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Note that $2^\mathfrak{m}<2^\mathfrak{n}\Rightarrow\mathfrak{m}<\mathfrak{n}$ follows from the axiom of choice. Is it equivalent to the axiom of choice?

A similar question.

Remark. Lindenbaum and Tarski assert in ``Communication sur les recherches de la théorie des ensembles'' without proof that $\mathfrak{p}^\mathfrak{m}<\mathfrak{p}^\mathfrak{n}\wedge\mathfrak{p}\neq0\Rightarrow\mathfrak{m}<\mathfrak{n}$ is equivalent to the axiom of choice (see page 186, No. 82). This is not difficult to prove, but I do not know whether the statement for $\mathfrak{p}=2$ is also equivalent to the axiom of choice.

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    $\begingroup$ @YCor: You were taught that in the context of ZFC. In the context of ZF it is not helpful to restrict yourself to $\aleph$ numbers if you want to talk about cardinal arithmetic in general. $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 17, 2021 at 11:49
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    $\begingroup$ This is the type of question that is either deceptively easy once you know a trick; or incredibly hard (which implies the answer may very well be negative; but we do not have any actual tools for constructing counterexamples). $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 17, 2021 at 12:05
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    $\begingroup$ @YCor: Well, that's a shame. Many people stick to this definition quite religiously for some reason, even though it doesn't make any sense. If you want to talk about cardinal arithmetic like that, then in ZF exponentiation does not exist, which is kind of silly. On the other hand, Scott's trick provides us with a very elegant solution as to why the cardinals are themselves sets and why their arithmetic is well-defined. It's almost as shameful as how people insist to not teach symmetric systems and just focus on $L(A)$ and such for the proof that AC is not provable from ZF. But I digress. $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 17, 2021 at 12:44
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    $\begingroup$ @Michal: Let $X$ be any non-empty set; $A=X^\omega$ and $\kappa=\aleph(2^A)$ is the Hartogs number of $2^A$. Note that $A^2=A$. Take $\frak p$ to be $2^A$, we get: $$(2^A)^A=2^{A^2}=2^A<2^{A\cdot\kappa}=(2^A)^\kappa,$$ and therefore $A<\kappa$, so $A$ can be well-ordered and therefore $X$ can be well-ordered. $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 17, 2021 at 16:04
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    $\begingroup$ @Michal: No, because $=$ is not the same as $<$. $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 18, 2021 at 10:51

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