# Direct proofs $2m < 2^m$ and $2^m \not\le m^2$ for infinite cardinals

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

• 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 $P(K)\sim K^+.$ – DanielWainfleet Jan 9 '18 at 15:46
• @Daniel: They turn out to be all equivalent, as they all imply the axiom of choice... What's Tarski-infinite, though? – Asaf Karagila Jan 10 '18 at 6:28
• @AsafKaragila. Tarski-infinite is "not Tarski-finite". – DanielWainfleet Jan 10 '18 at 6:37
• @Daniel: Yes. I can imagine that would be the definition. Let's go one step further. What is "Tarski-finite"? – Asaf Karagila Jan 10 '18 at 6:38
• Probably means Dedekind-finite? – Mark Kortink Jan 11 '18 at 6:45