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Consider the function $f : (0, 1] → R$ defined by $f(x) = 1/x$.

Prove that, for any $0 < r < 1$, f is uniformly continuous on $[r, 1]$.

I was trying to use the theorem if $f$ be continuous on [a,b] then $f$ is uniformly continuous. But I don't know how to relate and explain closed interval use the given $0< r < 1$.

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  • $\begingroup$ Do you see that $[r,1]$ is also a interval "like" $[a,b]$ and $f$ is continuous on $[r,1]$? $\endgroup$ – Arpit Kansal Mar 3 '17 at 3:29
  • $\begingroup$ yes I know, but I don't know how to relate 0 < r < 1 $\endgroup$ – bingyan zhu Mar 3 '17 at 3:33
  • $\begingroup$ Well once you fix a $r$ then $[r,1]$ is a compact interval and $f$ is continuous on $[r,1]$ as i said hence $f$ is uniformly continuous. $\endgroup$ – Arpit Kansal Mar 3 '17 at 3:34
  • $\begingroup$ It is just the range of values of $r$ that you need to prove that $f$ is uniformly continuous over $[r;1]$ $\endgroup$ – Graham Kemp Mar 3 '17 at 3:34
  • $\begingroup$ I totally understand, but the interval keeps change. so how to prove f on a variable interval continuous? $\endgroup$ – bingyan zhu Mar 3 '17 at 3:38
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Let $\epsilon > 0$, you need to find a $\delta(\epsilon)$ such $|x-y|<\delta \implies |f(x)-f(y)|<\epsilon$

Let $x,y\in[r,1]$ then $|f(x)-f(y)| = |\frac1x-\frac1y|=|\frac{y-x}{xy}|\le|y-x|\frac1{r^2}=|x-y|\frac1{r^2}<\epsilon$

Let $\delta=\epsilon*r^2$, and that concludes your proof.

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  • $\begingroup$ i use [r 1] is subset of R and then f is uniformly continuous on R and then f is uniformly continuous on [r,1] since [r,1] is subset of R. Is the way correct ? $\endgroup$ – bingyan zhu Mar 3 '17 at 19:14
  • $\begingroup$ Since the function is only defined from domain $(0,1]$ then no, however it is uniformly continuous for all $a,b > 0$ $\endgroup$ – Rab Mar 3 '17 at 22:56
  • $\begingroup$ Why are you allowed to say $\frac{1}{xy} \leq \frac{1}{r^{2}}$ exactly? Edit: Nevermind, I get it - since $r$ is the lower bound then $\frac{1}{r}$ is the largest possible value for both $\frac{1}{x}$ and $\frac{1}{y}$. $\endgroup$ – Alena Gusakov May 1 '19 at 16:35

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