I want to prove that the function $g:(0,\infty)\to (0,\infty),\ g(x):=x^{1/n} \ (n\in\mathbb{N}, n\geq 1)$ is continuous on $(0,\infty)$.
Now, if the domain was a finite interval, it would suffice to apply the following result:
Thm. ($f\colon [a,b]\to\mathbb{R}$ continuous and monotone increasing)$\Rightarrow$ ($f$ is a bijection from $[a,b]$ to $[f(a),f(b)]$ and the inverse $f^{-1}\colon [f(a),f(b)]\to [a,b]$ is also continuous and strictly monotone increasing)
since if $n$ is a positive integer and $R>0$ then $x^n$ is strictly increasing on $[0,R]\Rightarrow x^{1/n}$ is a continuous function from $[0,R^n]$ to $[0,R]$.
Now, my question is: how do I extend this reasoning to the entire interval $(0,\infty)$?
Best regards,
lorenzo
Continuity is a local property. If $x>0$, we know $x\in(0,R^n)$ for large enough $R$ and that $g$ is continuous on this interval, so in particular $g$ is continuous at $x$. Since this argument works for every $x$, we may conclude that $g$ is continuous on all of $(0,\infty)$.