So let || || be any norm on $\mathbb{R}^m$ and let $B = \{ x \in \mathbb{R}^m:$ || $x$ || $\leq 1 \}$. I want to prove that B is compact.

So from the Heine-Borel Theorem, I know that every closed and bounded subset of $\mathbb{R}^m$ is compact. So it suffices to show that B is closed and bounded with respect to the Euclidean metric.

To show boundedness, I must show $B \subset \mathbb{R}^m \ _r(p)$ for some $p \in \mathbb{R}^m$ and some $r \gt 0$.

However, because || || is any norm, what $r$ would I choose? I'm assuming $p = (0, 0, ... , 0)$, where there are $m$ zeroes.

This problem that || || is any norm also messes with showing that B is closed with respect the Euclidean metric. How do I show that B contains all its limits in the Euclidean space if I don't even really know what the norm is defined as?


  • 1
    $\begingroup$ Show that on $\mathbb{R}^n$ all norms are equivalent. For that, it may be useful to use the compactness of the closed unit ball and the unit sphere with respect to the Euclidean norm. $\endgroup$ – Daniel Fischer Sep 24 '13 at 11:32
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    $\begingroup$ Closedness comes from the fact that it is the inverse image of the closed set $[0, 1]$ by the function $||\cdot ||:\Bbb R^m \to \Bbb R$. $\endgroup$ – Arthur Sep 24 '13 at 11:33

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