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Must all models of ZFC (in a standard formulation) be at least countable?

  1. May 24, 2012 #1
    Must all models of ZFC (in a standard formulation) be at least countable?

    Why I think this: there are countably many instances of Replacement, and so, if a model is to satisfy Replacement, it must have at least countably many satisfactions of it.

    Does my question only apply to first-order formulations of ZFC, or are there second-order formulations of ZFC that can be finite?

    Thanks.
     
  2. jcsd
  3. May 25, 2012 #2

    AKG

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    Yes, they're all at least countable, because they all have an empty set, a set containing just the empty set, a set containing just the set containing just the empty set, etc. Your argument using Replacement doesn't work, since there's no guarantee that two different instances of Replacement generate different sets.
     
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