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Homework Help: Basic proof about inverse images of sets

  1. Jan 1, 2008 #1
    [SOLVED] basic proof about inverse images of sets

    1. The problem statement, all variables and given/known data

    let f map X onto Y, and let A be a subset of X, B subset of Y

    1) Prove that if y is in B, then f^(-1) (y) is in f^(-1) (B)

    3. The attempt at a solution

    1) I think I'm missing some two-line proof, because I'm getting nowhere and this is supposed to be simple. I just don't know where to start, so just a hint would be appreciated.
    Last edited: Jan 1, 2008
  2. jcsd
  3. Jan 1, 2008 #2


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    The statement, as given, is not true. Since f is only given as a "map", it may not have an inverse function and so f-1(y) may not exist. Your question, however, is about "inverse images", which exist whether or not f has an inverse. What is true is that f-1({y}) is a subset of f-1(B). {y} is, of course, the set containing the single point y. That statement follows directly from the definition of "inverse image". What is the definition of f-1(B)?
    Last edited by a moderator: Jan 1, 2008
  4. Jan 1, 2008 #3
    f^-1 (B) := {x in X| f(x) in B}

    The book told us that strictly speaking one should write f({y}), but in order to keep simplicity, they just use the notation f(y). What I'm trying to prove is what you have written down. I am sorry I didn't bring that part over correct.

    but, back to the problem. this would make

    f^-1 (y) = {x in X| f(x) in {y}}
    f^-1 (B) = {x in X| f(x) in B}

    because y is in B, this implies that f^-1 (y) is in f^-1 (B), is that right?

    Anyway, thanks for your help... I'm not even sure if this is 'calc&beyond'... I'm studying physics and took mathematics as a major too since the start of this year, and this class is given at the same time as the first calculus and linear algebra classes.
  5. Jan 1, 2008 #4


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    Hmmph! Well, go by your text book- but the part about "is a subset of" rather than "is in" is important.

    Again, "is a subset of" not "is in" which means "is a member of". [itex]f^{-1}(y)= f^{-1}(\{y\})[/itex] is a subset of X, not a member of X! If [itex]x\in f^{-1}(\{y\})[/itex], then, by definition, [itex]f(x)\in \{y\}[/itex] which is a subset of B. Therefore, if [itex]x\in f^{-1}({y})[/itex], [itex]f(x)\in B[/itex].
  6. Jan 1, 2008 #5
    haha, I love the fact you could respond so quickly.

    The textbook used 'is a subset of' instead of 'is in' on many places (I think in all the places it is necessary), I just went over it too fast so I didn't copy it right because I hasn't realized it made such a difference. I get what you're explaining now, though.

    Thanks, and SOLVED!
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