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Let $f(x) = \frac{1}{x}$ for $x > 0$ and take our set at which the function act on $(0,1]$. This function is continuous but not uniformly continuous on $A$.

To prove this consider $\epsilon = 10$,and suppose we have $\delta$, $0 < \delta < 1$. Taking $x = \delta$ and $p = \frac{\delta}{11}$, we get $|x - p| < \delta$ and $|f(x) - f(p)| = \frac{11}{\delta} - \frac{1}{\delta} > 10$.

In the following example from my book shows this is continuous but not uniform continuous but the example is missing correct ? since we also need to find counter example for $\delta \geq 1$ right ? If someone could present that example too that would be nice.

Also my question is different since the possible duplicate above shows very different question to the question I am asking here since I am asking why we must consider $\delta \geq 1$ so please read carefully my question.

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The authors use the condition $0 < \delta < 1$ so that $x = \delta$ will be within $(0, 1]$. As other answers have shown, you don't actually need to consider larger $\delta$ (briefly, if $\delta_1 < \delta_2$ and $x,y$ are such that $|x-y| < \delta_1$, then $|x-y| < \delta_2$). However, here's an quick re-write of the book's argument that doesn't rely on the condition $\delta < 1$ (changes in bold):

Consider $\epsilon = 10$,and suppose we have $\delta > 0$. Let $\mathbf{\eta = \min\{\delta, \frac{1}{2}\}}$. Taking $x = \mathbf{\eta}$ and $p = \frac{\mathbf{\eta}}{11}$, we get $|x - p| < \mathbf{\eta \le} \delta$ and $|f(x) - f(p)| = \frac{11}{\mathbf{\eta}} - \frac{1}{\mathbf{\eta}} > 10$

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The case $\delta\geq 1$ is not interesting here because you're on $(0,1]$, so if $\delta > 1$, you always have $\vert x-y\vert <\delta$.

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  • $\begingroup$ yes but my we will always have $|x - y| < \delta$ but we can have that it satisfies the $\epsilon$ definition because all we need is just one delta that works for every $\epsilon$ maybe I am wrong here $\endgroup$ – Dude Aug 1 '15 at 19:54
  • $\begingroup$ If $\delta > 1$, you can take $x=1$ and $y=0.01$ for example. Then you have $\vert x-y\vert < \delta$ and $\vert f(x)-f(y)\vert \geq 10$. $\endgroup$ – Augustin Aug 1 '15 at 19:59
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The definition of uniform continuity requires that given $\epsilon > 0$ we can pick $\delta > 0$ such that for all $y$ and for all $x$ within distance $\delta$ of y, $|f(x) - f(y)| < \epsilon$. If we've shown that we can't pick a suitable $\delta_1 < 1$ then we certainly can't pick $\delta_2 \geq 1$ since given $y$, $(y - \delta_1, y+ \delta_1) \subset (y - \delta_2, y + \delta_2)$ and there is a counterexample in $(y - \delta_1, y+ \delta_1)$.

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  • $\begingroup$ Okay perfect thanks alot ! $\endgroup$ – Dude Aug 1 '15 at 19:55

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