# Why is $\{ x: \|x\|=1\}$ compact?

Let $K$ be either $\mathbb{R}$ or $\mathbb{C}$. Let $\mathcal{X}$ be a finite-dimensional normed vector space on $K$. More specifically, let $e_1,\cdots,e_n$ be a basis of $\mathcal{X}$. The norm is defined as $\|\sum a_j e_j\| := \sum |a_j|$.

I would like to show that $S = \{x\in\mathcal{X}:\|x\|=1\}$ is compact. I have managed to show that $S$ is sequentially compact, but I'm not sure how to show that it is compact. I know Heine–Borel theorem but I thought it only works in $\mathbb{R}^n$.

Some solutions online use the fact that all norms are equivalent in finitely-dimensional normed space. But ultimately I want to prove that claim, and I suppose showing $S$ is compact is part of the proof.

• $\mathbb{C}^n=\mathbb{R}^{2n}$ and the norms are the same, so you can use Heine-Borel. – Spenser Mar 11 '17 at 9:12
• @Spenser That's one thing that I'm not sure about. I know that if two vector space have same dimensions, then they are isomorphic to each other as vector space. But that doesn't seem to say anything about their topology? – 3x89g2 Mar 11 '17 at 9:14
• Yes they have the same topology since that topology is defined by the norm, and these two norms are the same. – Spenser Mar 11 '17 at 9:16
• @Spenser Exactly. But "two norms are the same" is something that I want to prove, right? – 3x89g2 Mar 11 '17 at 9:17
• Yes, just note that $|x_1+iy_1|^2+\cdots+|x_n+iy_n|^2=x_1^2+y_1^2+\cdots+x_n^2+y_n^2$. – Spenser Mar 11 '17 at 9:18

Sequential compactness is equivalent to compactness in metric (so also normed) spaces. So you were actually done already.

• One more reason to review my topology! – 3x89g2 Mar 13 '17 at 15:05

For $a=(a_1,\ldots,a_n) \in \mathbb{K}^n$ define

$$f(a) := \sum_{j=1}^n a_j e_j \in \mathcal{X}.$$

As $f:\mathbb{K}^n \to \mathcal{X}$ is a linear function, it is not difficult to see that $f$ is continuous and bijective. Since

$$S = \{x \in \mathcal{X}; \|x\| = 1\} = f \bigg( \underbrace{\left\{ (a_1,\ldots,a_n) \in \mathbb{K}^n; \sum_{j=1}^n |a_j| =1 \right\}}_{\text{compact subset of \mathbb{K}^n}} \bigg)$$

we find that $S$ is compact as an image of a compact set under a continuous mapping.

This answer is more or less in response to your uncertainty in the comments. We use the following lemma:

Lemma: Let $x_1,\dots,x_n$ b a collection of linearly independent vectors in a normed space $X$. Then there exists some $c>0$ such that for every choice of $\alpha_1,\dots,\alpha_n$ we have $$\|\alpha_1 x_1+\dots+\alpha_n x_n\| \geq c(|\alpha_1|+\dots|\alpha_n|).$$

A full proof of this Lemma can be found in Chapter $2$ of Kreyzwig's introductory functional analysis.

Theorem: On a finite dimensional vector space $X$, all norms are equivalent (they generate the same topology since there exist positive scalars so that $$a\|x\|_0 \leq \|x\| \leq b\|x\|_0.$$

proof: Specify a basis $e_1,...,e_n$. Then every $x=\alpha_1 e_1+ \dots+\alpha_n e_n$ By the lemma we can find $c$ so that $$\|x\| \geq c)|\alpha_1|+\dots+|\alpha_n|).$$

But then we take the maximum: $k:=\mathrm{max} \|e_j\|_0$, and apply the triangle inequality:

$$\|x_0\| \leq \sum_{i=1}^{n}|\alpha_i| \|e_i\|_0 \leq k\sum_{i=1}^{n} |\alpha_i|.$$

Hence, $\frac{c}{k} \|x\|_0 \leq \|x\|$. The other inequality is the same process, switching the roles of the norms.

A different proof can be found here.

Anyhow, we know that we can already think of any norm topologically as euclidian, so Heine-Borel applies.