True or False:-
There is a continuous function $f$ from {$x\in \mathbb{R}^n :||x||\le 1$} onto $\mathbb{R}^n$. where $||x||=(x_1^2+...+x_n^2)^{1/2}$.
I think it is not possible as the domain is compact set and $f$ is continuous , so range set will also be compact.But $\mathbb{R}^n$ is not compact.
Am I right?
You are absolutely right. If you know the characterization of compact subsets of $\Bbb R^n$ as those that are closed and bounded, then there's nothing more to do! Otherwise, you'll probably want to show that the closed unit ball is compact and that $\Bbb R^n$ is not.
You can do this directly if you want, if you know that every bounded sequence has a convergent subsequence (Bolzanno-Weierstrass). If $f$ takes the closed unit ball (ie. $\|x\|\le 1$) onto $\mathbb{R}^n$, then there is some sequence $x_n$ in the closed unit ball (Bolzanno-Weierstrass lets us assume without loss of generality assume that it converges in $\mathbb{R}^n$) for which $f(x_n)\to\infty$ as $n\to\infty$. But since the closed ball is, well, closed, it contains its limit points and so $x_n\to x$ where $x$ is in the closed ball. But then $f(x)<\infty=\lim\limits_{n\to\infty}f(x_n)$, so that $f$ is not continuous.
But yes, you are right.
