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S Apr 26, 2021 at 6:04 history bounty ended CommunityBot
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S Apr 18, 2021 at 4:35 history bounty started Alec Rhea
S Apr 18, 2021 at 4:35 history notice added Alec Rhea Draw attention
Apr 13, 2017 at 12:58 history edited CommunityBot
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Jan 8, 2017 at 14:33 comment added Franka Waaldijk Or to put it the other way round: the fact that a class-sized linear order A is embeddable in No apparently does not give a way to Ord-enumerate A. If one wants to somehow use the property `all class-sized linear orders are embeddable in No', then again, we trivially know that all class-sized linear orders isomorphic to a subclass of No are embeddable...so you would have to come up with a class-sized linear order B defined outside of No, and from B's embeddability in No deduce that No is Ord-enumerable. [Well, I hope I'm making some sense here, feel free to disregard my question/comments].
Jan 8, 2017 at 13:34 comment added Franka Waaldijk The question intrigues me although I have extremely little knowledge of such matters (but I'm pondering on variations of surreal numbers for constructive math). If I understand correctly, No itself is Ord-enumerable iff Global AC holds? (I don't know the first thing about this, but read it in another post). If that is correct, then intuitively I'd say: global AC is not necessary for UNo, because one can embed No in No without global AC... This may seem a strange argument, but it is meant to illustrate that one can have weaker conditions which already give embeddability.
Jan 7, 2016 at 7:41 comment added Joel David Hamkins Keep in mind that the surreal numbers No themselves are definable in L[G], and as elements these define every object in L[G].
Jan 7, 2016 at 6:34 comment added Asaf Karagila Also, when you appeal to UNo, do you have to shout UNo when you have one line left in your proof, or else you have to take four more lines to finish?
Jan 7, 2016 at 6:32 comment added Asaf Karagila Joel, what I'd try to do is to show that every linear ordering comes from a bounded part.
Jan 7, 2016 at 6:29 comment added Joel David Hamkins If you think about $L[G]$ in the model in my answer to your question about linear orders of $V$, then the question is whether UNo holds there. We know there is no linear order of all of $L[G]$ there, so perhaps there is a paucity of definable linear orders there (always allowing parameters). If so, then we might expect UNo to hold. But I can't quite prove it.
Jan 7, 2016 at 6:27 comment added Asaf Karagila I just opened my eyes ten minutes ago, it's my pleasure to have choice related questions to think about when I wake up. :-)
Jan 7, 2016 at 6:25 comment added Joel David Hamkins @AsafKaragila But let me say that I'm very glad you're looking at this question. I'd appreciate any insight you might have.
Jan 7, 2016 at 6:22 comment added Joel David Hamkins I think of global-AC as an assertion in second-order set theory, as in Goedel-Bernays set theory, so one doesn't need to worry about definability. That is, one simply asserts that there is a proper class choice function on all nonempty sets, or equivalently, that there is an Ord-enumeration of V. In particular, global AC definitely holds after adding a Cohen real.
Jan 7, 2016 at 6:20 comment added Asaf Karagila When you say global choice, does it come with parameters? Specifically, if you add a Cohen real to $L$, you have global choice with parameters, but not global choice itself. It stands to reason that in such model you might have what you were looking for.
Jan 7, 2016 at 6:18 comment added Joel David Hamkins I'll post my slides on Friday, and you can read all about it!
Jan 7, 2016 at 6:14 comment added Todd Trimble Hypnagogic digraph! Quite an intriguing title. Will there be forthcoming discussions of the hypnopompic digraph as well? :-)
Jan 7, 2016 at 6:09 history edited Joel David Hamkins CC BY-SA 3.0
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Jan 7, 2016 at 6:01 history asked Joel David Hamkins CC BY-SA 3.0