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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Jul 10, 2014 at 1:09 answer added Paul Larson timeline score: 3
Nov 1, 2013 at 10:42 history edited Everett Piper CC BY-SA 3.0
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Nov 1, 2013 at 10:28 comment added Everett Piper let us continue this discussion in chat
Nov 1, 2013 at 10:27 comment added Asaf Karagila What do you mean "this"?
Nov 1, 2013 at 10:27 comment added Everett Piper @Asaf. This I've never heard. Perhaps I'm just dense at this hour, but could you elaborate on your last comment?
Nov 1, 2013 at 10:25 comment added Asaf Karagila Well, you can note that this assertion follows from "$\mathcal P(\lambda^+)$ can be well-ordered" (and if not, then probably taking another power set, or five, would suffice). But I would think it's unlikely that this would be equivalent to Solovay's theorem. Therefore Solovay's theorem is weaker than this principle, which itself is vastly weaker than the axiom of choice. My point was that weakest being "proves less things", and the only thing which proves only Solovay's theorem (and its consequences) would be... Solovay's theorem! :-)
Nov 1, 2013 at 10:25 comment added Everett Piper @I guess I am looking for choice principles (weaker than full AC at $\lambda^+$ (or Solovay's theorem itself) so far considered in the literature. I'm not interested (at the moment) in getting into a general discussion concerning what constitutes a "choice principle" in general. My aim is for something like "Given a set indexed by I, there is another set (not constructed by the other usual axioms of ZF)." I realize this is vague, which is why I included the infinite exponent partition part of the question. Perhaps there is a $\lambda$-complete filter or ultrafilter idea yeilding the theorem?
Nov 1, 2013 at 10:18 comment added Everett Piper @Asaf: I think your suggestion is the strongest possible (that does not prove full AC), since it is trivially equivalent to "Solovay's theorem holds at $\lambda^+$. Am I missing something obvious?
Nov 1, 2013 at 9:09 comment added Asaf Karagila Also, what do you mean when you say a choice principle? In mathoverflow.net/questions/104016/… there are two equally reasonable definitions. One based on syntactical construction of the statement $\Phi$ as asserting the existence of a choice function for families which satisfy some condition; and the other is just any statement not provable from $\sf ZF$, but provable from $\sf ZFC$ (or sometimes even more to include statements which imply the axiom of choice, e.g. $\sf GCH$).
Nov 1, 2013 at 9:01 comment added Asaf Karagila The weakest $\Phi$ is trivially "Solovay's theorem holds at $\lambda^+$".
Nov 1, 2013 at 2:59 history asked Everett Piper CC BY-SA 3.0