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Number of choice functions?

  1. Jun 11, 2013 #1
    Is the following a theorem from ZFC?
    Given a collection C of non-empty sets that includes at least one infinite set, the cardinality of the collection of distinct choice functions on C (as defined in AC) equals the cardinality of the largest element of C.
    My feeling that this is true is from generalizing the case when the largest cardinality is [itex]\aleph[/itex]0, where it seems that a simple proof is possible, but I am not sure whether it is true and, if so, provable (from ZFC) for higher cardinalities.
     
  2. jcsd
  3. Jun 11, 2013 #2
    Won't the claim fail when the cardinality of C is greater than the cardinality of the largest element of C?

    That is, if C is a collection of aleph1 sets of aleph0 elements, there will be at least aleph1 choice functions.
     
  4. Jun 11, 2013 #3
    yossell, thanks for the answer. Good point. So, if I were to amend it, would the theorem be that the number of choice functions is max (|C|, |S|) with S being a set with the largest cardinality? Whereas this seems intuitively clear, is it provable in ZFC?
     
  5. Jun 11, 2013 #4

    micromass

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    Given sets ##(X_i)_{i\in I}##, what you want is the cardinality of

    [tex]\prod_{i\in I} X_i[/tex]

    For convenience, we set ##I## to be equal to a cardinal number, so we put ##I = |I| = \lambda##. It is a theorem in ZFC that if ##\lambda## is infinite and if ##|X_i|## are nondecreasing and nonzero, then

    [tex]\prod_{i<\lambda} |X_i| = (\sup_{i<\lambda} |X_i| )^\lambda[/tex]

    See "Set Theory" by Jech for a proof.
     
  6. Jun 11, 2013 #5
    Thanks, micromass. Lemma 5.9, to be exact.
     
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