Let $\alpha>0$. Show that the function $f(x)=1/x$ given by $f:[\alpha,\infty)\rightarrow \mathbb{R}$ is uniformly continuous.

Question

I encountered this question on my textbook but I think that the function $f(x)$ is not uniformly continuous for all $\alpha$. For example let us say $\alpha=10^{-10}$ and fix $\epsilon=1$. Then $\exists \delta>0$ s.t. $\forall x,y\in\mathbb{R}$ with $|x-y|<0$, $|\frac{1}{x}-\frac{1}{y}|<1$. Now, pick $x\in[10^{-9},10^{-8}]$ with $x<\delta$ and set $y=x/10$. Then $$|x-y|=|\frac{9x}{10}|<\delta, \text{but}$$ $$\lvert \frac{1}{x}-\frac{1}{y}\rvert=|-\frac{9}{x}|>1\quad \text{since $x\in[10^{-9},10^{-8}]$}$$ However, this contradicts with $f$ being uniformly continuous on its domain.

If there is a mistake, please help me to find out and to solve the exercise.

Answer 1

Choose $\epsilon\gt0$. Then for any $x,y\in[\alpha,\infty)$ we have that, for $\delta=\alpha^2\epsilon\gt0$, \begin{align} |x-y|\lt\delta &\implies|x-y|\lt xy\epsilon\\ &\implies\frac{|x-y|}{xy}\lt\epsilon\\ &\implies\left|\frac{x-y}{xy}\right|\lt\epsilon\\ &\implies\left|\frac1y-\frac1x\right|\lt\epsilon\\ &\implies\left|\frac1x-\frac1y\right|\lt\epsilon\\ \end{align} as required.