Proving that $x^{\frac{1}{n}}$ is uniformly continuous over $[0, \infty)$ with the usual metric As the title says, I'm attempting to prove that the function $f(x) = x^\frac{1}{n}$ from $R$ to $R$, equipped with the usual metrics, is uniformly continuous over $[0, \infty)$. I believe that I have a proof of this. However, my final result seems a little too simple, so I was hoping someone could check/correct my proof. 
So, a function $f: E \rightarrow E'$ is uniformly continous if $ \forall \epsilon > 0$, $\exists \delta >0$, s.t for any $x,y \in E$, $d(x,y) < \delta$ implies $d'(f(x),f(y)) < \epsilon$.
In this particular case, f is given by $f(x) = x^\frac{1}{n}$, and $E = [0, \infty)$ and $E' = R$.
So, for $|f(x) - f(y)| = |x^{1/n} - y^{1/n}| = |x^{1/n} - y^{1/n}| \cdot \frac{x^\frac{n-1}{n} + y^\frac{n-1}{n}}{x^\frac{n-1}{n} + y^\frac{n-1}{n}} = \frac {|x-y|}{x^\frac{n-1}{n} + y^\frac{n-1}{n}} < \frac {\delta}{x^\frac{n-1}{n} + y^\frac{n-1}{n}} < \delta$
Therefore, as long as $\delta \leq \epsilon$, we will have that $d(x,y) < \delta$ implies $d'(f(x),f(y)) < \epsilon$. Does this seem correct, or did I make some small/major mistake?
 A: Your proof is not correct because for $n>2$,
$$|x^{1/n} - y^{1/n}| \left(x^\frac{n-1}{n} + y^\frac{n-1}{n}\right)\not=|x-y|.$$
However you can show the required property by noting that $x^{1/n}$ is u. c. in $[0,1]$ (continuous function on a compact set) and for $1\leq x<y$, by the Mean Value Theorem, there is $t\in (x,y)$ such that
$$|x^{1/n} - y^{1/n}|=\frac{t^{1/n-1}}{n}|x-y|\leq \frac{1}{n}|x-y|.$$
A: For $a>0$ and $X\ge 0$ the function $X^a$ is $\nearrow$ 
So for $\quad 0\le x\le y\quad$ we have $\quad 0\le x^a\le y^a$.
Let set $$\phi(y)=(y^a-x^a)-(y-x)^a$$

For $a-1\le 0$ and $X\ge 0$ the function $X^{a-1}$ is $\searrow$
So $\phi'(y)=ay^{a-1}-a(y-x)^{a-1}=a\left[(y-x)^{a-1}-y^{a-1}\right]\le 0\quad$  since $0\le y-x\le y$.

Finally for $a\in]0,1]$ the function $\phi$ is $\searrow$, this means $0\le y^a-x^a\le (y-x)^a$

It means that for $a\in]0,1]$ the function $X^a$ is $a$-Hölderian thus uniformly continuous.

Apply for $a=\frac 1n$ and $n\ge 1$.
Note: 
The uniform continuity is easily deduced from Hölder property: $|f(x)-f(y)|\le C|x-y|^a$
$\forall \varepsilon>0$ let $\delta=(\frac{\varepsilon}C)^{\frac 1a}$ then $|x-y|<\delta\implies|f(x)-f(y)|\le C|x-y|^a<C{(\frac{\varepsilon}C)^{\frac aa}}\le\varepsilon$
