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The axiom of choice one a finite family of sets.

  1. Aug 7, 2013 #1
    The axiom of choice on a finite family of sets.

    I just been doing some casual reading on the Axiom of CHoice and my understanding of the is that it assert the existence of a choice function when one is not constructable. So if we have a finite family of nonempty sets is it fair to say we can assume the existence of a choice function because it is always possible, in theory, to manually pick an element of each set?
     
    Last edited: Aug 7, 2013
  2. jcsd
  3. Aug 7, 2013 #2

    mfb

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    Exactly.
     
  4. Aug 7, 2013 #3
    Thanks. Do you know if this can be proved?
     
    Last edited: Aug 7, 2013
  5. Aug 7, 2013 #4

    mathman

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    The proof is trivial. If there are n (non-empty) sets, pick a member from the first, then from the second. This requires n steps - so it can be done.
     
  6. Aug 7, 2013 #5
    It seemed too trivial, to be true. Thanks.
     
  7. Aug 21, 2013 #6
    It depends.
    Only if you have an explicitly finite family of nonempty sets, that you can list : E1,...En then you can use a proof whose length is proportional to n :
    Let x1 in E1,
    Let x2 in E2,
    ....
    Let xn in En
    then (x1,...,xn) is in the product, which is thus nonempty.

    But for the mathematical statement of the general case "Any finite family of nonempty sets has a choice function" it needs a different proof, namely it can be done by rewriting the claim as "For any natural number n, any family of n nonempty sets has a choice function" to be proven by induction on n.
     
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