Consider the function $f : (0, 1] → R$ defined by $f(x) = 1/x$.
Prove that, for any $0 < r < 1$, f is uniformly continuous on $[r, 1]$.
I was trying to use the theorem if $f$ be continuous on [a,b] then $f$ is uniformly continuous. But I don't know how to relate and explain closed interval use the given $0< r < 1$.
Let $\epsilon > 0$, you need to find a $\delta(\epsilon)$ such $|x-y|<\delta \implies |f(x)-f(y)|<\epsilon$
Let $x,y\in[r,1]$ then $|f(x)-f(y)| = |\frac1x-\frac1y|=|\frac{y-x}{xy}|\le|y-x|\frac1{r^2}=|x-y|\frac1{r^2}<\epsilon$
Let $\delta=\epsilon*r^2$, and that concludes your proof.