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Math Help - Undecidability and models of ZFC

  1. #1
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    Undecidability and models of ZFC

    This is question that probably comes up a lot here, but every time I think I have it straight, I find I don't. So allow me to ask it in the following form. Four elementary facts:


    (1)ZFC \rightarrow[ if \kappa is an inaccessible cardinal, then ( < V_{k} , \epsilon> \models ZFC) ]

    (2) The existence of an inaccessible cardinal is consistent with ZFC


    (3)not [ZFC \rightarrow ( ZFC is consistent)].


    (4) A theory is consistent iff it has a model.


    At first glance, (1) and (4) would seem to imply


    (*) ZFC  \rightarrow( ZFC is consistent) which of course is rubbish. One possibility for the problem is that perhaps one cannot assert


    (4') ZFC \rightarrow (A theory is consistent iff it has a model),


    but I am not sure whether one cannot (since the model concept is formalizable in the language of ZFC), and even if one cannot, I am not sure that that would be the main fissure in the faulty conclusion. I suspect there is something much more basic here that I am missing.
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  2. #2
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    (3)not [ZFC \rightarrow ( ZFC is consistent)].
    What does your arrow stand for? Provability? Yes, if ZFC is consistent, then it is not the case that ZFC proves that ZFC is consistent. (We usually drop the antecedent "if ZFC is consistent", as you did, as we may take that as implicitly assumed in our discussion.)

    At first glance, (1) and (4) would seem to imply

    (*) ZFC  \rightarrow( ZFC is consistent)
    How so? I don't see that (1)-(4) imply that ZFC proves ZFC is consistent. (1)-(4) show that ZFC+"there exists an inaccessible cardinal" proves that ZFC is consistent. That is not the same as showing that ZFC proves ZFC is consistent.
    Last edited by MoeBlee; March 1st 2011 at 09:05 AM.
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  3. #3
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    Thanks, MoeBlee. You are right, there was that omission on my part. This solves the problem. Perhaps I should get more sleep before I post.
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