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Set Theory, Logic, Probability, Statistics
Does one need V not equal to L in order to do Cohen's forcing?
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[QUOTE="SSequence, post: 6436383, member: 601088"] [USER=155756]@Citan Uzuki[/USER] I don't understand your post#4 but I have few questions. In post#4, when given a model ##M##, one writes ##L^M##, I have two basic questions: (i) Will ##L^M## also always be guaranteed to be a model (of set-theory) given that we know ##M## to be a model of set-theory. (ii) Second question, is related to what I want to understand (to see if I have a misunderstanding here). Is ##L^M## "completely determined" by the ordinals contained in ##M## or not? [USER=112452]@nomadreid[/USER] To be honest, I don't quite understand the discussion in preceding posts. However, it seems that there should be many models (of set-theory) that will contain less sets than ##L##. For example, there are countable models ##M## of set-theory (given its congruity). It seems to me, that any such model ##M## will not contain many of the ordinals contained in ##L## (in particular, many countable ordinals of ##V##). That's what it seems like to me (I could be wrong). [/QUOTE]
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Does one need V not equal to L in order to do Cohen's forcing?
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