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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.

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  • $\begingroup$ Which definition of compactness are you using? $\endgroup$ – BigbearZzz Jul 13 '17 at 2:57
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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.

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  • $\begingroup$ thank you Ramiz Karaeski $\endgroup$ – seyfullah Jul 14 '17 at 3:27
  • $\begingroup$ you are welcome $\endgroup$ – Marios Gretsas Jul 14 '17 at 10:11

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