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Proof of Axiom of Choice equivalent.

  1. Dec 1, 2009 #1
    I'm trying to prove that the axiom of choice is equivalent to the following statement:

    For any set [tex]X[/tex] and any function [tex]f:X\to X[/tex], there exists a function [tex]g:X\to X[/tex] such that [tex]f\circ g\circ f=f[/tex].

    I was able to prove that the AoC implies this, but I'm having a harder time going the other direction. It seems like if you define an equivalence relation on [tex]X[/tex] where [tex]x\sim y[/tex] iff [tex]f(x)=f(y)[/tex], then the composite function [tex]g\circ f[/tex] must map everything in each equivalence class to one of its members.

    This seems like it's important, but we're still only choosing points for a very specific collection of subsets of [tex]X[/tex], namely the equivalence classes induced by the function. Is there a way to extend this to any collection of subsets, or am I heading in the wrong direction?
     
  2. jcsd
  3. Dec 1, 2009 #2
    Hmm... this is kind of a weird way of doing it, but I think it works: Let A be a pairwise disjoint family of nonempty sets. Then Let X be (UA)U(A)U{null}. Then define f:X->X by letting f(x)=B if x is in UA and B is in A and x is in B, and letting f(B)=null for B in A, and letting f(null)=null.

    Then the equivalence relations include elements of A, since for every element B of A, everything in B and nothing else maps to B.
     
  4. Dec 1, 2009 #3
    That looks very clean. Thank you very much. It was the fact that the domain and codomain had to be equal that was causing most of the trouble it seems.
     
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