Compactness- Euclidean metric Hi Could you help me to solve this question?
IF  $||.||$  be any norm on   $\mathbb{R}^m$  and let   $B = \{ x \in \mathbb{R}^m  : ||x||≤ 1 \}$ . Prove that   $B$  is compact. Hint: It suffices to show that   $B$  is closed and bounded with respect to the Euclidean metric.

The euclideian metric induces the norm $d_2(x,0)=||x||_2=\sqrt{x_1^2+x_2^2+...+x_m^2}$.
Now in a finite dimensional space $X$ all the norms are equivalent thus are equivalent with $||.||_2$ i.e: $\exists C_1,C_2>0 $ such that $C_1||x||_2 \leqslant||x|| \leqslant C_2||x||_2, \forall x \in X$.
Now $B$ is bounded with respect to the norm $||.||_2$  thus form the equivalence of norms  it is bounded with respect to $||.||$
Also let $x_n \in B$ such that $x_n \rightarrow x$.
We have that $||x_n|| \leqslant 1$ and also we have that $||x_n|| \rightarrow ||x||$(the  norm is a continuous function).
Thus $||x|| \leqslant 1 \Rightarrow x \in B$.
We proved that $B$ is closed.
