I have some problems in understanding the following proof for equivalence of norms for finite dimensional space.
Theorem : let $v$ be a finite dimension space. If $||.||$ and $|||.|||$ are two norms on $v$, then there are $c_1$ and $c_2$ s.t. for all $x \in v$ , $c_1||x|| \leq|||x||| \leq c_2||x||$.
for showing $|||x||| \leq c_2||x||$, it was simple. The norm of $|||.|||_{\infty}$ and $ ||.||_{2}$ are compared, but for showing $c_1||x|| \leq|||x|||$ it was written:
$ S =\{x \in v :||x|| =1\} = \{(\alpha_1,...\alpha_n)^{T} \in R^{n}, ||\alpha_1e_1+...\alpha_n e_{n}|| =1 \}$
$ S\subseteq R^{n}$ ,and S is closed and bounded => S is compact and since $0 \notin S => distanace (0, S) =d>0$ ==> $ \forall x \in v: \quad \frac{1}{d} |||x|||| \leq ||x|| \quad $
I cannot understand the last part of the proof:
$ S\subseteq R^{n}$ ,and S is closed and bounded => S is compact and since $0 \notin S => distanace (0, S) =d>0$ ==> $ \forall x \in v: \quad \frac{1}{d} |||x|||| \leq ||x|| \quad $
Can anyone help please?
Thanks
P.S.
I have not took a course in real analysis. I read the very basic facts about compact and closed set.
-As an example, I understand how rational numbers are dense in real numbers set.
-I understand if a set contains its limit points, it is considered as a closed set.