Answer to Question #213036 in Functional Analysis for smi

Question

Show that R and C and, more generally, R^n and C^n are locally compact.

Answer 1

First of all, let us remark that for all $n\in \mathbb{N}$, $\mathbb{C}^n$ has the topology of $\mathbb{R}^{2n}$ (with the canonical homeomorphism $(z_1,...,z_n)\to (\Re z_1, \Im z_1, ..., \Re z_n, \Im z_n)$), so we can just study the case of $\mathbb{R}^n$.

Let $x\in \mathbb{R}^n$, we consider an open neighbourhood $x\in B(x,1)$ (an open ball of radius 1) and we shall prove that the unit cube $C_x =\{y=(y_1,...,y_n)\in \mathbb{R}^n \;|\; \max_{1\leq i \leq n} (|y_i-x_i| ) \leq 1\}$ is compact. First we will remark that $B(x,1)\subseteq C_x$, as $y\in B(x,1) \Rightarrow \sum (y_i-x_i)^2<1$. Secondly we remark that a cube $C_x$ is a product of $n$ closed intervals $C_x=\prod_{1\leq i\leq n} [x_i-1; x_i+1]$. Closed bounded intervals are compact (by Bolzano-Weierstrass theorem and the equivalence between sequential compactness and compactness for metric spaces). From this we conclude that $C_x$ is compact by using one of the following arguments :

• by Tychonoff's theorem, product of compact spaces is compact,
• as product is finite, we can apply a weaker result - we can just use Bolzano-Weierstrass theorem for each coordinate,
• we could have used a general result that closed bounded subsets of $\mathbb{R}^n$ are compact.

Therefore, $x$ admits a compact neighbourhood and thus for all $n\in \mathbb{N}$, $\mathbb{R}^n$ (and therefore $\mathbb{C}^n$) are locally compact.