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There is a weak choice principle called $DC_\lambda$ which holds in $L(V_{\lambda+1})$ under the assumption of a non-trivial elementary embedding $$j:L(V_{\lambda+1})\prec L(V_{\lambda+1})$$ and it is known that this choice principle is not sufficient to split the ordinals below $\lambda^+$ (a regular cardinal) which have cofinality $\omega$ into disjoint stationary sets.

  1. Is there a choice principle $\Phi$, which, when augmented with $DC_\lambda$ and strictly weaker than full AC that suffices to prove Solovay's Theorem on Partition of Stationary Sets? A little more specifically, what is the minimal $\Phi$ such that $T=ZF + DC_\lambda + \Phi$ where $$T\vdash \text{All stationary subsets of }\lambda^+\text{ have a disjoint partition into stationary sets}?$$ One such candidate could be $Unif(V_{\lambda+1}\times V_{\lambda+1})$: given any $R\subseteq V_{\lambda +1}\times V_{\lambda +1}$ there exists some function $f\subset R$ with the same domain as $R$. (This question is related to both A proposed axiom of Laver (updated) and Model of ZF + $\neg$C in which Solovay's Theorem on stationary sets fails? .)

  2. Are there other weak choice principles $\Phi$ that could be considered?

  3. Are there (perhaps) some partition properties with infinite exponents that would prohibit Solovay's Theorem?

EDIT: While I am interested in the more general question (which I believe Asaf Karagila has addressed in the comments and chat), I am really specifically interested in the context that Woodin's Axiom $I_0$ holds. Specifically, any assertion $\Phi$ I'm looking for can't imply that $[\lambda]^\omega$ is well-ordered (in conjunction with $DC_\lambda$).

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The weakest $\Phi$ is trivially "Solovay's theorem holds at $\lambda^+$". –  Asaf Karagila Nov 1 '13 at 9:01
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. $\sf GCH$). –  Asaf Karagila Nov 1 '13 at 9:09
@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? –  Everett Piper Nov 1 '13 at 10:18
@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? –  Everett Piper Nov 1 '13 at 10:25
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! :-) –  Asaf Karagila Nov 1 '13 at 10:25
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