So the following statement seems to be obvious but I don't see how to prove it:


Q: How does one prove that a closed totally disconnected subgroup of a connected real Lie group is discrete?


Note that it is essential to take into account the group structure since the Cantor set is a closed totally disconnected subset of $\mathbf{R}$ which is not discrete.



On closed totally disconnected subgroups of connected real Lie groups 
Let's just give a quick argument:


Let $G$ be a Lie group (with Lie algebra $\mathbb g$) and $H \subset G$ be a closed subgroup. If $H$ is not discrete, then there exists a sequence $(g\_n)\_{n \in \mathbb N}$ of elements of $H$, which converges to $1\_G$. Write $g\_n = \exp( \alpha\_n \xi\_n)$ (for $n$ large enough), where $\xi\_n \in \mathbb g$ are unit vectors (with respect to some innerproduct) and $\alpha\_n \in \mathbb R\_{> 0}$. Since $g\_n \to 1\_G$, we get $\alpha\_n \to 0$. Let $\delta \in \mathbb R$ and find integers $k\_n(\delta)$, such that $|\alpha\_n \cdot k\_n -\delta|< \alpha\_n$.


Let $\xi$ be some accumulation point of the set $\lbrace \xi\_n \mid n \in \mathbb N \rbrace$. Now, it is easy to see that $$g\_n^{k\_n(\delta)} = \exp(\alpha\_n k\_n(\delta) \xi\_n) \to \exp(\delta \xi)$$ on a subsequence. This implies that the whole 1-parameter subgroup generated by $\xi$ lies in $H$. In particular, $H$ is not totally disconnected.


