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Math Help - Real Analysis Proof

  1. #1
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    Real Analysis Proof

    Can someone check my work? I've attached it. Thanks!
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  2. #2
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    It is wrong. Becauase y\not = f(x), y is some other point in the set. So |x-y|<\delta are all points in the set so that this is true and not |x-f(x)|<\delta.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    It is wrong. Becauase y\not = f(x), y is some other point in the set. So |x-y|<\delta are all points in the set so that this is true and not |x-f(x)|<\delta.
    Ok, change all the f(x)'s to y's. Then, I think, it is correct.
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    TPH, how does it look now? Attached the new one.
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    Good proof.

    In the second part where you show (0,1] is not unfiormly continous, trying using the definition, i.e. show (without Cauchy sequences) that you can violate the definition of uniform continuity.
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