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What is the weakest axiom $X$ that we can add to $ZFC$ to enable $GCH$ to be proven? Obviously we could just add $GCH$ but because $ZF\vdash GCH\implies AC$ that makes $AC$ redundant. So $X$ would have the properties that $ZF\vdash AC+X\implies GCH$ but $ZF\nvdash GCH\implies X$. Is such an $X$ known?

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  • $\begingroup$ You think that in ZF that GCH implies AC (and thus also that every set can be well ordered ) . Where did you get that from ? $\endgroup$ – StuartMN Feb 21 '18 at 6:27
  • $\begingroup$ Should the last part be $ZF \nvdash X\implies GCH$? Because I don't see how $GCH\implies X$ would imply $X\implies AC$. And if $X$ does not imply $AC$ then $X$ does not make $AC$ redundant. $\endgroup$ – celtschk Feb 21 '18 at 6:27
  • $\begingroup$ @StuartMN: Correct me if I'm wrong, but I think GCH implies that the power set of a well-orderable set is well-orderable. But the von Neumann hierarchy is built on power sets, starting with a finite (and thus well-orderable) set. And clearly any subset of a well-orderable set is also well-orderable. Thus unless I'm missing something, GCH should indeed imply the well-ordering theorem, which is equivalent to AC. $\endgroup$ – celtschk Feb 21 '18 at 6:34
  • $\begingroup$ @StuartMN This stackexchange question gives good explanations of the fact that $\mathsf{GCH}$ implies $\mathsf{AC}$. $\endgroup$ – Hayden Feb 21 '18 at 6:44
  • $\begingroup$ Thanks ,Celtschk and Hayden , I didn't know this $\endgroup$ – StuartMN Feb 21 '18 at 9:01
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I think that your question is inherently problematic. We can't quite say "weakest".

It can be $V=L[A]$ for some $A\subseteq\omega_1$, but that's not right since we have models of $\sf GCH$ which are not of this form (e.g. start with $L$ and add a single Cohen subset to some regular $\kappa>\omega_1$, or to all of them).

Requiring $\sf HOD$ is too weak, since it is known that $\sf HOD$ need not imply $\sf GCH$, but we can do all sort of crazy shenanigans and still have $\sf GCH$ around.

We can also always force $\sf GCH$ by collapsing cardinals. Perhaps a proper class of them. Which can completely mess up most "nice definability axioms".

So you end up with $X$ being a very nondescript axiom, which may or may not be just $\sf GCH$ itself again, or something equivalent to it (e.g. $\lozenge$ exists for all regular cardinals+$\sf SCH$ or something like that).

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  • $\begingroup$ Thanks Asaf, enlightening as always. $\endgroup$ – Mark Kortink Feb 21 '18 at 21:21

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