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Jan 14, 2013 at 2:27 history edited 喻 良 CC BY-SA 3.0
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Jan 14, 2013 at 1:31 history edited 喻 良 CC BY-SA 3.0
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Jan 13, 2013 at 21:33 comment added 喻 良 I fixed an error in the proof. To show that $x$ is not hyperarithemtic in $N$, we really need Leo's proof.
Jan 13, 2013 at 21:28 history edited 喻 良 CC BY-SA 3.0
I fixed an error in the proof.
Jan 13, 2013 at 21:16 comment added Peter Gerdes Ok, I see how this works. Thanks, Now to figure out what the heck Leo meant in his notes to see if this is the same thing (he couldn't remember what he had meant in his notes when I asked him about it a few days ago). I think this is probably what he meant even if the notes sound more like they only use the nonstandard model to produce a tree with special paths.
Jan 13, 2013 at 9:23 history edited 喻 良 CC BY-SA 3.0
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Jan 13, 2013 at 0:49 history edited 喻 良 CC BY-SA 3.0
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Jan 12, 2013 at 23:54 comment added 喻 良 $\Pi^0_1$-singletoness is not a absoluteness notion among the $\omega$-models. Also you have to apply Gandy's basis to get a model not the singleton. I added more details.
Jan 12, 2013 at 23:53 history edited 喻 良 CC BY-SA 3.0
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Jan 12, 2013 at 23:29 comment added Peter Gerdes Well, it's not quite appropriate to say that Leo obtained a non-standard $\Pi^0_1$ singleton that is not hyperarithmetic. Rather in transitioning to the non-standard one obtains a single tree all of whose paths are not hyperarithmetic (indeed subgeneric so any degree computable from $0^{(\alpha)}$ and $g^{(\beta)}$ is also computable from $0^{(\beta)}$). By basic results one couldn't have a $\Pi^0_1$ singleton that failed to he hyperarithmetic. But I see what got me! I failed to use Gandy Basis theorem to yield a path $f$ through $T$ which leaves $\omega_1^f=\omega_1^{ck}. More later.
Jan 12, 2013 at 14:43 history edited 喻 良 CC BY-SA 3.0
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Jan 12, 2013 at 14:38 history answered 喻 良 CC BY-SA 3.0