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In his answer to user33038's mathoverflow question "What axioms are stronger than the Axiom of choice?", Prof. Hamkins writes:

"What's more, the axiom of choice is equivalent over $ZF$ to the assertion "there are unboundedly many strong limit cardinals."

My question is simply this: Is there an analogous assertion (meaning stated in terms of strong limit cardinals) to "there are unboundedly many strong limit cardinals" that is equivalent to Global Choice over $ZF$?

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  • $\begingroup$ Why do you expect that there will be such a way? $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 22, 2015 at 20:15
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    $\begingroup$ Without having any expertise in this area, I'd guess that such a statement cannot exist, because ZF+gAC is conservative over ZFC so that any statement which only speaks about sets cannot be equivalent to gAC, only to AC. But of course, this relies on my interpretation that the clause "stated in terms of strong limit cardinals" refers to a statement not containing any class terms. $\endgroup$
    – Johannes Hahn
    Commented Aug 22, 2015 at 20:31
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    $\begingroup$ If one thinks of global AC in the Godel-Bernays sense, then this is a second-order assertion, and Johannes's remark about conservativity is correct. But in the first-order ZFC context, where one has only definable classes, another reasonable interpretation for what "global AC" could mean is that "$\exists A\subset\text{Ord}\ V=\text{HOD}[A]$," which holds just in case there is an actual definable well-ordering of the universe using set parameters. In this case, there would be no conservativity result. $\endgroup$
    – Joel David Hamkins
    Commented Aug 22, 2015 at 20:53
  • $\begingroup$ Here is a link to the answer that Thomas mentions: mathoverflow.net/a/117807/1946. $\endgroup$
    – Joel David Hamkins
    Commented Aug 22, 2015 at 20:54
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    $\begingroup$ Yes, because it has a definable well-ordering of the universe, and no parameters needed. The assertion V=HOD is equivalent to saying that there is a parameter-free definable well-ordering of the universe. $\endgroup$
    – Joel David Hamkins
    Commented Aug 23, 2015 at 7:33

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