How to show that $x/|x|$ is continuous? This is from p.156 of Topology, Munkres: The unit sphere $S^{n-1}$ in $\mathbb{R}^n$ is path-connected, since it is the continuous image of the surjective function $g: \mathbb{R}^n -0\to S^{n-1}$ by $g(x)=x/|x|$. (Note that the punctured euclidean space $\mathbb{R}^n -0$ is path-connected for $n>1$.)
But how can I show that $g$ is continuous? I have an approach but this looks not cool: Since the domain and codomain are metric spaces, we can use epsilon-delta method. So 
$$\begin{align*}
\left|\frac x {|x|} - \frac y {|y|}\right| &= \frac1{|x||y|}\Big|\big(x|y|-|x|y\big)\Big|\\
&\le \frac1{|x||y|}\Big(\Big|\big(x|y|-|y|y\big)\Big|+\Big|\big(|y|y-|x|y\big)\Big|\Big)\\
&\le \frac1{|x||y|}\Big(|y||x-y|+||y|-|x|||y|\Big)\\
&\le \dfrac{|x-y|}{|x|}+\dfrac{|y-x|}{|x|}\\
&<2\dfrac{\delta}{|x|}\;.
\end{align*}$$
So at first by choosing delta to be smaller than $2\dfrac{\delta}{|x|}<\epsilon$, this function is continuous.


*

*Is it right?

*Are there any cool method?


The map $g$ is the projection map when the unit sphere is taken to be the quotient space of $\mathbb{R}^n-0$ under the equivalence relation that identifies point on the same ray from the origin.
In fact, you can prove that $g$ is continuous directly from the definition using open sets: A basis for the induced topology of the sphere is the intersection of balls from the space with the sphere. The inverse images under $g$ of these sets are open cones in the space.
