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$f: (0,+\infty) \to (0,+\infty)$ $f(x) = 1/x$, prove that f is not uniformly continuous.

Firstly, I negated the definition of uniform convergence obtaining:

$\exists \epsilon > 0 $ s.t. $\forall \delta > 0 $ with $|x-y| < \delta$ & $x,y \in (0, + \infty)$ and $|f(x) - f(y)| = \left|\dfrac{x-y}{xy}\right| \geq \epsilon$

so I choose $\epsilon = 1$ and $ x = \delta/2 \in (0,+\infty)$ and $y = \delta /4 \in(0,+\infty)$ so $|x-y| = \delta/2 < \delta$ and $|f(x) - f(y)| = |2/\delta|$ How do I show that this is greater than or equal to epsilon?

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2 Answers 2

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The condition you want is $|2/\delta|\ge \epsilon = 1$, which holds true as long as $|\delta|\le 2$. So you're set for small $\delta$. Can you see a way to manage the case when $\delta>2$ accordingly?

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    $\begingroup$ I'll give it a go - but is there something specifically wrong with mine? $\endgroup$
    – Warz
    Jan 20, 2014 at 19:40
  • $\begingroup$ @Warz See my edit. $\endgroup$ Jan 20, 2014 at 19:48
  • $\begingroup$ Could we not just set $\epsilon = 1/\delta $ then we would still get $ |2/\delta| > 1/\delta = \epsilon$? $\endgroup$
    – Warz
    Jan 20, 2014 at 19:53
  • $\begingroup$ No, you don't get to set $\epsilon$ in terms of $\delta$. $\endgroup$ Jan 20, 2014 at 19:55
  • $\begingroup$ hmm, I don't really see what we could do then - I always thought of delta as small so didn't really consider the other case. $\endgroup$
    – Warz
    Jan 20, 2014 at 19:56
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Does the function take Cauchy sequences to Cauchy sequences? What does it take the sequence of numbers $x_n = n^{-1}$ to ?

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