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Finding a bijection

  1. Nov 16, 2012 #1
    1. The problem statement, all variables and given/known data

    Let X be a set. Suppose that f is a bijection from p(X) to p(X) such that [itex]f(A)\subseteq f(B)[/itex] iff [itex]A\subseteq B[/itex] for all subsets A,B of X.
    Show that there is a bijection g from X to X such that for all [itex] A\subseteq X [/itex] one has f(A)=g(A).

    2. Relevant equations

    p(X) is the power set of X.

    3. The attempt at a solution

    This seems too elementary and I doubt that there is something to prove. Can't I just take f=g?
     
  2. jcsd
  3. Nov 16, 2012 #2

    micromass

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    But f is a function from P(X) to P(X). And g is a function from X to X. So taking f=g makes no sense. The arguments of f should be subsets of X. The arguments of g should be elements of X.
     
  4. Nov 16, 2012 #3
    I can't believe I didn't see it:D thanks
     
  5. Nov 16, 2012 #4
    Define g(x) as the unique element satisfying {g(x)}=f{x}.show that g(A)=U{g(a)}=Uf{a}=
    f(A)
     
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