# Every (general) finite dimensional banach space has compact unit ball

I was trying to prove that the closed unit ball $$B$$ is compact if and only if the space is finite dimensional. I have already shown that if the space is infinite dimensional, then the closed unit ball is not compact. My trouble is in proving the reverse direction. My idea was to use the fact that if V is the fin-dim Banach space, then there is an isomorphism $$T:V\rightarrow \mathbb{R}^{n}$$. If I can show that T is a homeomorphism, then I could use the fact that $$T$$ is bounded to show that $$T(B)$$ is bounded. $$T(B)$$ would also necessarily be closed meaning we can use Heine-Borel to say that $$T(B)$$ is compact. And then $$T^{-1}T(B)=B$$ must be compact as well. However my trouble lies in showing $$T$$ is bounded. Most proofs that I saw invoke equivalence of norms but to prove equivalence of norms in an arbitrary fin-dim Banach space I would need to somehow show that the closed unit sphere is compact. I know in $$\mathbb{R}^{n}$$ the closed unit sphere is compact by Heine-Borel but I'm not sure if that generalizes to arbitrary finite dimensional banach spaces without using the fact that there is an homeomorphism to $$\mathbb{R}^{n}$$ which is what I am trying to prove in the first place.

• @JulioPuerta I guess in some sense that is what I am trying to prove no?, that in this case T is a homeomorphism. Could you refer to me a proof of the existence of such a topological isomorphism? Mar 11 at 20:14

Here are two options. The first option is to prove that if $$(V, \|\cdot\|_{V})$$ is a real finite-dimensional Banach space of dimension $$n$$ with a basis $$\{v_{1}, \ldots , v_{n}\}$$, then $$\|\cdot\|_{2}:V\to\mathbb{R}$$ defined by

$$\left\| \sum_{i=1}^{n}\alpha_{i}v_{i} \right\|_{2} := \left( \sum_{i=1}^{n}|\alpha_{i}|^{2} \right)^{1/2}$$

is a norm on $$V$$ equivalent to $$\|\cdot \|_{V}$$. To show that there is some $$C\in (0, \infty )$$ such that $$\|x\|_{2} \leq C\|x\|_{V}$$ for all $$x\in V$$, this can be done by showing that if $$\{e_{1}, \ldots , e_{n}\}$$ is the standard basis for $$\mathbb{R}^{n}$$, the map $$\phi :(V, \|\cdot\|_{2})\to (\mathbb{R}^{n}, \|\cdot\|_{\mathbb{R}^{n}})$$ given by

$$\phi \left( \sum_{i=1}^{n}\alpha_{i}x_{i} \right) = \sum_{i=1}^{n}\alpha_{i}e_{i}$$

is a vector space isomorphism which is continuous in both directions. You can apply the Heine-Borel theorem in $$(\mathbb{R}^{n}, \|\cdot\|_{\mathbb{R}^{n}})$$ to continue the proof. This proves that all norms on $$V$$ are equivalent, and from there you can show that any linear map $$T:V\to\mathbb{R}^{n}$$ is continuous.

A second option is to first show that all norms on $$\mathbb{R}^{n}$$ are equivalent in a method similar to the one above. Then if $$T: \mathbb{R}^{n} \to V$$ is a vector space isomorphism, for any norm $$\|\cdot\|_{V}$$ on $$V$$, the map $$\rho :\mathbb{R}^{n}\to\mathbb{R}$$ defined by $$\rho (x) := \|T(x)\|_{V}$$ is a norm on $$\mathbb{R}^{n}$$ which is equivalent to the standard norm on $$\mathbb{R}^{n}$$. So there are $$C_{1}, C_{2} \in (0, \infty )$$ such that for all $$x\in \mathbb{R}^{n}$$,

$$C_{1}\|x\|_{\mathbb{R}^{n}} \leq \|T(x)\|_{V} \leq C_{2}\|x\|_{\mathbb{R}^{n}}.$$

From there, it can be shown that $$T$$ is a homeomorphism, which allows you to conclude your proof.