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Math Help - Real Analysis - Sequences #2

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    Real Analysis - Sequences #2

    Hey, I have been having trouble ever since I took part 1 of this class. So, here is question 2:

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

    Any help would be greatly appreciated. Thanks.
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    Quote Originally Posted by ajj86 View Post
    Hey, I have been having trouble ever since I took part 1 of this class. So, here is question 2:

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

    Any help would be greatly appreciated. Thanks.
    Here is the idea of the proof.

    Since f_n \to f uniformly there exits an N \in \mathbb{N} such that for n > N |f_n-f|< \frac{\epsilon}{3} Now fix  M > N

    Since each f_n is uniformly continous there exists a delta such that |x-y|< \delta \implies |f_M(x)-f_M(y)|< \frac{\epsilon}{3}

    Now if |x-y|<\delta then

    |f(x)-f(y)|= |f(x)-f_M(x)+f_M(x)-f_M(y)+f_M(y)-f(y)|

    <|f(x)-f_M(x)|+|f_M(x)-f_M(y)|+|f_M(y)-f(y)|< \frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsil  on}{3}=\epsilon
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