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Math Help - Distributivity of Cardinal Numbers

  1. #1
    Super Member Deadstar's Avatar
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    Distributivity of Cardinal Numbers

    If \kappa, \lambda, \mu be three cardinals. Prove that \kappa(\lambda + \mu) = \kappa \lambda + \kappa \mu.

    Thoughts...
    My basic problem is that I have never really understood how to construct in/sur/bijections so all this cardinal number stuff is proving to be a bit hard...
    But my thoughts on this are that...
    Let \kappa, \lambda, \mu be the cardinal numbers of the sets A,B and C respectively such that B and C are disjoint. Then A \times B and A \times C are disjoint and A \times (B \cup C) = A \times B \cup A \times C.

    I'm not sure whether I'm allowed to assume the disjoint part and how to construct a bijection between the two...
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  2. #2
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    Hi

    Let \lambda and \mu be two cardinal numbers, then \lambda +\mu is, by definition, the cardinal of the disjoint union of two sets whose cardinalities are \lambda and \mu respectively.

    If in the definition of the cardinal addition the sets weren't assumed to be disjoint, that would not make sense: for instance we would have 1+1=card(\{0\}\cup\{0\})=card(\{0\})=1 !

    If this is true, then you've finished: the identity map will be a bijection between these two sets, which means, under the assumptions and the remark:

    Let be the cardinal numbers of the sets A,B and C respectively such that B and C are disjoint. Then and are disjoint
    that \kappa (\lambda +\mu )=\kappa\lambda + \kappa\mu

    Therefore all you have to do is to prove the equality. By the way, this equlity holds even if B\cap C\neq\emptyset (the case B\cap C\neq\emptyset does not interest us for this problem, but we don't have to suppose B and C are disjoint to prove the equality)

    The proof is quite simple, it can be done with two inclusions; the only things you have to use are the definitions of union and cartesian product of sets.
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