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One-to-one function

  1. Feb 7, 2008 #1
    [SOLVED] One-to-one function

    Let f: X -> Y. Suppose f(X-A) = Y-f(A) for any A in X. Prove that f is one-to-one.

    Suppose f(a)=f(b). I tried letting A = {a,b}, A= X-{a,b}, but I can't manage to prove that a=b. Can someone help?
     
  2. jcsd
  3. Feb 7, 2008 #2

    NateTG

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    You need to clean up your notation. What are [itex]X[/itex] and [itex]Y[/itex]? Sets?
     
  4. Feb 7, 2008 #3
    X is the domain of f and Y is the range of f, so they are sets. A is any subset of X. So the question is asking to prove that if f(complement of A)= complement of f(A), then f is one-to-one.
     
  5. Feb 7, 2008 #4

    morphism

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    Let a be in X and set A=X-{a}. What's f(X-A) now?
     
  6. Feb 7, 2008 #5

    HallsofIvy

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    You might want to try proving the contrapositive: if f is not one to one, then there exist a set A such that f(complement of A) is not complement of f(A).

    If f is not one to one, then there exist x1 and x2 such that f(x1)= f(x2). What if A= {x1}?
     
  7. Feb 7, 2008 #6

    NateTG

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    In situations like this, you can always try assuming that [itex]a \neq b[/itex] by contradiction. The proof might not require it, but you can sometimes find direction that way.
     
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