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Sep 17, 2020 at 12:21 history edited Mirco A. Mannucci CC BY-SA 4.0
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Sep 17, 2020 at 9:24 vote accept Mirco A. Mannucci
Sep 17, 2020 at 4:01 answer added jonasreitz timeline score: 5
Sep 15, 2020 at 13:07 comment added Mirco A. Mannucci @GabeGoldberg thanks for chiming in! So, looks like geology is a complicated science... I would still love if you and/or Asaf posted a partial answer: listing what is known and what is not known (example: your counterexample cited above, but hopefully examples where the meet exists in GROUNDS(M), etc. We need to build a list of what is known...
Sep 15, 2020 at 4:28 comment added Gabe Goldberg @AsafKaragila Mathematics is not ready for such problems
Sep 15, 2020 at 0:57 comment added Asaf Karagila @Gabe: Although in that case $L(\Bbb R)$ is in fact a ground... Just not a ZFC one.
Sep 14, 2020 at 22:50 comment added Gabe Goldberg @MircoA.Mannucci Another thing you can do is get a situation where you have two grounds whose intersection is $L(\mathbb R)$, and in this case assuming there is no $\omega_1$-sequence of reals in $L(\mathbb R)$, there cannot be a largest common ground, since any inner model of ZFC contained in $L(\mathbb R)$ has countably many reals, and hence has a forcing extension contained in $L(\mathbb R)$.
Sep 14, 2020 at 22:39 comment added Asaf Karagila See, now that's why I didn't post an answer. Thanks, @Gabe, I felt like this might "too easy".
Sep 14, 2020 at 22:35 comment added Gabe Goldberg @Asaf It's not true that the intersection of two grounds is a ground since it might not satisfy ZFC. You can get an example using the technique from my answer here by changing the word "generic" in the second sentence to "Cohen generic" and working in a big collapse extension: mathoverflow.net/questions/297756/…
Sep 13, 2020 at 17:48 comment added Mirco A. Mannucci LOL! Menial works has always priority. I will be waiting then, no worries.
Sep 13, 2020 at 17:47 comment added Asaf Karagila No good deed goes unpunished. I'm sure someone will come soon that can write a much better answer to all of your questions. I, on the other hand, need to clean the kitchen and cook the hell out of some chicken.
Sep 13, 2020 at 17:46 comment added Mirco A. Mannucci apologies, I had not read your full comment. THAT is interesting! Why don't you put it into a formal answer? It does not address all points, but it is very good start
Sep 13, 2020 at 17:44 comment added Mirco A. Mannucci I suspect so. You are a (very) smart guy Asaf, and I do greatly esteem your set theory first-class competence, I would appreciate if you take your time and come up with an answer. I can guarantee you that you will get my vote, and if you come up to an answer to all of the above, even a negative one, you have my GREEN. L'hitraot, haveri
Sep 13, 2020 at 17:42 comment added Asaf Karagila Hmm. I could be misremembering, then. But if $W$ and $W'$ are both grounds, and $W\cap W'$ is a model of ZFC which contains a ground $U$, then it is a model of ZFC between $U$ and $V$, where $V=U[G]$ for some set-generic filter $G$. So by Vopenka's theorem $W\cap W'$ is a ground.
Sep 13, 2020 at 17:36 comment added Mirco A. Mannucci what about sups, or even iarbitrary sups?
Sep 13, 2020 at 17:36 comment added Mirco A. Mannucci If this is true, you have answered my first question: the partial order of grounds is a infinite meet semilattice. But, is it true? I was under the impression that it was only directed, in general, not a semilattice. If this is true, then I doubt that the logic of forcing is simply S4.2, but rather a somewhat stronger modal logic. Secondly, I do not think you read through my question carefully enough, that is ONE question
Sep 13, 2020 at 17:31 comment added Asaf Karagila I mean, the intersection of any set-many grounds is a ground, is there anything left to say?
Sep 13, 2020 at 16:38 history edited Mirco A. Mannucci CC BY-SA 4.0
added 8 characters in body; edited title
Sep 13, 2020 at 16:31 history asked Mirco A. Mannucci CC BY-SA 4.0