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I just want to make sure I understand certain things well. I believe my questions are quite simple. Are the following statments true ?

1/ One cannot prove Con(ZFC) in ZFC. However ZFC might not be consistant (and in this case its inconsistancy would be 'provable' in a some sense).

2/ ZFC + existence of an inaccessible cardinal proves Con(ZFC). However it is possible that ZFC is consistant but 'ZFC + existance of an inaccessible cardinal' is not consistant.

3/ If ZFC + existence of an inaccessible cardinal proves a statment about natural numbers, then it never implies that ZFC proves this statment for all actual natural number. (The statment is true in all $\omega$-model).

(Not like for example if a $\Pi^0_1$ statement is provable in ZFC + CH then Con(ZFC) $\rightarrow$ Con(ZFC + CH) $\rightarrow$ For at least a model of ZFC the statment is true $\rightarrow$ Its negation is not provable in ZFC $\rightarrow$ The statment is true for actual natural numbers.)

Thanks for any answer.

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closed as not a real question by Andres Caicedo, Ricky Demer, Chris Godsil, Simon Thomas, Asaf Karagila Apr 24 '12 at 23:12

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4  
What exactly are your questions? –  Péter Komjáth Apr 24 '12 at 19:09
    
I think the OP is asking if 1,2 and 3 are true. –  David Roberts Apr 24 '12 at 20:52
    
Sorry, David Roberts is right, but it was not clearly stated that way. –  Archimondain Apr 24 '12 at 22:20

1 Answer 1

up vote 7 down vote accepted

Your statements are nearly correct, but none of them is fully exactly right, in that there are in each case missing consistency hypotheses. So let me explain how each of them can be improved in some small way.

For the first part of statement (1), the correct thing to say is that if ZFC is consistent, then ZFC does not prove Con(ZFC). This is an immediate consequence of the 2nd incompleteness theorem. For the second part, although it is widely believed that ZFC is consistent, we don't actually know that ZFC might be consistent; it is an extra assertion going beyond the ZFC axioms. That is, the assertion that ZFC is consistent can legitimately be made only under the assumption that it is true or under a stronger assumption, such as the existence of large cardinals, which implies that ZFC is consistent. Similarly, if ZFC really is consistent, then many would say it is wrong to say that it "might not be consistent", since that wouldn't be the case. However, a closely related correct statement is that if ZFC is consistent, then so is the theory ZFC+$\neg$Con(ZFC). That is, if ZFC is consistent, then it is consistent with ZFC to hold that it isn't consistent.

For the first part of the second statement, yes, one can prove in ZFC that if there is an inaccessible cardinal, then Con(ZFC) holds. This is simply because it is relatively easy to see that if $\kappa$ is inaccessible, then the rank initial segment $V_\kappa$ satisfies all the ZFC axioms, and so there is a model of ZFC and consequently Con(ZFC) holds. The correct thing to say for the second part of statement (2) is that if ZFC is consistent, then so is the theory ZFC+$\neg$Con(ZFC+"there is an inaccessible cardinal"). This follows from the incompleteness theorem applied to the theory asserting that there is an inaccessible cardinal. Since ZFC+I, if consistent, does not prove Con(ZFC+I), then either every model of ZFC is a model of $\neg$Con(ZFC+I) or else there is a model of ZFC+I+$\neg$Con(ZFC+I). Something a little closer to your statement is that if ZFC+Con(ZFC) is consistent, then so is the theory asserting ZFC+Con(ZFC)+$\neg$Con(ZFC+I).

Statement three is not true in the generality in which you make it. For example, one of the things that you can prove in ZFC+I is that 1=1, but it is not correct to say that this "never implies that ZFC proves 1=1 in the actual natural numbers", since in fact ZFC does prove that statement. But what you probably mean to say instead is that the arithmetic consequences of ZFC+I is a strictly larger theory than the arithmetic consequences of ZFC. This is true, provided that we assume ZFC is consistent, since one of the arithmetic consequences of ZFC+I is Con(ZFC) itself, which is not provable in ZFC, provided that ZFC is consistent.

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thank you for your answers. –  Archimondain Apr 25 '12 at 9:19

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