Timeline for Code universal arithmetical sets by a hyperarithmetical set?
Current License: CC BY-SA 3.0
5 events
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| Jul 25, 2011 at 15:29 | comment | added | Cole Leahy | Indeed, I am doubly silly. Of course the R from the original question is hyperarithmetical; we've shown that there is a hyperarithmetical Q meeting my description, and Q and R are coextensive! Please cut me some slack here, as I'm just learning. And I haven't had coffee today. | |
| Jul 25, 2011 at 15:10 | comment | added | Cole Leahy | That was silly of me! I corrected the original posting. Thanks for the tip. | |
| Jul 25, 2011 at 14:59 | comment | added | Ali Enayat | @Cole: I am not sure what you mean by Borel here; all subsets of $\omega$ are Borel if Borel is interpreted in its usual way. | |
| Jul 25, 2011 at 14:54 | comment | added | Cole Leahy | Does this mean that the set R from the original question is indeed hyperarithmetical, and not just Borel? It was obtained by stacking, after all. | |
| Jul 25, 2011 at 12:06 | history | answered | Andreas Blass | CC BY-SA 3.0 |