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Peter Gerdes's user avatar
Peter Gerdes's user avatar
Peter Gerdes's user avatar
Peter Gerdes
  • Member for 12 years, 7 months
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Are Cohen Generics Minimal Covers?
@JoelDavidHamkins Thanks, I was getting a bit confused because it ends up being true that exactly the opposite is true effectively (every n generic is a Turing minimal cover of an n-generic..n >1) but I presume this is basically the non-atomless case ruled out by the stronger closure conditions in full set theoretic forcing.
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Are Cohen Generics Minimal Covers?
I was thinking that if you knew why that claim was true off the top of your head you'd respond and otherwise you'd ignore it. Sorry if it felt like I was expecting you to solve my problem. Anyway, I finally managed to download something from libgen that was labeled right and figured it out. Thanks for the citatoin
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Are Cohen Generics Minimal Covers?
Unfortunately, the part I'm having trouble with is the very first theorem. That if a is cohen generic and 0 < x < a then x is of cohen generic degree. The paper cites to Jech but I only have the 3rd edition and I don't think it preserved the numbers since the cited results don't seem obviously related.
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Complexity of |a| < |b| for ordinal notations?
Thanks, that's what I needed to understand your answer.
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Complexity of |a| < |b| for ordinal notations?
So I looked up that lemma and I don't see where it says the bound is infinitely often correct. Also we can write the relation and it's negation on O^2 in a $\Sigma^1_1$ way so whether there is a $\Delta^1_1$ way to write it comes down to the question of whether we can ensure those two sets are disjoint outside off of O^2 and it's not clear to me why that's impossible. I guess the problem is for two notations that look like 'infinite' ordinals so intuitively I tend to agree but I'm not totally sure.
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Name For Effective Cantor-Bendixsonish Derivitive
In $\omega^{< \omega}$ the result won't be fully pruned (tho when I first published the Tex wasn't rendering). We don't have compactness. But maybe $n$- pruning isn't bad.
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Direct construction of an arithmetically high degree below $0^{(\omega)}$
Clarified sentence in 2nd to last paragraph. Balanced parens
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In what respect are univalent foundations "better" than set theory?
Thanks, that's interesting. However, I don't think anyone doubts it's interesting mathematics and like most interesting math it can help motivate/understand proofs in other areas but I think that's different than answering the question of whether it's offering something valuable as an alternate foundations or if it's just another one of the many interesting mathematical areas of study.
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In what respect are univalent foundations "better" than set theory?
Mathematicians are utterly comfortable with the idea that one can view some objects differently or abstract away from particular questions of representation when they don't matter. Practically, what's important for the average mathematician for foundations is that in the rare situation they need to use it it's as intuitive as possible and comprehension and sets are intuitive while keeping track of types gets ugly in exactly the cases you need to turn to foundations to verify your reasoning.
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Properties of all relatively computable branches
But I guess you can save the argument but instead extending them with a perfect tree whose even levels branch and odd levels encode 0'.
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Properties of all relatively computable branches
Wait, I'm confused. The tree T you built doesn't seem to be perfect since the leaf nodes get extended with a single path.
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