$f: \mathbb{R}^{n+1}-\{0\}\rightarrow S^{n}$ via $x \mapsto\frac{x}{\|x\|}$ is continuous

Question

I'm having trouble understanding why a map $f: \mathbb{R}^{n+1}-\{0\}\rightarrow S^{n}$ (unit $n$-sphere, $n\ge 1$) via $x \mapsto\frac{x}{||x\|}$ is continuous. Since Unit n-Sphere under Euclidean Metric is Metric Subspace of Euclidean Real Vector Space, I think that continuity of $f$ can be proved by $(\epsilon-\delta)$-definition of limit. That is for every real number $\epsilon > 0$ there exists $\delta > 0$ such that for every $x, y \in \mathbb{R}^{n+1}-\{0\}$ with $||x-y|| < \delta$, we have that $||\frac{x}{||x\|}- \frac{y}{||y\|}|| < \epsilon$. How $\delta$ can be chosen?

Any help would be much appreciated.

Answer 1

Hint: reverse triangle inequality will be helpful.

$$ \left\Vert \frac{x}{\Vert x \Vert} - \frac{y}{\Vert y \Vert} \right\Vert \leq \frac{\Vert x-y\Vert + \vert \Vert x \Vert-\Vert y \Vert \vert}{\Vert x \Vert} \leq \frac{2}{\Vert x \Vert} \Vert x - y \Vert $$

You get the first inequality by adding and subtracting $y/\Vert x \Vert$ then applying the usual triangle inequality and doing some manipulations. The second inequality is the reverse triangle inequality. Now take $\delta < \epsilon\Vert x \Vert/2$.

Answer 2

To start, you may assume without loss that $\|x\| > \|y\|$. Now let $\epsilon>0$ be given then if $\|x-y\|< \delta$ where $\delta = \|y\| \epsilon$ we have; $$\left\|\frac{x}{\|x\|} - \frac{y}{\|y\|}\right\| = \frac{1}{\|x\| \|y\|} \cdot \left\|\ x\|y\|- y\|x\| \ \right\| \leq \frac{1}{\|x\| \|y\|} \cdot \left\|\ x\|x\|- y\|x\| \ \right\|$$

$$=\frac{1}{\|x\| \|y\|} \cdot \left\|\ x\|x\|- y\|x\| \ \right\| = \frac{1}{\|x\| \|y\|} \cdot \left\|\|x\|(x-y)\| \ \right\| = \frac{1}{\|y\|} \cdot \|x-y\|< \epsilon$$