A group generated by a closed subset of a closed subgroup $G$ of $GL(n,\mathbb R)$ that is neither open nor closed in $G$. I am looking for an example of A group generated by a closed subset of a closed subgroup $G$ of $GL(n,\mathbb R)$ that is neither open nor closed in $G$.
My first thought is to take $G$ as $SL(n,\mathbb R)$, but how to come up with a closed subset $U$ of it that generates a group which is neither open nor closed?
 A: One example is as follows:
One has pretty naturally $\mathbb{R} \subseteq GL_n(\mathbb{R})$ as a closed subgroup by sending 
$$r \mapsto\begin{pmatrix}
1 & 0 &0 & \dots & r \\
0 & 1 & 0 & \dots & 0 \\
\vdots &  \vdots & \vdots & \dots & \vdots \\
0 &  0& \dots &1 &0 \\
0 &  0& \dots &0 &1 \\ \end{pmatrix}.$$
Then it is enough to find a non-closed subgroup of $\mathbb{R}$ generated by a closed set. This should be possible in more that one ways: One can take e. g. $\langle \{ \frac{1}{n}\;|\; n \in \mathbb{N}\} \cup \{0\} \rangle=\mathbb{Q} \subseteq \mathbb{R}$ non-closed, even though the given set of generators is clearly closed. 
A: Let $G = SO(2) \in GL(2, \mathbb R)$, which is homeomorphic to $S^1$ via
$$
J : S^1 \to SO(2) : t \mapsto \begin{bmatrix} \cos t & -\sin t \\ \sin t & \cos t \end{bmatrix}.
$$
That's a closed subgroup. 
Now let $S = \{ J(0), u = J(1) \}.$, and consider the group $H$ generated by $S$. 
That, being a finite discrete subset of $G$, is closed. But since $\pi$ is not rational, $H$ doesn't contain $J(\pi)$. On the other hand, the powers of $u$ form a dense subset of the circle. Its closure is the whole circle, so this set isn't closed. And its interior is empty, since it's a subset of a $S^1$, whose interior (in $GL(2, \mathbb R)$) is also empty, because it's a 1-manifold nicely embedded in a 4-manifold. 
