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Proving chains identical or disjoint ?

  1. Oct 13, 2012 #1
    Using Schroder-Bernsten Theorem. Assume there exists a 1-1 function f:X→Y and another 1-1 function g:Y→X. If we define f−1(y)=x, then f−1 is a 1-1 function from f(X) onto X. Similarly, g-1: g(Y)→Y. Follow the steps to show that there exists a 1-1, onto function h:X→Y.


    Let x be in X be arbitrary. Let the chain Cx be the set consisting of all elements of the form

    ....,f−1(g−1(x)),g−1(x),x,f(x),g(f(x)),f(g(f(x))),....

    (a) Explain why the (distinct) number of elements to the left of x in the above chain may be zero, finite, or infinite. <-- I already got this part.

    (b) Show that any two chains are either identical or disjoint.

    I need help with part a, I'm stuck on showing how they can be infinite, I'm thinking it would have to be bijective, is that right? I got the zero and finite part.

    For part b, I don't know how to think about it. Any ideas?

    Any help is appreciated.

    Thanks
     
  2. jcsd
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