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Math Help - Group Theory - Find a bijection F: Xo -> Xe (Xo-odd, Xe-even)

  1. #1
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    Group Theory - Find a bijection F: Xo -> Xe (Xo-odd, Xe-even)

    Hi, I have the following problem. I have solved the second part, but I don't know how to solve the first one.

    Let X_o be the set of all odd cardinality subsets of the set X, and let X_e be the set of all even cardinality subsets of the set X. Do the following:
    - Find a bijection F:X_o -> X_e,
    - Express |X_o| and |X_e| in terms of the number |X|.

    So the part I don't know is the first one, but here is what I got for the second one. Is it OK? And do you know a bijection for the first part? Thanks!


    "There is a bijection between X_o and X_e because |X_o|=|X_e|.

    Also, we know that if |X|=2^n, we have |Xe|=2^(n-1) and |X_o|=2^(n-1). But we also know that in every set X, we only have odd and even subsets, so if |X_o|=|X_e| then we can say

    |X|=|X_o|+|X_e|, so |X_o|=(1/2)*|X| and |X_e|=(1/2)*|X|"
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Basically, you just need to show that |X_o| = |X_e|.

    You can, if you wish, note that if |X| is odd then the complement of every even set us a unique odd set, and your done (taking the complement of X with X to be the empty set, and even set). This is just an observation, it isn't needed to prove the result per se.

    Now, if |X|=n is then induct on n. n=1 can be checked manually (note that the empty set is take to be even).

    Let Y=X\setminus\{x\} where x \in X is some arbitrary element. The result holds for Y. Can you conclude that the result thus holds for X? (The subsets of X are the subsets of Y together with all the subsets of Y with x (as in, if S \subset Y the S\cup \{x\} \subset X)).
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