Answer to Question #213034 in Functional Analysis for smi

Question

Show that R^n and C^n are not compact.

Answer 1

First of all let us show that $\mathbb{R}^n$ is not compact. There are two possible definitions of compactness for metric spaces (Bolzano-Weierstrass sequence property and Borel-Lebesgue open cover property), let us disprove both of them for $\mathbb{R}^n$.

• Let us consider a sequence $u_n=\begin{pmatrix} n \\0 \\ ... \\ 0 \end{pmatrix}$ for $n\in \mathbb{N}$. This sequence does not admit a converging subsequence, as for any subsequence the first coordinate diverges to infinity. Therefore, $\mathbb{R}^n$ does not admit the Bolzano-Weierstrass property and thus is not compact (as it is metric).
• Let us consider an open cover $U_n=B(0,n)$ by the open balls centered at origin with a growing radius. This open cover does not admit a finite subcover, as for any finite subcover we have $\bigcup_{n\leq N} U_n \subseteq B(0,N+1)\neq \mathbb{R}^n$. Therefore, $\mathbb{R}^n$ does not admit the Borel-Lebesgue property and thus is not compact.

To prove that $\mathbb{C}^n$ is not compact, we can either use the former approach or use a simpler argument : if $\mathbb{C}^n$ were compact, then $\mathbb{R}^n\subseteq \mathbb{C}^n$ being a closed subset of a compact space, would be compact too. However we have just proven the contrary, and thus $\mathbb{C}^n$ is not compact.