Timeline for Moschovakis Coding Lemma
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 21, 2011 at 17:02 | vote | accept | Rachid Atmai | ||
| Oct 10, 2011 at 2:03 | comment | added | Rachid Atmai | Oh, hold on minute. I can't use transfinite inductions here, we're under AD :) Only your case will work. | |
| Oct 10, 2011 at 1:48 | comment | added | Rachid Atmai | Ok. That is what I had in mind but I was not sure about it. After running through this argument I would kept telling myself that the situation would be the same with a limit ordinal: say I have choice sets of length $\delta<\beta$ then by transfinite induction (maybe over wellfounded relations) this should also hold for $\beta$. Something must be wrong with what I just said this but I don't know what it is. | |
| Oct 9, 2011 at 20:26 | comment | added | Noah Schweber | The reason it's a limit ordinal is - again, unless I'm missing something - just the argument above. Think about it this way: if I have a (definable) choice set for an initial segment of length $\beta$, I can extend it to a (definable) choice set of length $\beta+1$ just by adding to the definition of the choice set for length $\beta$ a single clause describing its behavior at one more term. | |
| Oct 9, 2011 at 18:33 | comment | added | Rachid Atmai | I am sorry I am not understanding your answer. You say the above proof works. I think the point of the coding lemma is that we have definable choice sets given some conditions. Also my question is about the following point: when we take $\delta$ to be the least such that the theorem fails, why does it have to be a limit ordinal? | |
| Oct 9, 2011 at 3:06 | history | answered | Noah Schweber | CC BY-SA 3.0 |