if $\kappa; \lambda$ are infinite Scott cardinals, then: $$2^\kappa = 2^\lambda \leftrightarrow \kappa \leq \lambda < 2^\kappa \lor \lambda \leq \kappa < 2^\lambda $$ Would adding the above to axioms of $\sf ZF$ imply $\sf AC$?
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asked Jun 4 at 13:18
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$\begingroup$ By "Scott cardinal" do you mean you are identifying cardinalities with their representatives of minimal rank, or something else? $\endgroup$– Andrés E. CaicedoCommented Jun 4 at 18:37
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1$\begingroup$ Nitpick: the phrase “Scott cardinals” makes your questions unnecessarily reliant on Foundation. It’s fine here to just have $\kappa, \lambda$ be arbitrary sets. $\endgroup$– Elliot GlazerCommented Jun 4 at 18:53
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$\begingroup$ @AndrésE.Caicedo, Yes, the Scott cardinality of $x$ is the set of all sets equinumerous with $x$, of minimal rank. So a Scott cardinal is the Scott cardinality of some set. $\endgroup$– Zuhair Al-JoharCommented Jun 4 at 19:15
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$\begingroup$ @ElliotGlazer, True! But, after all I'm speaking within $\sf ZF$. To make it foundation independent we may take $\kappa, \lambda$ to be any arbitrary infinite sets, define exponentiation, and the cardinal comparison's directly on sets in terms of sets of all functions from the index (exponent) to the indexed, and in terms of injections\bijections in the usual manner. $\endgroup$– Zuhair Al-JoharCommented Jun 4 at 19:17
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