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Undecidable statement made into axiom?

  1. Feb 24, 2006 #1
    i am a little bit confused on 1 thing. can a statement that is undecidable in a axiomatic system just be added as a axiom to the original system and never lead to a contridiction?

    for example godel and cohen showed that the continuum hypothesis is independent of ZFC. does this mean then that if we add the CH or its negation as an axiom to ZFC we will still have a system that is still consistent (assuming ZFC is already consistent)? Does this mean then that if I add the negation of the CH i can show that there is some cardinal x such that aleph_0<x<c?

    this is the wiki article that is confusing me-
    http://en.wikipedia.org/wiki/Axiom_of_choice#Independence_of_AC

    quote from article: "By work of Kurt Gödel and Paul Cohen, the axiom of choice is logically independent of the other axioms of Zermelo-Fraenkel set theory (ZF). This means that neither it nor its negation can be proven to be true in ZF. Consequently, assuming the axiom of choice, or its negation, will never lead to a contradiction that could not be obtained without that assumption."

    so if the CH( or ~CH) can be added to ZFC(since it was proven independent just like AC), then why is it still unsolved?
     
    Last edited: Feb 24, 2006
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  3. Feb 24, 2006 #2

    Hurkyl

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    You are correct.

    If [itex]\Phi[/itex] is some collection of statements, and [itex]\phi[/itex] is a statement that is "undecidable" in [itex]\mathcal{TH}(\Phi)[/itex]... that is, neither [itex]\phi[/itex] nor [itex]\neg \phi[/itex] follow from [itex]\Phi[/itex]... then [itex]\Phi \cup \{ \phi \}[/itex] is a consistent collection of statements. (And so is [itex]\Phi \cup \{ \neg \phi \}[/itex])


    (P.S. anybody know how I can get a nice, fancy lower-case script letter? If I try to turn [itex]Th[/itex] into the mathcal font, I get [itex]\mathcal{Th}[/itex])
     
    Last edited: Feb 24, 2006
  4. Feb 24, 2006 #3

    mathwonk

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    i think it is decidable, in fact i have decided the answer is no.
     
  5. Feb 24, 2006 #4

    Ok, I still don't understand why then people are still trying to prove the falsity of the continuum hypothesis. If the continuum hypothesis is independent of ZFC, then it can be added as an axiom. But if it is an axiom shouldn't it be universally true, why would people still bother trying to negate it?
     
  6. Feb 24, 2006 #5

    Hurkyl

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    If the negation of the continuum hypothesis is independent of ZF, then it could be added as an axiom.

    One might not want to assume the CH. Just as people might not want to assume the AC.

    And then, one might want to classify other statements in terms of how they relate to the CH. In other words, one might find it useful to know that P=>~CH. Among other things that would prove that P is independent of ZF.

    (You sure the CH is independent of ZFC? I know it is independent of ZF, but I can't remember ZFC for sure)
     
  7. Feb 24, 2006 #6

    But doesn't the fact that the CH is independent of ZFC already imply that its negation is independent of ZFC (or is the article i am reading simply wrong)? I also am pretty sure that CH has already been proven independent of ZFC.
     
  8. Feb 24, 2006 #7

    Hurkyl

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    Yes. Incidentally, that wiki article is talking about the GCH, not the CH.
     
  9. Feb 24, 2006 #8

    I actually was reading that article and from what it said about the AC, I just made an analogy to the CH (cohen and godel showed that CH is independent of ZFC). So if we add ~CH to ZFC, it will be consistent. Doesn't this mean now that we can find a cardinal call it x such that aleph0<x<c?
     
  10. Feb 24, 2006 #9

    Hurkyl

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    Yes. ZFC~CH implies the existance of such a cardinal.
     
  11. Feb 24, 2006 #10
    alright thanks a bunch. i think i am out of questions now.
     
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