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Math Help - Uniformly continuous

  1. #1
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    Uniformly continuous

    Assume that (f_n) converges uniformly to f on A and that each f_n is uniformly continuous on A. Prove that f is uniformly continuous on A.

    f_n converges uniformly so |f_n(x)-f(x)|<epsilon.
    f_n is uniformly continuous so |x-y|<delta implies |f_n(x)-f_n(y)|<epsilon
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  2. #2
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    it's a standard trick (also known as "3-epsilon trick"):

    write |f(x)-f(y)| < |f(x)-f_n(x)| + |f_n(x)-f_n(y)| + |f_n(y)-f(y)| < epsilon+epsilon+epsilon

    just be careful with the order of all those epsilons, deltas and n's
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  3. #3
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    Oh, triangle inequality. Didn't think of that.
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