If $f:\mathbb{R}^d\to\mathbb{R}^d$ is a homeomorphism, then $\lim_{\| x \| \to \infty} \| f(x) \| = +\infty$.
My attempt:
I need to show that $\forall M\in\mathbb{R}: \exists R>0: \forall x\in\mathbb{R}^d:\| x\| \ge R \Rightarrow \| f(x)\| \ge M$. Suppose that this is not the case, i.e. $\exists M: \forall R: \exists x\in\mathbb{R}^d: \| x\| \ge R \quad\& \quad \| f(x)\|< M.$ Pick such an $M\in\mathbb{R}$. Then $$\forall n\in\mathbb{N}: \exists x_n: \|x_n\|\ge n \quad \&\quad \| f(x_n)\|<M.$$ Then we have that $x_n\not\to 0$. I thought of using the bijectivity of $f$, i.e. $\exists y_n\in\mathbb{R}^d: f(x_n)=y_n \iff x_n = f^{-1}(y_n)$. Then we would have that $\| f^{-1}(y_n)\|\ge n \quad \& \quad \| y_n\| < M$.
I want to show that $f^{-1}(x_n)\not\to f^{-1}(0)$ as this would imply that $f^{-1}$ is not continuous in $0$, a contradiction. How can I do this?
Thanks.
Starting from here:
$\forall n\in\mathbb{N}. \exists x_n: \|x_n\|\ge n \quad \&\quad \| f(x_n)\|<M$
I would use the fact that every sequence which lies in a compact set of $\mathbb{R}^d$ has a convergent subsequence. Notice this applies to the sequence $f(x_n)$, which lies in the compact set $\{x: ||x|| \leq M\}$. Let $(f(x_{n_m}))_{m=1}^\infty$ be such a subsequence, with $f(x_{n_m}) \to L$ as $m \to \infty$.
By continuity of $f^{-1}$, we must have $f^{-1}(f(x_{n_m})) = x_{n_m} \to f^{-1}(L)$ as $m \to \infty$. But this is a contradiction, since $||x_{n_m}|| \to \infty$.