# Thread: Compact Unit Ball Proof

1. ## Compact Unit Ball Proof

Question:
Let $\displaystyle \Vert \cdot \Vert$ be any norm on $\displaystyle R^m$ and let $\displaystyle B = \{ x \in R^m : ||x|| \le 1 \}$. Prove that $\displaystyle B$ is compact. Hint: It suffices to show that $\displaystyle B$ is closed and bounded with respect to the Euclidean metric.

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I don't think I'd have any difficulty doing this problem. I know that a subset of $\displaystyle R^m$ is compact if and only if it's closed and bounded, but I don't get why the hint says that it's sufficient to show that it's closed and bounded ONLY under the Euclidean metric. Isn't it possible that there may be other norms on $\displaystyle R^m$ such that the set is not closed or not bounded with respect to the other norm?

2. Originally Posted by JG89
Question:
Let $\displaystyle \Vert \cdot \Vert$ be any norm on $\displaystyle R^m$ and let $\displaystyle B = \{ x \in R^m : ||x|| \le 1 \}$. Prove that $\displaystyle B$ is compact. Hint: It suffices to show that $\displaystyle B$ is closed and bounded with respect to the Euclidean metric.

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I don't think I'd have any difficulty doing this problem. I know that a subset of $\displaystyle R^m$ is compact if and only if it's closed and bounded, but I don't get why the hint says that it's sufficient to show that it's closed and bounded ONLY under the Euclidean metric. Isn't it possible that there may be other norms on $\displaystyle R^m$ such that the set is not closed or not bounded with respect to the other norm?
No! That's the point of the problem : to show that any norm induces the same topology on Euclidean space.