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Math Help - Function with Darboux's Property

  1. #1
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    Function with Darboux's Property

    Let f: [0,1] \rightarrow \mathbb{R} be a function with Darboux's property. \forall \ y \in \mathbb{R}, we have f^-1({{y}}) is closed. Prove that f is continuous.

    So I have an idea of how to approach this, but I'm not entirely sure on the closed part - does that necessary entail that the function is surjective? (If not, the pre-image wouldn't exist?)
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    Quote Originally Posted by h2osprey View Post
    Let f: [0,1] \rightarrow \mathbb{R} be a function with Darboux's property. \forall \ y \in \mathbb{R}, we have f^-1({{y}}) is closed. Prove that f is continuous.

    So I have an idea of how to approach this, but I'm not entirely sure on the closed part - does that necessary entail that the function is surjective? (If not, the pre-image wouldn't exist?)
    Read this.... What you want is a corollary on the 2nd page.

    http://tt.lamf.uwindsor.ca/314folder...ns/Darboux.pdf
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    Quote Originally Posted by TheEmptySet View Post
    Read this.... What you want is a corollary on the 2nd page.

    http://tt.lamf.uwindsor.ca/314folder...ns/Darboux.pdf
    Sorry I misread your question, The answer is not in this
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  4. #4
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    Ah I think I've solved it, but I'm not entirely sure if this is correct. Basically we prove by contradiction: assume that the function is not continuous, then for some x_0 \in\ [0,1],\ \exists\ \epsilon_0 > 0 such that \forall\ \delta > 0, \exists\ x_\delta such that |x_\delta\ - x_0| < \delta\ but  |f(x_\delta)\ - f(x_0)| \ge \epsilon_0 . Then we can find a sequence \{x_n\}\ which converges to x_0, but f(x_n) = f(x_0) - \epsilon/2,\ \forall\ n \ge 1 (this is due to Darboux's property, and we're assuming without loss of generality that  f(x_\delta)\ - f(x_0) \ge -\epsilon_0).

    Now this sequence is in the pre-image of f(x_0) - \epsilon/2, which is closed by assumption. Hence it contains all its accumulation points  \implies f(x_0) \in\ \{f(x_0) - \epsilon / 2\}, which is a contradiction. Hence, f is continuous.

    Is that okay?
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