Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
@newaccount: No, since it is provable from ZFC that $\omega_1$ does not have the tree property. But you're asking me about what I had in mind over 8 years ago, so... (But I was aware of Ioanna's answer to my own previous question, so it's not like it I didn't know this.)
@Joel: Your argument in the blog post kinda reminds me of François' answer on my [quite] old question. mathoverflow.net/questions/151234/…, not the same, of course, but kind of a flavour of "here's a case we can prove the Rasioa–Sikorski lemma for".
@Mike: Suppose that I picked a well order whose first $\omega$ segment is actually a cofinal sequence of ordinals. This would break Replacement on sets, so you can't have that. The idea to add a generic well-order is to ensure this sort of problem doesn't happen.
$L(\Bbb R)$ seems like a very illogical place for such field to exist, if it can exist at all. I don't see why (or why not, though) a field could have countably many automorphisms (other than the finite case).
@Emil: $(\Bbb C,+,\times)$ is $2^{\aleph_0}$-categorical. And the only elements stable under all automorphisms are the rationals, which are definable from $\Bbb Z$ anyway. This is my point. Anyway, I had a chat with one of our model theorist who said that to an extent, my argument isn't far off. All that it's missing is to show some QE over the rationals. Which, again, homogeneity should be helpful to get.
@Emil: But the reals, as a field, are not categorical in their cardinality, it's not the same at all. Move to an ultra power, or a sufficiently saturated model.
@Emil: I don't see why every real number is fixed by all automorphisms of $\Bbb C$. At least assuming the Axiom of Choice, which I'm assuming is the case here. Otherwise, work in $L$ and use the fact that the whole thing is $\Pi^1_2$ or so.
@Mike: That should work. I'd say you probably also want to make sure that your well-ordering is not "too long". Since there will only be $\kappa$ many ordinals definable from $V_\kappa$, but $\kappa^+$ ways to well-order it, if you picked a well-ordering which is "too long" you might run into some non-obvious problems (although, again, I don't know enough off-hand to say that's definitely the case). But picking a well-ordering that is amenable is probably enough. Namely, you want $x\cap V_\alpha$ to be a well-ordering of $V_\alpha$ for all $\alpha<\kappa$.
