Timeline for Definition of HYP in $L_{\omega_1^{CK}}[a]$?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 18, 2013 at 3:50 | comment | added | Joel David Hamkins | Yes, every set in $L_{\omega_1^{CK}}$ is countable, and is coded by a hyperarithmetic real code. Clearly every ordinal there is countable there, since these ordinals are all the order types of computable relations on $\omega$. It follows that the corresponding $L_\alpha$'s are also countable there. | |
| Sep 18, 2013 at 2:30 | comment | added | catrincm | @Joel, do you mind explaining your response. I understand this as that it is true in $L_{\omega_1^{CK}}$ that each set is countable. | |
| Sep 17, 2013 at 13:41 | comment | added | Joel David Hamkins | Those other sets are nevertheless countable in $L_{\omega_1^{CK}}$ and (hence) are coded by hyperarithmetic reals. | |
| Sep 17, 2013 at 13:06 | comment | added | catrincm | You claim that $L_{\omega_1^{CK}}$ consists of only HYP sets. I want to make it clear what is meant by this. The reals (i.e. subsets of $\omega$) which are in $L_{\omega_1^{CK}}$ are exactly the hyperarithmetic sets, but $L_{\omega_1^{CK}}$ of course contains other things. | |
| Jan 13, 2013 at 1:34 | comment | added | Andrés E. Caicedo | (@Asaf: Fixed.) | |
| Jan 13, 2013 at 1:03 | history | edited | Andrés E. Caicedo | CC BY-SA 3.0 |
edited body; edited title
|
| Jan 12, 2013 at 15:00 | comment | added | Asaf Karagila♦ | I think you may have confused the subscript and the superscript. | |
| Jan 12, 2013 at 14:38 | answer | added | 喻 良 | timeline score: 4 | |
| Jan 12, 2013 at 5:18 | comment | added | Wei Wang | If $M = L_\alpha[a]$ then $(L_\beta)^M = L_\beta$ for $\beta < \alpha$. So, if $\alpha = \omega^{CK}_1$, then $HYP$ is the set of reals in $L^M$. Is this sufficient? | |
| Jan 12, 2013 at 4:35 | history | asked | Peter Gerdes | CC BY-SA 3.0 |