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Cardinality proof by indicating a bijection

  1. Feb 9, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove that |AB[itex]\cup[/itex]C|=|ABx AC| by demonstrating a bijection between the two sets.



    2. Relevant equations

    Two sets have equivalent cardinality if there is a bijection between them/

    3. The attempt at a solution

    Essentially I can prove that there is a function from AB[itex]\cup[/itex]C to ABx AC, defined by <f restricted to B, f restricted to C> but it's only onto if B[itex]\cap[/itex]C=0.

    Or I can prove a function the other way but it's only one-to-one if the same condition holds.
    Other solutions I've seen online also say that this condition is necessary, but it's not included in the homework question. Any ideas?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 9, 2013 #2

    Dick

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    If B and C are not disjoint and all of the sets A, B and C are finite, then the statement is definitely false. If they didn't give you any information about the sets, then you can't prove it. If some of the sets are infinite, then it might be true. It depends on the details.
     
  4. Feb 9, 2013 #3
    thanks for responding!
    If A, B, and C are infinite, I think the statement holds (just by results of cardinal arithmetic that I've seen in the textbook), but I'm still not able to find a bijection.
     
  5. Feb 9, 2013 #4

    Dick

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    You need a bijection between the union of B and C and the 'disjoint union' of B and C. And, yes, if B U C is infinite then its cardinality is the max(card(B),card(C)). Same for the disjoint union, so there's definitely a bijection. But that's just an indirect conclusion. I think to actually write down the bijection easily you need to assume B and C are disjoint. If not, the point is not to write it down but to prove it exists.
     
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