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Uniform continuous proofs

  1. Oct 27, 2013 #1
    1. The problem statement, all variables and given/known data
    Show that the function ##x^2## is not uniformly continuous on ##\mathbb{R}##


    2. Relevant equations

    Delta - Epsilon Definition:

    ##\exists \epsilon > 0, \ \forall \delta >0, \exists x \in S [|x-x_0|< \delta \text{and} |x^2 - x_0^2| \ge \epsilon ].##

    3. The attempt at a solution

    I am confused on how they got ##\epsilon## and ##\delta##. It seems they found it through thin air.

    What I did was:

    Let ##\delta >0## and ##|x-x_0| < \delta## then for a given ##\epsilon >0## we have that ##|f(x) - f(x_0)| = |x^2 - x_0^2| = |x - x_0||x+x_0| = \delta|x+x_0| = \delta|x-x_0 +2x_0| = \delta[|x-x_0| + |2x_0|] = \delta^2 + \delta|2x_0| ... ##

    I googled this and find out they chose ##\epsilon = 1## and ##x= x_0 + \delta/2## and ##x_ 0 = 1/\delta##. I am not sure how they got that? And this happens a lot in real analysis, they choose an ##\epsilon## and ##\delta## that I have no idea how they got. It's like they got it from thin air. I know those ##\epsilon## and ##\delta## work but they never provide information on how they obtained it.

    Can anyone help me rigorously show how to solve problems like this?
     
  2. jcsd
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