Timeline for A question about the first Cohen model
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 17, 2014 at 23:09 | comment | added | Asaf Karagila♦ | Carlos, ah yes. That is a doozy. Jech uses some partial interpretation functions. | |
| Jan 17, 2014 at 23:03 | comment | added | Carlos | Thanks. Yes, working with sufficiently large sets is clearly enough in this context.I think that the results that Jech proves to show that the Cohen model satisfies the ordering principle will be enough to "translate" the Halpern and Lévy proof to a modern language. Now I am jammed in a step in the proof that every set has a minimal support, but the problem has nothing to do with the present question. | |
| Jan 14, 2014 at 20:56 | comment | added | Asaf Karagila♦ | Carlos, I forgot to give you a short update. After thinking about it much, I've asked my advisor. He pointed out that it is still open, and he felt there might be a counterexample. I, on the other hand, feel that symmetric class names are likely to be classes of the symmetric extension. In either case, in 99% of the time, the fact the class (in the full generic extension) intersected every set in the symmetric extension is in a set in the symmetric extension suffices. | |
| Jan 7, 2014 at 0:30 | vote | accept | Carlos | ||
| Jan 5, 2014 at 19:15 | comment | added | Asaf Karagila♦ | Carlos, yes. Symmetric classes are classes of the generic extension. To be fair, I can't recall the proof for that right now. It seems reasonable like some reflection-like argument (i.e. each initial segment is definable in some large enough portion of the universe), but I'm relying on Jech and others to have checked that more thoroughly. I'll get back to you with that, though. | |
| Jan 5, 2014 at 19:06 | comment | added | Carlos | Anyway, I will try to check if working witn large enough $\alpha$ suffices to follow the proof of Halpern and Lévy. | |
| Jan 5, 2014 at 19:05 | comment | added | Carlos | Thanks again, Asaf! I had never seen class names, and so I have a doubt. Does the class name $f$ really define a class in $N$? If the generic ultrafilter would be available in $N$ the answer would be clearly yes. For instance, we can assign an ordinal rank to each name, and restrict $f$ to a set name $f_\alpha$ defined for names in $D$ of rank $\leq \alpha$. Then the sequence $\{f_\alpha\}_{\alpha\in ON^M}$ is definable in $M$ and each $(f_\alpha)_G\in N$, but, is the map $\alpha\mapsto (f_\alpha)_G$ definable in $N$? I cannot see this. | |
| Jan 5, 2014 at 17:39 | comment | added | Asaf Karagila♦ | Carlos, I've added an answer to your actual question. Let me know if it helps you out. | |
| Jan 5, 2014 at 17:34 | history | edited | Asaf Karagila♦ | CC BY-SA 3.0 |
added 1094 characters in body
|
| Jan 5, 2014 at 16:11 | comment | added | Asaf Karagila♦ | Oh, I didn't say that you do try to avoid it. I just pointed out that if one wants to write a full exposition on the independence of the Boolean Prime Ideal theorem from the Axiom of Choice, then it's going to be hard to avoid formal arguments at one point of another. | |
| Jan 5, 2014 at 16:06 | comment | added | Carlos | Thank you very much! I will think carefully about all you have said. I did not pretend to avoid the Halpern-Lauchli lemma, but just the specific way Halpern and Lévy describe the sets of $N$. | |
| Jan 5, 2014 at 15:56 | history | answered | Asaf Karagila♦ | CC BY-SA 3.0 |