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 $\displaystyle f(0,1]\to\mathbb{R}$ were uniformly continuous, then we could extend it to some uniformly continuous map $\displaystyle \tilde{f}:[0,1]\to\mathbb{R}$. But, this is just nonsense since we'd have to have that $\displaystyle f(0)=\lim f\left(\frac{1}{n}\right)=\lim n$
The negation of uniform continuity:
For some $\displaystyle \epsilon > 0 $ and for each $\displaystyle \delta > 0$
there exist $\displaystyle x_0, x_1$ such that $\displaystyle |x_0 - x_1| < \delta $ but $\displaystyle |f(x_0) - f(x_1)| \geq \epsilon$.
Assume that there is exist$\displaystyle \delta <\frac{1}{n}$ $\displaystyle n\in \left \{\mathbb{N} / 0}{ \right \}$.
Now let $\displaystyle x_0,x_1 \in\mathbb{R}^+ x_0=x_1+\frac{\delta }{2} and x_1=\frac{\delta }{2}$
$\displaystyle \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 |
$
$\displaystyle =\frac{\frac{2}{n}}{x_0(x_0+\frac{\delta }{2})}\frac{\delta }{2}=\frac{\frac{2}{n}}{\delta^2}\frac{\delta }{2}=\frac{\frac{1}{n}}{\delta }>1$
Another much better solution:
Choose $\displaystyle x_n=\frac{1}{n} $and $\displaystyle y_n=\frac{1}{n+1} $for any $\displaystyle n\in\mathbb{N}$.
$\displaystyle \left | x_n-y_n \right |<\frac{1}{n}$, but $\displaystyle \left | f(x_n)-f(y_n) \right |=1
$.