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Set function proof

  1. Oct 13, 2011 #1
    I need to prove f(f^-1(Y'))[itex]\subseteq[/itex]Y' for some f: X -> Y and Y' in Y.

    So far, I've been able to figure this much out:

    Let y[itex]\in[/itex]f(f^-1(Y')). Then, f^-1(Y') = x' for some x' in X such that f(x') = y' for some y' in Y'. Then, f(x') = y'. Thus, f(f^-1(Y'))[itex]\subseteq[/itex]Y'.

    I feel like there's something wrong with my proof. Any ideas on where I went wrong?
  2. jcsd
  3. Oct 13, 2011 #2
    Any ideas?
  4. Oct 13, 2011 #3
    You probably mean Y' subset of Y.

    What does that even mean?? Y' is a set, so [itex]f^{-1}(Y^\prime)[/itex] is a set. But x' is an element. So you're saying that a set is equal to an element????

    Start with [itex]y\in f(f^{-1}(Y^\prime))[/itex]. Write out the definitions. What is the definition for [itex]y\in f(A)[/itex]??
  5. Oct 13, 2011 #4
    Thanks for the input micromass.

    I've revised the proof as thus:

    Let y∈f(f^-1(Y')). Since f^-1(Y') = X' s.t. X'⊆X and f(X')=Y'. Since y∈f(X'), y∈Y'. Thus, f(f^-1(Y'))⊆Y'.

    I still feel like I'm missing a step or two.
  6. Oct 13, 2011 #5

    This would imply


    which does not always hold.
  7. Oct 13, 2011 #6
    You're right, I can't assume f is bijective. Should I instead say something like "there exists an x∈X s.t. x∈f^-1(Y')?
  8. Oct 13, 2011 #7


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    Science Advisor

    Maybe a good comment to make is that, in the nicest-possible case, you have:


    Nicest possible is, of course, f is 1-1 and onto. Try to see why identity above fails when f is either not 1-1 or not onto.

    TMFKAB (The Mathematician* Formerly Known as Bacle)

    *In training.
  9. Oct 13, 2011 #8
    Not sure how that helps, Bacle2. Basically, if the function isn't surjective there could exist a b in B' such that f^-1(B') doesn't exist, and I'm not sure how to handle this case.
  10. Oct 13, 2011 #9
    Bacle wasn't giving you a hint to solve the problem. He gave you another problem which could be rewarding to look at to expand your knowledge.

    To solve your problem. What does it mean that [itex]y\in f(A)[/itex]. Apply this to [itex]y\in f(f^{-1}(y))[/itex].
  11. Oct 13, 2011 #10
    I think I follow. You mean something like:

    Let y∈f(f^-1(Y')). Then, there exists an x in f^-1(Y') s.t. f(x) = y. Since x in f^-1(Y'), f(x) = y for some y in Y'. Thus, f(f^-1(Y'))⊆Y'.
  12. Oct 13, 2011 #11
    Seems ok.
  13. Oct 13, 2011 #12


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    Science Advisor

    Yes, Autre, sorry if my post was confusing; just trying to give some insight and some related results, as Micromass said.
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