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Thread: Distributivity of Cardinal Numbers

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

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

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

    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
    Senior Member
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    Hi

    Let $\displaystyle \lambda$ and $\displaystyle \mu$ be two cardinal numbers, then $\displaystyle \lambda +\mu$ is, by definition, the cardinal of the disjoint union of two sets whose cardinalities are $\displaystyle \lambda$ and $\displaystyle \mu$ 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 $\displaystyle 1+1=card(\{0\}\cup\{0\})=card(\{0\})=1$ !

    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 $\displaystyle \kappa (\lambda +\mu )=\kappa\lambda + \kappa\mu$

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