begingroup$ How do you deduce $2^k=2^{k^+}$ from $k^++2^k=2^{k^+}$? $\endgroup$
    – Wojowu
    Commented Aug 17, 2021 at 8:16
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    $\begingroup$ math.stackexchange.com/a/30339/622 $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 17, 2021 at 8:17
  • $\begingroup$ I believe this argument appears in Levi's "Basic Set Theory" and there are some historical remarks too. $\endgroup$
    – Pace Nielsen
    Commented Aug 17, 2021 at 14:26
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    $\begingroup$ @PaceNielsen I also like Azriel's book very very much. There is only the proof of GCH${}\Rightarrow{}$AC in that book, although the proof given by Asaf is similar to that proof. $\endgroup$
    – Guozhen Shen
    Commented Aug 18, 2021 at 2:25
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    $\begingroup$ @SarcasticSully: In ZF the statement "The power set of an ordinal can be well-ordered" implies the Axiom of Choice. Yes. It is not the naive and "obvious" proof that one can think of, and in fact the proof relies on the Axiom of Regularity in a nontrivial way. $\endgroup$
    – Asaf Karagila ♦
    Commented Jun 7 at 23:02

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