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Does one need V not equal to L in order to do Cohen's forcing?
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[QUOTE="SSequence, post: 6437457, member: 601088"] [USER=112452]@nomadreid[/USER] It seems to me that there is something that needs to be pointed out w.r.t. what I wrote in post#8. I am not sure it is worth bumping the thread for, but I wasn't aware of it at least. It seems that when I wrote in post#8: there are some further conditions that are required. ========== I was looking at the MSE question linked in previous post. Looking at one of the suggested questions: [URL]https://math.stackexchange.com/questions/85196/how-many-countable-models-of-zfc-are-there[/URL] One thing that the answer states is that for a model in which the statement ~con(ZFC) is true, inside that model there would be no countable model of ZFC at all (admittedly, I don't know why). After that the answer states a stronger condition (for a model) inside which a countable model is guaranteed. I am guessing you would be aware of this already anyway. But it seems like useful enough point to add. [/QUOTE]
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Does one need V not equal to L in order to do Cohen's forcing?
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