Timeline for A Question Regarding the Relation Between 0-sharp and Koepke's Bounded Truth Predicate.
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Nov 29, 2014 at 3:24 | comment | added | Jesse Elliott | Done. mathoverflow.net/questions/188339/… | |
| Nov 29, 2014 at 2:31 | comment | added | Joel David Hamkins | @JesseElliott Interesting! I'm not sure where the boundary is, and I'd suggest that you post that as an actual MO question. | |
| Nov 29, 2014 at 2:21 | comment | added | Jesse Elliott | I'm curious about a related question: What is the simplest non-constructible set of integers (say, in the analytical hierarchy) that is compatible with the nonexistence of $0^\sharp$? In particular, can there still be a non-constructible $\Delta^1_3$ set of integers in the absence of $0^\sharp$? | |
| Jun 27, 2013 at 0:01 | vote | accept | Thomas Benjamin | ||
| Jun 10, 2013 at 6:39 | comment | added | Thomas Benjamin | Actually, it should be "Koepke-Koerwien's system SO". Sorry. | |
| Jun 9, 2013 at 10:00 | comment | added | Thomas Benjamin | @Prof. Hamkins: Interesting. As regards Koepke's ordinal computability, can one define a notion of productive set in Koepke's system SO (assuming SO not-=L) analogous to the definition of productive set found in ordinary recursion theory? I ask because if one could, in analogy with ordinary recursion theory, such a set would not be able to be generated by one of Koepke's ordinal turing machines and thus a non-constructible set. How would such a non-constructible set be related to 0-sharp? | |
| Jun 7, 2013 at 12:14 | comment | added | Joel David Hamkins | And then the point would be that these $L$-generic Cohen reals are nonconstructible, but definitely much simpler than $0^\sharp$, for these reals would have lower Turing degree, lower constructibility degree, lower consistency strength and so on. | |
| Jun 7, 2013 at 10:36 | comment | added | Joel David Hamkins | With $0^\sharp$ as an oracle, you can compute $L$-generic Cohen reals, since $0^\sharp$ allows you to enumerate the dense subsets of this forcing in a countable sequence, which you can then descend through by diagonalization, building the generic real. Similarly, any other definable forcing notion in $L$, including the forcing notion I had mentioned (using the least inaccessible cardinal of $L$, which will be countable in $L[0^\sharp]$, admit $L$-generic filters that are Turing computable from $0^\sharp$. | |
| Jun 7, 2013 at 7:31 | comment | added | Thomas Benjamin | In fact, what would 0-sharp 'look like' from the perspective of ordinal computability? | |
| Jun 7, 2013 at 7:18 | comment | added | Thomas Benjamin | (ramified) forcing 'look like' from the perspective of ordinal computability? | |
| Jun 7, 2013 at 7:14 | comment | added | Thomas Benjamin | @Professor Hamkins: Thanks for the counterexamples--they are very nice! Regarding forcing: since Cohen used forcing to 'create' (would 'create' be the proper term?) nonconstructible sets, could one use forcing to 'create' nonconstructible sets that are in some sense 'simpler' than 0-sharp (I guess for want of a better definition of 'simpler', simpler in this case would mean not implying the consistency of a proper class of inaccessible cardinals)? Also, given Koepke's main theorem: a set S is constructible iff S is ordinal computable from finitely many ordinal parameters, what would | |
| Jun 6, 2013 at 23:43 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Jun 6, 2013 at 23:27 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Jun 6, 2013 at 22:19 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |