Assuming initially the background set theory ZF, what is the exact status of the existence of a well-ordering on the power set of $\omega$? How much needs to be added to guarantee this?
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asked Jun 22, 2016 at 9:10
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More or less the only you need to require it explicitly.
It is consistent with any amount of Dependent Choice that the reals cannot be well ordered. Just blow the continuum to be large enough, and preserve the amount of required Dependent Choice.
I am not sure, however that this appears explicitly anywhere in print.
The sketch of the proof is as follows:
Start with a model of $\sf ZFC+GCH$, e.g. $V=L$, fix a regular $\kappa$, and consider the forcing $\Bbb P=\operatorname{Add}(\omega,\kappa^+)$. For a permutation $\pi$ of $\kappa^+$, $\pi$ acts naturally on the conditions of $\Bbb P$ as follows $$\pi p(\pi\alpha,n)=p(\alpha,n),$$ taking the normal filter of subgroups to be generated by groups which are the identity on some set of size $\leq\kappa$.
The fact that the reals cannot be well-ordered is proved via standard symmetric arguments. To see that $\sf DC_\kappa$ holds as well, note that any name for a sequence of hereditarily symmetric names of length $\kappa$ must be symmetric as well, simply due to chain conditions and the fact that $\kappa^+$ is regular of cofinality $>\kappa$.
Therefore the symmetric extension is closed under $\kappa$-sequences in the full (choice-y) extension, and it is not hard to show this implies $\sf DC_\kappa$ holds in the symmetric extension; and as we remarked, the reals cannot be well-ordered there.
Boolean Prime Ideal theorem holds in Cohen's first model, so certainly it does not prove a well ordering of the continuum.
answered Jun 22, 2016 at 9:16
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$\begingroup$ I could add a short sketch of a proof later if you're interested. $\endgroup$– Asaf Karagila ♦Commented Jun 22, 2016 at 9:29
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$\begingroup$ Also if you're interested in other choice principles, let me know specifically. $\endgroup$– Asaf Karagila ♦Commented Jun 22, 2016 at 9:32
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2$\begingroup$ Conversely, any choice principle I can think of can fail without directly impacting the reals - just have failures occur at high rank. $\endgroup$ Commented Jun 22, 2016 at 11:56
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1$\begingroup$ Well, yes, except things like "there is a Dedekind finite set of reals", or other things more or less directly implying that the continuum is not well orderable... :-) $\endgroup$– Asaf Karagila ♦Commented Jun 22, 2016 at 12:06
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1$\begingroup$ @Andreas: And also because given any "general failure of choice" that we can [symmetrically] force, we can always [symmetrically] force it to happen above rank $\omega+1$, therefore ensuring that the reals are the same as in the ground model, and therefore can be well-ordered. $\endgroup$– Asaf Karagila ♦Commented Jun 22, 2016 at 16:06