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Do you believe in the Axiom of Choice?

  1. Yes

    9 vote(s)
    81.8%
  2. No

    1 vote(s)
    9.1%
  3. Undecided

    1 vote(s)
    9.1%
  1. Mar 16, 2008 #1
    Feel free to give your reasons.

    I voted yes, because too many useful theorems are thrown out the window if Axiom of Choice is rejected. I believe that these useful theorems outweigh the surprising (strange?) results that also arise from AC (e.g. every set can be well-ordered). Also, if AC is truly a failure (based on what I have no idea), then shouldn't it have failed by now, over 100 years later? I'm assuming that there has been no physical experiment available to disprove the Axiom of Choice, am I right? Will there ever be such a physical experiment?
     
    Last edited: Mar 16, 2008
  2. jcsd
  3. Mar 16, 2008 #2

    Hurkyl

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    It's not really something to "believe in". ZFC is one theory, ZF-C is another theory.
     
  4. Mar 16, 2008 #3
    Then I guess I'm asking which theory you prefer: ZFC or ZF without AC?
     
  5. Mar 16, 2008 #4

    Hurkyl

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    Models of ZF that satisfy choice are generally simpler objects than models of ZF that do not; you can usually say more interesting things about models of ZFC.
     
  6. Mar 18, 2008 #5
    Can someone explain to me what Zorn's lemma is?
     
  7. Mar 18, 2008 #6

    Hurkyl

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  8. Mar 19, 2008 #7
    another axiom used to prove the linear ordering of the cardinals i.e. for any two sets A and B there exists f: A->B or f:B->A , f an injection. this of course non-exclusive or. if you want i can type out the proof in my book for you.
     
  9. Mar 19, 2008 #8
    :smile:Well I would like to take a look if its not too long!
     
  10. Mar 19, 2008 #9
    Proving the Law of Dichotomy is easier using the axiom of choice. The proof using Zorn's lemma is also very elegant too:
    Let K be the set of all bijections from a subset of A to a subset of B. Then every totally ordered subset L of K contains an upper bound in K (the union of the bijections in L) (showing that the upper bound in K is the most crucial part). So there is a maximal function F, which you then show has either domain A or image B.


    Here's a Zorn's Lemma problem that I posed earlier:
    (Show that if r partially orders X, then there exists a total order relation m such that m contains r and m totally orders X.)
    https://www.physicsforums.com/showthread.php?t=208395
    with my full solution typed out. Anyone wanting to add to improve my solution please feel free to do so.
     
    Last edited: Mar 20, 2008
  11. Mar 20, 2008 #10

    Hurkyl

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    I've personally always preferred invoking the well-ordering theorem for my axiom of choice needs. This particular proof is especially simple this way. :smile:
     
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