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Bijection for a power set function

  1. Aug 20, 2011 #1
    1. The problem statement, all variables and given/known data

    Let N={1,2.....n} .Define the Power set of N,P(N).
    a) show that the map f:P(N)->P(N)
    defined by taking A to belong to P(N) to N\A is a bijection.
    b)C(n,k)=C(n,n-k).



    3. The attempt at a solution
    Now the power set is defined by P(N)=2^n and a bijection is a one-to-one function that is both injective and surjective . The domain and range have the same cardinality since we are using the power set of N.and f:N->N is bijective maybe a logical consequence will be that f:P(N)->P(N) is also bijective
    Also |N\A|=n-|A| where | | represents the number of elements in a set.
    Ineed help in solving a ) and b)
     
  2. jcsd
  3. Aug 20, 2011 #2

    micromass

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    OK, so for a you must show both injectivity and surjectivity

    For injectivity: You must show that if [itex]\mathbb{N}\setminus A=\mathbb{N}\setminus B[/itex], then A=B

    For surjectivity, you must show that for all [itex]A\subseteq \mathbb{N}[/itex], there is a [itex]B\in \mathbb{N}[/itex] such that [itex]\mathbb{N}\setminus B=A[/itex].

    Do you agree that this is what you need to show?
     
  4. Aug 20, 2011 #3
    Yes but for part b)I have no ideea :frown:
     
  5. Aug 20, 2011 #4

    micromass

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    Well, for part (b) you must show first that the bijection restricts to

    [tex]f:\{A\subseteq N~\vert~|A|=k\}\rightarrow \{A\subseteq N~\vert~|A|=n-k\}[/tex]
     
  6. Aug 20, 2011 #5
    We assume that N \ A = N \ B. We show A=B by showing that A ⊆ B and also B ⊆ A. To prove A ⊆ B, let x be an element of A. Then x is not an element of N \ A, and thus x is not an element of N \ B. Thus x is in B. (Showing B ⊆ A is similar.)

    For surjectivity: Let B = N \ A.=> |B|=n-|A|=>|A|=n-|B|=>A=N\B

    For part b)Using par a) we see that and restricting the doman of F to k elements then the image under this function F will be n-k since again the function is defined on N\A.

    Is this ok?
     
  7. Aug 20, 2011 #6

    micromass

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    Well, B=N\A is the right thing to choose. But saying |A|=n-|B| doesn't necessarily imply A=N/B.
    Can you prove directly that B=N\A implies A=N\A?

    The rest looks ok.
     
  8. Aug 20, 2011 #7
    Cand I do something like this? B=N\A=> N\B=N\(N\A)=>N\B=A?
     
  9. Aug 20, 2011 #8

    micromass

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    Yes, certainly!! That's ok!
     
  10. Aug 20, 2011 #9
    Thank you
     
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