Prove the set of vectors of norm one is compact I'm trying to prove the set of vectors of norm one, $\{v\in\mathbb{R}^n:v_1^2+\cdots+v_n^2=1\},$ is compact. I think I could use Heine-Borel since it's in $\mathbb{R}^n.$ But, how do I show that this set is closed and bounded in terms of norm? Thanks.
 A: Boundedness is obvious since the set is contained in the ball of radius $2$. It is also closed since it is the inverse image of the closed set $\{1\}$ under the continuous function $\mathbb R^n \to \mathbb R$ given by $x\mapsto \|x\|$ (both given the usual metric). 
A: This might not be the most efficient solution to closedness, but when quick answers don't jump to mind, I usually just check the definition:
Let $A=\{v\in\mathrm{R}^n:v_1^2+\ldots+v_n^2=1\}$ and show that $A^C$ is open. Let $x\in A^C$, thus $||x||\not=1$. 


*

*Case 1: $||x||<1$. Let $\epsilon=1-||x||$, thus $||x||=1-\epsilon$ and $\epsilon>0$. Define $B=B(x,\epsilon)$, then:
$$y\in B \implies ||x-y||<\epsilon,$$
thus
$$||y||=||x-(x-y)||\leq||x||+||x-y||<1-\epsilon+\epsilon=1.$$

*Case 2: $||x||>1$. Let $\epsilon=||x||-1$, thus $||x||=\epsilon+1$ and $\epsilon>0$. Define $B=B(x,\epsilon)$, then:
$$y\in B \implies ||x-y||<\epsilon,$$
thus
$$||x||=||y+(x-y)||\leq||y||+||x-y||\implies ||y||\geq||x||-||x-y||.$$
Now $||x-y||<\epsilon$, thus $-||x-y||>-\epsilon$, such that
$$||y||\geq||x||-||x-y||>\epsilon+1-\epsilon=1.$$


Showing that if $x\in A^C$, there is an $\epsilon>0$, such that $B(x,\epsilon)\subset A^C$, i.e. $A^C$ is open, i.e. $A$ is closed.
