Not so much, I don't think. But that's a good question!

@Lorenzo Send me an email. This question has a long answer.

Thanks, Ali! I've fixed the link to the preprint in the last paragraph, it somehow ended up mangled.

The Feferman–Levy model does not satisfy $V=L(\Bbb R)$, whereas the model in Farmer's answer does. Truss' model, mentioned in Farmer's answer, is $L(\Bbb R)$ of the Feferman–Levy model, but it is distinct from it. In both $\omega_1$ is singular, but only in one of them (the Feferman–Levy model) the reals are a countable union of countable sets.

@Ali: The Truss model is of the form $L(\Bbb R)$, whereas the Feferman–Levy is not, in short.

@Ali: That's a misconception I had for year. But I've realised the key difference is that Truss takes all the bounded reals, and in that model the reals are not a countable union of countable sets, whereas the Feferman–Levy model has the decomposition of the reals in the model. They are very close and fairly similar, but they are different.

But what counts as "2023"? Is it the time of announcement? arXiv preprint? Submission? Revision? Acceptance? Online publishing? Publication in print? These can all be quite different from each other.

@KConrad: Yes, I know several other versions of this without ultrapowers. Compactness (for countable languages, to make it choice free), absoluteness arguments to move to $L$ and do it there, which generally isn't technically an ultrapower argument in some cases, or so on. My point in this argument is to show interesting and unusual uses for logic and model theory. Doing it "without" kinda defeats the purpose...

This works for me $\&$ ($\&$). I suspect that the reason is that you've copy-pasted it out of some PDF or whatnot, so the symbols are actually not the ASCII ones, but somewhere up there in unicode-land instead, so MathJax isn't sure what to make of the escaped symbol. If you re-type it, it seems to work.

Sure. John Truss, Models of set theory containing many perfect sets. Ann. Math. Logic 7, 197-219 (1974) (ZBL0302.02024).

The model you described is due to Truss, by the way.

You seem to have misread my comment, Joel. :-)

Feels a bit convoluted, to be honest. "A set can be linearly ordered if and only if for every two points in the set we can choose a linear ordering which puts one above the other", although admittedly not as contrived.

karagila.org/2013/infinite-dimensions-and-the-axiom-of-choic‌​e and math.stackexchange.com/questions/300494/…

Something needs to be clarified here. Just a real which is not set generic? Take any real which codes a generic for $\operatorname{Col}(\omega,V)$ and that's not going to be set generic. Why is there one? Well, why is there one which codes the universe in the Jensen case? We're working over set-models (otherwise working in $V=L$ there's no real which is not generic for the trivial forcing over $L$), so those set-models might as well be countable.

The link is dead. Is there an alternative source?