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Math Help - Uniform Continuity for 1/x

  1. #1
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    Uniform Continuity for 1/x

    I'm trying to prove the following statement:

    Prove that the function f(x)=1/x is continuous on (0,1] but it is not uniformly continuous on this interval.

    Thanks a lot.
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  2. #2
    Senior Member Tinyboss's Avatar
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    For the "not uniformly continuous" part: do you know how to formally negate a logical proposition, i.e. exchange "for-all's" and "there-exists's" and reverse (in)equalities? Try that with the definition of uniform continuity.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by AKTilted View Post
    I'm trying to prove the following statement:

    Prove that the function f(x)=1/x is continuous on (0,1] but it is not uniformly continuous on this interval.

    Thanks a lot.
    I actually discuss this in my post here. In essence, (we'll speak less generally now) if (0,1]\to\mathbb{R}" alt="f(0,1]\to\mathbb{R}" /> were uniformly continuous, then we could extend it to some uniformly continuous map \tilde{f}:[0,1]\to\mathbb{R}. But, this is just nonsense since we'd have to have that f(0)=\lim f\left(\frac{1}{n}\right)=\lim n
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by AKTilted View Post
    I'm trying to prove the following statement:

    Prove that the function f(x)=1/x is continuous on (0,1] but it is not uniformly continuous on this interval.

    Thanks a lot.

    The negation of uniform continuity:

    For some \epsilon > 0 and for each \delta  > 0
    there exist x_0, x_1 such that |x_0 - x_1| < \delta  but |f(x_0) - f(x_1)| \geq  \epsilon.

    Assume that there is exist  \delta <\frac{1}{n} n\in \left \{\mathbb{N} / 0}{  \right \}.

    Now let x_0,x_1 \in\mathbb{R}^+ x_0=x_1+\frac{\delta }{2} and x_1=\frac{\delta }{2}


    \left | f(x_0)-f(x_1) \right |=\left | \frac{1}{x_0}-\frac{1}{x_1} \right |=\left |\frac{ x_0-x_1}{x_0x_1} \right |<br />

    =\frac{\frac{2}{n}}{x_0(x_0+\frac{\delta }{2})}\frac{\delta }{2}=\frac{\frac{2}{n}}{\delta^2}\frac{\delta }{2}=\frac{\frac{1}{n}}{\delta }>1
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    Another much better solution:


    Choose x_n=\frac{1}{n} and y_n=\frac{1}{n+1} for any n\in\mathbb{N}.


    \left | x_n-y_n \right |<\frac{1}{n}, but \left | f(x_n)-f(y_n) \right |=1<br />
.
    Last edited by Also sprach Zarathustra; November 4th 2010 at 06:29 AM.
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