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Math Help - Cardinality

  1. #1
    Super Member Deadstar's Avatar
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    Cardinality

    Since so much of cardinal numbers involve creating in/bijections I'm not sure if that's what I should be doing here instead...

    Prove that the addition and multiplication is well defined for cardinal numbers.

    My answer was... (roughly, more to see if I'm on the right track)

    Let A and B be two sets with cardinality \kappa and \lambda respectively.

    Want to show that if card(A) = card(A') and card(B) = card(B') then...

    card (A \cup B) = card (A' \cup B')

    So we have card (A' \cup B') = card(A') + card(B') = \kappa + \lambda = card(A) + card(B) = card (A \cup B).

    Hence addition is well defined... Seems wrong looking back on it. Do I even need to the \kappa, \lambda bit?
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  2. #2
    Junior Member
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    Zagreb
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    maybe...

    dim(A \cup B) = dim(A) + dim(B) - dim(A \cap B)

    dont know for multiplication...

    do we look AXB = \{(a,b)| a \in A, b \in B \} ?
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  3. #3
    Senior Member
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    card = card(A') + card(B') = = card(A) + card(B) = card
    Doing that you assume the addition for cardinal numbers is well-defined.

    Moreover, cardA=cardA' and cardB=cardB' does not mean card(A\cup B)=card(A'\cup B') ; you must assume A\cap B=A'\cap B'=\emptyset to be sure of that (and that's the way cardinal addition is defined).

    What you have to prove is: If cardA=cardA' and cardB=cardB', then card(A\oplus B)=card(A'\oplus B') , where X\oplus Y := X\times\{0\}\cup Y\times\{1\} for any two sets X and Y.

    cardA=cardA' means there exists a bijection f:A\rightarrow A'<br />
    cardB=cardB' means there exists a bijection g:B\rightarrow B'

    Question: Can you define a bijection between A\oplus B and A'\oplus B' ?

    For the product, the hypotheses and the question are the same with the cartesian product \times instead of \oplus.


    @josipive: The question is about sets in general, you cannot apply such result about vector subspaces of finite dimension.
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