Lemma 1

— Any compact topological vector space $E$ over a non-discrete valued division ring, is just the set $\{0\}$.

Since E is complete we can suppose K is complete (I, p. 6). If E $\neq \{0\}$ then E contains a line that is closed in E (I, p. 14, cor. 1) and therefore compact. This line is isomorphic to $K_s$ (I, p. 12, prop. 2) and hence K must be compact. Now the mapping $\xi \mapsto |\xi|$ of K in $\mathbf{R}$ is continuous and thus the image of K must be bounded, on the other hand there exists $\gamma \in K$ with $|\gamma| > 1$, and the set $|\gamma^n| = |\gamma|^n, n \in \mathbf{N}$, is unbounded. This contradiction shows that $E = \{0\}$.

To prove the second assertion in the theorem, if $E \neq \{0\}$ then from the first part of the theorem E is isomorphic to $K_s^n$ with $n > 0$; now K is complete, hence so is E, and thus E is locally compact. But $K_s$ is isomorphic to a line in E (I, p. 12, prop. 2) which is necessarily closed in E (I, p. 14, cor. 1); it follows that K is locally compact.