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I do not fully understand the proof for even roots so I cannot apply it for odd.

The proof for even $n$ uses Bolzano-Weierstrass theorem which I understand.

What I don't understand is the restriction of domain of $f(x) = x^n$ to $[\ 0, \infty)$. One can then show that the codomain is $[\ 0, \infty)$. Why one does that?

After that the whole proof goes by setting the grounds for using the Bolzano-Weierstrass theorem to prove that $x^n$ is surjective, continuous.

Since we know that $x^n$ for even $n$ is injective on the picked interval we have proved that function is bijective which means that its inverse is also continuous.

I was thinking of using the $x^{2n+1}$ notation for odd $n$, provide a proof for even and then, using the fact that product of continuous functions is continuous, prove that $x^{2n}x$ is continuous. I was considering this only because I don't know how to apply the proof for even roots to odd(for odd $n$ function is injective on the entire domain).

But then I got stuck on proving that $x$ is continuous.

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$x\mapsto x$ is shown to be continuous by immediately applying the definition of continuous. Then $x\mapsto x^n$ is the product of several continuous functions, hence continuous. The fact that $x\mapsto x^n$ is continuous, howevr, should probably already be well-known at this stage as we are actually focussing on $x\mapsto \sqrt[n]x$. –  Hagen von Eitzen Dec 31 '12 at 14:28

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The $n$-th root function is usually defined on $[0,+\infty)$. Otherwise you would have to define $n$-th roots when $n$ is odd and when $n$ is even differently.

Now why does he show the condomain is $[0,+\infty)$? To show that $f$ is surjecitive. In effect he shows that the range of $f$ is $R_f=[0,+\infty)$.

This is necessary in order to define $f^{-1}:R_f\to [0,+\infty)$. After proving that $f$ is bijective, the inverse ($n$-th root) is defined $[0,+\infty)$. Because $f$ is continuous so is $f^{-1}$.

Finally, the continuity of $x$ (identity map) can be easily shown by the $\epsilon-\delta$ definition of continuity (take $\delta=\epsilon$). I don't know that you are trying to accomplish in the end. Please elaborate.

Note that if you have been taught that every non negative real number has a unique $n$-th root then the only thing necessary is to prove that $f(x)=x^n$ is continuous in $[0,+\infty)$. (But this very easy to prove rigorously)

EDIT: Suppose we have shown $f(x)=\sqrt[n]{x}$ is continuous in $[0,+\infty)$. For odd $n$ let $$g(x)=\begin{cases}\sqrt[n]{x}&\mbox{if, }x\ge 0\\ -\sqrt[n]{-x}&\mbox{if, }x< 0\end{cases}$$ We want to show that $g$ is continuous. But $g(x)=f(x)$ for $x>0$ and $g(x)=-f(-x)$ for $x<0$ and so $g$ is continuous in $\mathbb{R}^*$. It remains to show continuity at $0$ but this is also simple:

$$\lim_{x\to 0^+}g(x)=\lim_{x\to 0^+}\sqrt[n]{x}=0$$ by continuity of $f$ in $[0,+\infty)$. Also, $$\lim_{x\to 0^-}g(x)=\lim_{x\to 0^-}-\sqrt[n]{-x}=-0=0$$ by continuity of $\sqrt[n]{-x}$ in $(-\infty,0]$. Thus, $$\lim_{x\to 0}g(x)=0=g(0)$$ which proves continuity at $0$.

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The point is that for odd $n$ one must show that n-th root is continuous on $\mathbb{R}$. Which I just cannot show by restricting the interval or I can but can't see how. –  Looft Dec 31 '12 at 14:47
    
@Leolinus As I said, it is a canonical spoince to define $n$-th roots only $[0,+\infty)$. But what you are trying to do is simple. I will edit it in my answer –  Nameless Dec 31 '12 at 14:49
    
I still don't understand how this makes things simpler. And I don't understand how can I show continuity in $0$ if I have no knowledge of root functions and their domains. I do, now, understand why we pick a domain of x^n but I don't think I should expand the inverse without expanding the original function. x^(2n+1) has domain \mathbb{R} and I should try proving it using that. –  Looft Dec 31 '12 at 16:59
    
@Leolinus I will prove continuity at $0$. I don't understand the rest of your comment though... –  Nameless Dec 31 '12 at 17:03
    
For example, you expand the g(x) so it becomes an odd function. You showed nothing stating that inverse of $x^n$ is odd(when $n$ is odd), it might be obvious but it made me think and prove that the inverse of odd function is odd. Your answer is fine. –  Looft Dec 31 '12 at 17:11

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