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I am after the simplest and most direct proofs available that $2\mathfrak{m} < 2^\mathfrak{m} $ and $ 2^\mathfrak{m} \not\le \mathfrak{m}^2 $ for infinite $ \mathfrak{m} $ given GCH but not AC.

This is in connection with proving GCH implies AC in a Kelley-Morse setting, and I am after the tersest most direct proof I can get assuming Foundation and that the theory of ordinals and cardinals have already been defined and developed to some elementary level without AC.

This is a result of Specker, and there are several articles on generalisations and more powerful results. See for example Andrés E. Caicedo which almost gives me what I am looking for.

I hope to avoid a lot of lemmas and theorems around choice and well-orders and get straight to these two results. I have not been able to figure out how to do this easily, although this paper seems close Kanamori and Pincus

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  • $\begingroup$ In ZF there is more than one "version" of GCH. One is that if $A$ is Tarski-infinite and $A\leq' B\leq' P(A)$ then $A\sim B$ or $A\sim P(A), $where $A\leq'B$ means there is an injection $f:A\to B,$ and $A\sim B$ means there is a bijection $g:A\to B,$ and $P(A)$ is the power-set of $A. $ Another is that if $K$ is an infinite cardinal ordinal then $P(K)\sim K^+.$ $\endgroup$ – DanielWainfleet Jan 9 '18 at 15:46
  • $\begingroup$ @Daniel: They turn out to be all equivalent, as they all imply the axiom of choice... What's Tarski-infinite, though? $\endgroup$ – Asaf Karagila Jan 10 '18 at 6:28
  • $\begingroup$ @AsafKaragila. Tarski-infinite is "not Tarski-finite". $\endgroup$ – DanielWainfleet Jan 10 '18 at 6:37
  • $\begingroup$ @Daniel: Yes. I can imagine that would be the definition. Let's go one step further. What is "Tarski-finite"? $\endgroup$ – Asaf Karagila Jan 10 '18 at 6:38
  • $\begingroup$ Probably means Dedekind-finite? $\endgroup$ – Mark Kortink Jan 11 '18 at 6:45

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