# proof verification: $f(x) = 1/x$ is not uniformly continuous on the open interval (0,1).

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.

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

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