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Is this true, if so is it obvious?

  1. Jun 12, 2008 #1
    There is no proper class such X such that every totally ordered set is isomorphic to a subclass of X.

    I'm using "proper class" and "isomorphic" rather liberally here, but you can assume them to be formulas in ZFC, or something.
  2. jcsd
  3. Jun 12, 2008 #2
    ? What about the class of all sets? Every set is a subclass of that, so certainly every totally ordered set is a subclass of it... Also, if you want something stricter; since the class of all sets is a class, you can consider the class of all totally ordered sets. Is your question whether or not that is a proper class? I'd think that it would be, but the way you asked your question, it sounds like the class of all sets should satisfy your requirement.

    Maybe I'm misunderstanding something here?
    Last edited: Jun 12, 2008
  4. Jun 12, 2008 #3
    No, you're right.
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