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I'm currently stuck on the following question

Let be $ \| \|$ be any norm on $\mathbb{R}^m$ and let $B = \{x \in \mathbb{R}^m : \| x \| \leq 1\}$. Prove that B is compact.

I've shown that this set is bounded but I can't seem to show that it's closed. Can someone give me a hint to start off?

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  • $\begingroup$ Apologies. Made an error typesetting it. I've fixed it now $\endgroup$ – marzg Sep 16 at 17:08
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Assuming that your set is $B=\{x\in\Bbb R^m\mid\|x\|\leqslant1\}$, then it is closed because, if $n\colon\Bbb R^n\longrightarrow\Bbb R$ is the norm, then $n$ is continuous, $B=n^{-1}\bigl([0,1]\bigr)$, and $[0,1]$ is a closed subset of $\Bbb R$.

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If $y\notin B$, then the ball of positive(!) radius $|y|-1$ around $y$ is disjoint from $B$. Hence the complement of $B$ is open and $B$ itself closed.

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