# Proving a set defined using the norm is compact

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?

• Apologies. Made an error typesetting it. I've fixed it now – marzg Sep 16 at 17:08

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 $$\Bbb R$$.
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.