Show that $f : \mathbb{R}^n \setminus \{0\} \to \mathbb{R}^n$ defined by $f(x) = \frac{x}{|x|}$ is continuous and that $f\left[\mathbb{R}^n \setminus \{0\}\right] = \mathbb{S}^{n-1}$
I managed to show that $f\left[\mathbb{R}^n \setminus \{0\}\right] = \mathbb{S}^{n-1}$
My proof: Pick $\alpha \in f\left[\mathbb{R}^n \setminus \{0\}\right]$, the $\alpha = f(x)$ for some $x \in \mathbb{R}^n$. Fix this $x \in \mathbb{R}^n$ then $\alpha = \frac{x}{|x|}$ for this $x$. Now $$d(0, \alpha) = |\alpha| = \left|\frac{x}{|x|}\right| = \frac{|x|}{||x||} = \frac{|x|}{|x|} = 1$$ hence $\alpha \in \mathbb{S}^{n-1} = \{y \in \mathbb{R}^n \ | \ d(0, y)=1 \}$. Thus $f\left[\mathbb{R}^n \setminus \{0\}\right] \subseteq \mathbb{S}^{n-1}$. Conversely pick $\beta \in \mathbb{S}^{n-1}$, then $d(0, \beta) = 1$ and hene $|\beta| = 1$. We need to show that $\beta = f(x)$ for some $x \in \mathbb{R}^n$. Since $\mathbb{S}^{n-1} \subseteq \mathbb{R}^n$ we have that $\beta \in \mathbb{R}^n$ hence $f(\beta) = \frac{\beta}{|\beta|} = \frac{\beta}{1} = \beta$. Thus $\beta \in f\left[\mathbb{R}^n \setminus \{0\}\right]$ and hence $\mathbb{S}^{n-1} \subseteq f\left[\mathbb{R}^n \setminus \{0\}\right]$, hence finally we may conclude that $f\left[\mathbb{R}^n \setminus \{0\}\right] = \mathbb{S}^{n-1}$ as desired. $\square$
To show that $f$ is continuous however I'm finding a bit more difficult. I don't want to show that $f$ is continuous using directly using the definition of continuity for metric spaces as I'm sure there are easier ways to show that $f$ is continuous.
For example I know that the composition of continuous functions are continuous (which holds in any topological space). I also know that $g : \mathbb{R}^n \setminus \{0\} \to \mathbb{R}^n$ defined by $g(x) = x$ is continuous and $h : \mathbb{R}^n \setminus \{0\} \to \mathbb{R}^n$ defined by $h(x) = |x| = d(0, x)$ is continuous (where $d$ is the standard metric on $\mathbb{R}^n$). From these two facts what's the easiest way to show that $f$ which we can now be defined as $f(x) = \frac{g(x)}{h(x)}$ is continuous?
Finally is my proof above correct? (Note that $d$ is the standard metric on $\mathbb{R}^n$)
Regarding the continuity proof, you can notice the fact that the application $$\begin{array}{l|rcl} \phi_4 : & \mathbb R \times \mathbb R^n & \longrightarrow & \mathbb R^n\\ & (\lambda,x) & \longmapsto & \lambda x \end{array}$$
is bilinear in finite dimensional spaces and therefore continuous. A more elementary way is to notice that all the coordinate functions are continuous.
The application $\phi_1 : x \mapsto \vert x \vert^2$ is polynomial and therefore continuous. $\phi_2 : x \mapsto \sqrt{x}$ is a continuous map (defined on $\mathbb R_+$). And $\phi_3 : x \mapsto 1/x$ is also continuous for $x \neq 0$.
You now have to write $f$ as a function composition of all those nice functions, namely: $$f(x)=\phi_4((\phi_3\circ \phi_2 \circ \phi_1)(x),x)$$