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Math Help - Uniform Discontinuous Proof

  1. #1
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    Uniform Discontinuous Proof

    Prove that f(x)=1/x is not uniformly continuous on (0, inf)?

    First of all I wanted to make sure that the negation of:
    for every epsilon > 0 there exists delta > 0 such that x,y element S and |x-y| => |f(x)-f(y)|< epsilon
    is:
    there exists epsilon > 0 such that for every delta > 0 there exists x, y element S such that |x-y|< delta and |f(x)-f(y)|>= epsilon

    So here is how I started:
    We can set epsilon = 1 so |1/x - 1/y|>= 1. Then |x-y|<delta so |x-y|=delta/2 which means y = x + delta/2. Then we can plug in to get
    |1/x - 1/(x+delta/2)|>=1. Simplifying we get delta/2 >= x(x+delta/2).

    Here I don't know what x I need to use so that the statement is true. Actually, I don't think my way is possible. I was wondering if there is another way to approach this but still use the definition.
    Thanks
    Last edited by tbyou87; October 15th 2007 at 11:49 AM.
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  2. #2
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    Scratch that, I actually have no idea how to do this problem.
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  3. #3
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    Note that \begin{array}{rcl}<br />
 f(x) = \frac{1}{x}\quad & \Rightarrow & \quad \left| {f\left( n \right) - f\left( {n + k} \right)} \right| \\ <br />
  & = & \quad \left| {\frac{1}{n} - \frac{1}{{n + k}}} \right| \\ <br />
  & = & k \\ <br />
 \end{array}.

    Consider the sequence \left( {\frac{1}{n}} \right) \to 0.
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  4. #4
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    Let sn = 1/n for n element N. Then (sn) is cauchy in (0,inf). Since f(sn) = n, (f(sn)) is not cauchy. Therefore f cannot be uniformly continuous on (0, inf).

    I think that proof works. I was just wondering how you would do a proof using the precise definition uniformly continuous and a contradiction

    for each eps >0 there exists delt>0 s. t. s, y ele S and |x-y| => |f(x)-f(y)|<eps

    THanks
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  5. #5
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    Oh yeah by the way how do you get the math symbols like epsilon, delta, subset and stuff?

    Thanks
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  6. #6
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    Theorem: If f is uniformly continous on a non-empty set S then any Cauchy sequence x_k in S we have f(x_k) is a Cauchy sequence.

    So consider x_k = \frac{1}{k} this is a Cauchy sequence in (0,\infty) but f(x_k) = k is not.
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