Proposition 2

Let $G$ be a Lie group. (i) If $K = \mathbf{R}$ or $\mathbf{C}$, $G$ is locally connected. (ii) If $K$ is distinct from $\mathbf{R}$ and $\mathbf{C}$, $G$ is zero-dimensional (General Topology, Chapter IX, § 6, Definition 5). (iii) Suppose that $K$ is locally compact. For $G$ to be locally compact, it is necessary and sufficient that $G$ be finite-dimensional. (iv) If $G$ is generated by a subspace whose topology admits a countable base, then the topology on $G$ admits a countable base.

Let $U$ be a neighbourhood of $e$. There exists an open neighbourhood $U_1$ of $e$ contained in $U$ and homeomorphic to an open ball of a normed space $E$ over $K$. If $K = \mathbf{R}$ or $\mathbf{C}$, $U_1$ is connected, which proves (i). Suppose that $K$ is ultrametric. There exists a neighbourhood $U_2$ of $e$ which is closed in $G$ and such that $U_2 \subset U_1$. Then there exists a neighbourhood $U_3$ of $e$ such that $U_3 \subset U_2$ and $U_3$ is open and closed relative to $U_1$. Then $U_3$ is closed relative to $U_2$ and hence $G$ and open relative to $U_1$ and hence $G$. This proves (ii). For $G$ to be locally compact, it is necessary and sufficient that $E$ be locally compact; if $K$ is locally compact, this amounts to saying that $E$ is finite-dimensional (Topological Vector Spaces, Chapter I, § 2, Theorem 3), whence (iii). Suppose that $G$ is generated by a subset $V$ and let $W = V \cup V^{-1}$; then

if there exists a sequence dense in $V$, we see that there exists a sequence dense in G and, as G is metrizable (Proposition 1), the topology on G admits a countable base.