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Oct 25, 2014 at 4:05 vote accept Asaf Karagila
Oct 20, 2014 at 19:09 comment added Joel David Hamkins I don't have one yet...
Oct 20, 2014 at 17:40 comment added Yair Hayut @Joel: You're right, our "proof" (for the "if" part) was wrong. Do you have a counter example for this implication?
Oct 19, 2014 at 21:33 comment added Joel David Hamkins I notice that you say "if and only if" in your statement 1. But is that right, if you don't require $V[G]=V[H]$ as in my answer? I don't think it is correct.
Oct 19, 2014 at 18:38 comment added Joel David Hamkins Agnostic is not the same as decisive, since he has no parameters. But my examples there work for your question if one used only ordinal parameters. So there can be Ord-decisive forcing notions that are not weakly homogeneous.
Oct 19, 2014 at 18:37 comment added Asaf Karagila @Joel: Agnostic is an interesting name, I'm not sure if it's better than $V$-decisive. But surely this is not a concept that was introduced for the first time on MathOverflow! :-)
Oct 19, 2014 at 17:38 comment added Joel David Hamkins See this related question mathoverflow.net/questions/154554/…
Oct 19, 2014 at 15:17 comment added Asaf Karagila @Ali: Yair insisted that this cannot be new, I said that it might be. I originally suggested "pseudo-homogeneous" and today Yair came up with this name instead, which I also think is a reasonable name.
Oct 19, 2014 at 15:16 comment added Ali Enayat @Asaf: I have not seen a name before; which is somewhat surprising given the naturality of the concept, your suggested name is well-chosen.
Oct 19, 2014 at 14:38 comment added Asaf Karagila @Ali, yes that is essentially the negation of a counterexample. Only stronger and quite a natural follow up. Do you happen to know the answer to the side question about the name?
Oct 19, 2014 at 13:15 comment added Ali Enayat @Asaf: Aha, I see that Joel retracted his claim about the Boolean completion, so your current version is fine (but if it has a negative answer, then one can start thinking about the more general version I suggested).
Oct 19, 2014 at 12:48 comment added Asaf Karagila @Ali: Yes, that was our essential question, and in light of Joel's answer we in fact noticed a mistake in the question (well, two of them really). Now it should be better, I hope. :-)
Oct 19, 2014 at 12:45 comment added Ali Enayat In light of Joel's nice answer, I am wondering whether one should ask: is every $V$-decisive poset forcing equivalent to some weakly homogeneous poset?
Oct 19, 2014 at 12:41 history edited Asaf Karagila CC BY-SA 3.0
added 19 characters in body
Oct 19, 2014 at 11:56 answer added Joel David Hamkins timeline score: 12
Oct 19, 2014 at 11:00 history edited Asaf Karagila CC BY-SA 3.0
added 4 characters in body
Oct 19, 2014 at 10:42 comment added Mohammad Golshani I think you mean a $V-$decisive which is not weakly homogeneous?
Oct 19, 2014 at 10:34 history asked Asaf Karagila CC BY-SA 3.0