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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.

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    $\begingroup$ Looks good to me. $\endgroup$
    – TonyK
    Jun 24, 2019 at 22:09
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    $\begingroup$ Refer here if you want to see an older discussion: math.stackexchange.com/questions/1371905/… $\endgroup$
    – scoopfaze
    Jun 24, 2019 at 22:10
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    $\begingroup$ I don't see this as a duplicate. The two questions concern the same theorem, but ask for specifics about different proofs of it. $\endgroup$ Jun 25, 2019 at 9:16

1 Answer 1

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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. $n\leq3$

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