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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.

Guozhen Shen
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  • What's the quantifier on $\mathfrak{p},\mathfrak{m},\mathfrak{n}$ in the quoted statement? And on the statement of the question what's the quantifier on $\mathfrak{m},\mathfrak{n}$? I guess it means $\forall\mathfrak{m},\mathfrak{n}$, but what are $\mathfrak{m},\mathfrak{n}$ supposed to denote? Cardinals (i.e. size of well-orderable sets)? – YCor Aug 17 '21 at 10:56
  • @YCor For all cardinals $\mathfrak{p}$, $\mathfrak{m}$ and $\mathfrak{n}$, $\dots$. Here by ``cardinal'' I mean the cardinality of any set (may be non-well-orderable). – Guozhen Shen Aug 17 '21 at 11:00
  • Ah, OK. (I was taught that a cardinal is an ordinal not in bijection with any smaller ordinal, so it's not the same convention.) – YCor Aug 17 '21 at 11:07
  • @YCor In the literature, well-ordered cardinals are always be denoted by $\kappa$, $\lambda$, and so on. – Guozhen Shen Aug 17 '21 at 11:12
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    @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. – Asaf Karagila Aug 17 '21 at 11:49
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    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). – Asaf Karagila Aug 17 '21 at 12:05
  • @AsafKaragila AC wasn't assumed, but indeed this allows to define cardinal as a certain collection of sets. The definition here seems to consider the quotient of the collection of sets by an equivalence relation, which seems more delicate to define properly. Of course in problems such as here, it's not important to make sense of this operation of taking quotient. – YCor Aug 17 '21 at 12:41
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    @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. – Asaf Karagila Aug 17 '21 at 12:44
  • @GuozhenShen: what is the proof of the statement from the paper? – Michal R. Przybylek Aug 17 '21 at 15:04
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    @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. – Asaf Karagila Aug 17 '21 at 16:04
  • (The reason the above doesn't generalise to answer the question is that $A^2=A$ implies that $A<A\cdot\kappa$, so there is not enough meat in the statement to deduce that $A<\kappa$.) – Asaf Karagila Aug 17 '21 at 16:06
  • @Asaf: thank you! But I think I am still missing something - why can't you use: $(2^A)^A=2^{A^2}=2^A= (2^A)^1$ to infer $A \approx 1$, i.e.~that $A$ is a singleton? – Michal R. Przybylek Aug 18 '21 at 10:49
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    @Michal: No, because $=$ is not the same as $<$. – Asaf Karagila Aug 18 '21 at 10:51
  • @Asaf, thanks, I've got it now :-) – Michal R. Przybylek Aug 18 '21 at 11:16
  • Does this answer your question? Does $2^X=2^Y\Rightarrow |X|=|Y|$ imply the axiom of choice? – Alex M. Feb 02 '23 at 16:06
  • @AlexM. I don't think that's the same question. It's about $=$ rather than $<$. – Jeremy Rickard Feb 03 '23 at 08:51

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