We define $ (\mathbb{R}^m, \|.\|)$ to be a finite dimensional normed vector space with $ \|.\|$ is defined to be any norm in $ \mathbb{R^m}$. Let $S = \lbrace x \in \mathbb{R}^m: \| x\| = 1 \rbrace.$ Prove that $S$ is compact in $ (\mathbb{R}^m, \|.\|).$

# [Math] The compactness of the unit sphere in finite dimensional normed vector space

compactnesslinear algebramatricesreal-analysis

## Best Answer

You can use induction on $m$ and properties of $\mathbb R$ to show compacity using

sequential compactness, which means the same thing for metric spaces. Now consider the norm induced on the space $\mathbb R \cong \mathbb R \times \{ 0 \} \times \dots \times \{ 0 \}$ viewed as a sub-metric-space of $\mathbb R^m$, and also consider the sequence $^1 x_n$ induced by putting all the other components but the first equal to $0$. Therefore the first component is a sequence of real numbers. Since in the reals, every metric is equivalent to the absolute value metric in the following sense $$ \forall (\mathbb R,d), \quad \exists c_1, c_2 > 0 \quad s.t. \quad \forall x,y \in \mathbb R, \quad c_1 d(x,y) \le |x-y| \le c_2 d(x,y). $$ One can deduce that the Bolzano-Weierstrass theorem also holds if we replace $| \cdot |$ by the induced metric from the norm in $\mathbb R^m$. Since the sequence $x_n$ is bounded, the sequence $^1x_n$ is also bounded in $\mathbb R$. Therefore there exists a subsequence of the sequence $x_n$ such that the first component converges. Repeat this procedure with the other components $^kx_n$ with $1 \le k \le n$, and you will get a subsequence that converges component by component, hence converges. This gives you for every sequence an element $x$ and a subsequence for which $x_n \to x$. Since $\| x_n \| = 1$ for every $n$, clearly $\|x \| = 1$, so that your subsequence converges in $S$ and we are done.That is

one wayto do it ; if you have seen theorems in class that might help, perhaps they might make this less complicated.Hope that helps,