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Math Help - closed interval continiuous

  1. #1
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    closed interval continiuous

    Prove the following: Let  a < b be real numbers. and let  f: [a,b] \to \bold{R} be a function which is continuous on  [a,b] . Then  f is also uniformly continuous.

    Assume for contradiction that  f is not uniformly continuous. Then there are two equivalent sequences  (x_{n})_{n=0}^{\infty} and  (y_{n})_{n=0}^{\infty} in  [a,b] such that the sequences  (f(x_{n}))_{n=0}^{\infty} and  (f(y_{n}))_{n=0}^{\infty} are not equivalent. Then we let  E = \{n \in \bold{N}: f(x_{n}) \ \text{and} \ f(y_{n}) \ \text{are not} \ \varepsilon \ \text{close} \} . Then use Bolzano-Weierstrass Theorem to obtain contradiction?

    Is this correct?
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  2. #2
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    If f is not uniformly continous then for there is \epsilon > 0 such that for any \delta > 0 we have |x-y|<\delta, x,y\in [a,b] but |f(x)-f(y)|\geq \epsilon.
    Let \delta = \frac{1}{n} for n\in \mathbb{Z}^+.
    Then we form two sequences \{x_n\} and \{y_n\} such that |x_n-y_n|<\tfrac{1}{n} and |f(x_n)-f(y_n)| \geq \epsilon. Since \{x_n\} is a bounded sequence by Bolzano-Weierstrass theorem there is a convergent subsequence X_n. Let X = \lim ~ X_n and note that X\in [a,b]. Let Y_n be the subsequence chosen in the same manner as X_n i.e. if X_n = x_{g(n)} for some choice function g then Y_n = y_{g(n)}. Now since |X_n - Y_n|<\tfrac{1}{g(n)}\leq \tfrac{1}{n} we see that X_n - \tfrac{1}{n} < Y_n < X_n + \tfrac{1}{n}. But each X_n converges to X and so by squeeze-theorem Y_n converges to X. Thus, \{X_n\} and \{Y_n\} are two convergent sequences to X\in [a,b]. By continuity it means \{f(X_n)\} and \{f(Y_n)\} converge both to f(X). However, |f(X_n)-f(Y_n)| \geq \epsilon and this is a contradiction.
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  3. #3
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    And the set  E is infinite? Because if it was finite, then there would be two sequences that are equivalent?
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