Compactness in $\mathbb R^n $

Question

I am not sure if I'm missing something subtle or I'm just totally wrong.

Is the following true?\

$K\subseteq \mathbb R^n$ is compact (usual topology with 2-norm) if and only if $K\subseteq \mathbb R^n$ is compact under a norm on $\mathbb R^n$?

I would think equivalence of norms would play a role here.

Answer 1

This is true. As you said this is due to the fact that in finite dimension all norms are equivalent. So here if $K$ is compact it just means :

$$\exists M, \forall x \in K, \|x\|_2 \leq M \text{ and } K \text{ is closed }$$

Now take any norm $N$ on $\mathbb{R}^n$. Since all norms are equivalent then there is such that : $\forall x \in \mathbb{R}^n, \| x \|_2 \leq C N(x)$. Hence $K$ is also bounded for the norm $N$.

You can check that "closeness" still holds on $K$ with $N$.

Answer 2

All norms on $\mathbb{R}^n$ are equivalent which means they yield the same topology and in particular the same compact subsets.