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Math Help - preimage of a continous function is an interval

  1. #1
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    preimage of a continous function is an interval

    To prove: a continuous real-valued function (f) defined on set S has an image (S') that is an interval then S must be an interval.

    Tools: definition of continuity, definition of a connected set, connected set in R is an interval,


    If S' is not a singleton; if f(a), f(b) \in S' such that f(a) < f(b) and f(c) \in S' such that f(a) < f(c) < f(b), can it be shown that a < c < b?
    I'm looking for something to get me started. Thanks.
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  2. #2
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    Re: preimage of a continous function is an interval

    Hi,
    I don't think you've stated your question correctly. Given what you said:
    Let f be a continuous function with domain S (subset of R) and image S' = f(S). If S' is an interval, then S is an interval.

    This is clearly false:

    Let S=[-2,-1]\cup[1,2] and f(x)=x2. Then f(S)=[1,4], but S is not an interval.

    Did you omit the hypothesis that f is one to one? But if so, the problem is trivial as you point out.
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  3. #3
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    Re: preimage of a continous function is an interval

    Thanks John, it's possible that I made a mistake. Here's the original problem statement:

    A non-empty set S of real numbers is an interval <=> every continuous real-valued function on S has an interval as its image
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  4. #4
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    Re: preimage of a continous function is an interval

    Quote Originally Posted by director View Post
    Thanks John, it's possible that I made a mistake. Here's the original problem statement:
    A non-empty set S of real numbers is an interval <=> every continuous real-valued function on S has an interval as its image
    It is certainly in one direction.
    Theorem: The continuous image of a connected set is a connected set.
    Now in the real numbers intervals are connected.
    Thus if \mathcal{I} is an interval and f is continuous then f(\mathcal{I}) is an interval.

    Now you have been given a counterexample for the other direction.
    Last edited by Plato; October 4th 2013 at 02:03 PM.
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  5. #5
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    Re: preimage of a continous function is an interval

    I understand. I didn't give much thought to the possibility of the second part being false. The counter example is clear. Thanks.
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  6. #6
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    Re: preimage of a continous function is an interval

    Hi again,
    As Plato points out, the main implication is: if S is an interval and f is continuous on S, then f(S) is an interval.
    However, the converse as written is trivially true: if every continuous function f on S has f(S) an interval, then S is an interval -- take f(x) = x.

    Kind of a funky statement, but it's true.
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