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I've written a proof that $f\left(x\right)=\frac{1}{x}$ is not uniformly continuous on the interval $(0,1)$ and would like to know if it is correct. Here's what I've got.

In order to show a function $f$ is not uniformly continuous on $A$, it suffices to show there exist two sequences $(x_n)$ and $(y_n)$ in $A$ and an $\epsilon_0>0$ satisfying $\lim(|x_n-y_n|)=0$ but $|f(x_n)-f(y_n)|\ge\epsilon_0$.

Let $x_n=\frac{1}{n}$ and $y_n=\frac{2}{n}$, with $n\ge3$, and set $\epsilon_0=\frac{3}{2}$. Then $\lim(|x_n-y_n|)=0$, but

$\left|\frac{1}{x_n}-\frac{1}{y_n}\right|=\left|n-\frac{n}{2}\right|=\frac{n}{2}\ge\epsilon_0=\frac{3}{2}$, as desired.

Wilhelm
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1 Answers1

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Proof is absolutely correct.

If you take $x_n$$=$$1/n$. And $y_n$$=$$1/(n+1)$ then you don't have to put extra thing I.e. $n\leq3$

Om Prakash
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