Let G be an open set in $R^n$, $f:G\to R^n$ a continuous injection. I need to show that $f:G \to f(G)$ is not a homeomorphism if and only if there exists a sequence $Xn$ that converges to some point on the boundary of G or to $\infty$ so that $b=limit_{n\to\infty}f(Xn)$ exists and belongs to $f(G)$.
I proved the second direction, but I'm stuck with the first direction. I know that if $f$ is not a homeomorphism then $f^{-1}$ is not continuous but I don't know how to proceed. Any hint would be appreciated.
A theorem (invariance of domain) says that if $G \subseteq \mathbb{R}^n$ is open and $f: G \to \mathbb{R}^n$ is continuous and injective, then in fact $f$ is a homeomorphism between $G$ and $f[G]$ and $f[G]$ is open in $\mathbb{R}^n$.
So the condition that such an $f$ exists which is not a homeomorphism does not occur. From a false stetement you can prove anything you like (ex falso quodlibet)..