Let $f(x)=x\sqrt{\frac{1+x}{1-x}}$. Is the function uniformly continuous over the interval $[-1,1)$ and $[0,1)$. I know that I have to use the Cantor's theorem but the interval is not closed and bounded? How to do this? And another question: If the interval includes $∞$ or $-∞$ how to prove is it uniformly continuous or not?
Maybe this question is trivial but I started studying it now. Thanks!
Suppose that $f$ is uniformly continuous. Then there is some $\delta > 0$ such that $$ | x - y | < \delta \implies | f(x) - f(y) | < 1. $$
Now $f$ is bounded on the compacta $[-1, 1 - \frac{\delta}{2}]$. But the formula above shows that, for all $x \in (1- \frac{\delta}{2}, 1)$, $$f(x) \in (f(1-\frac{\delta}{2}) -1,f(1-\frac{\delta}{2}) +1),$$
so that $f$ is bounded on $[-1,1)$. But an easy calculation shows that $\lim_{x \to 1} f(x) = \infty$. Contradiction!