Do you believe in the Axiom of Choice?

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Do you believe in the Axiom of Choice?

  • Yes

    Votes: 9 81.8%
  • No

    Votes: 1 9.1%
  • Undecided

    Votes: 1 9.1%

  • Total voters
    11
182
0

Main Question or Discussion Point

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?
 
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Answers and Replies

Hurkyl
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It's not really something to "believe in". ZFC is one theory, ZF-C is another theory.
 
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Then I guess I'm asking which theory you prefer: ZFC or ZF without AC?
 
Hurkyl
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Then I guess I'm asking which theory you prefer: ZFC or ZF without AC?
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.
 
Can someone explain to me what Zorn's lemma is?
 
1,703
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Can someone explain to me what Zorn's lemma is?
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.
 
:smile:Well I would like to take a look if its not too long!
 
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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.
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:
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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