TITLE: Is $SL(2, \mathbb{R})$ dense in saturated elementarily extensions of the reals?
QUESTION [5 upvotes]: Suppose $\mathcal{A} = \langle A, <, +, \cdot \rangle$ is a $\aleph_1$-saturated elementarily extension of the real field.
Is $SL(2, \mathbb{R})$ dense in $SL(2,A)$?
In compact groups one can find subgroups with infinitesimal entries, but is it still possible in $SL(2,A)$?

REPLY [3 votes]: It's the opposite - $SL(2, \mathbb{R})$ is closed in $SL(2, A)$ and has empty interior, so it is nowhere dense.
To see it is closed, first note that $\mathbb{R}$ is closed in $A$. Indeed: take any $\alpha \in A \setminus \mathbb{R}$. If $\alpha$ is unbounded (greater than $r$ for every $r \in \mathbb{R}$), then $(-\infty, \frac{\alpha}{2})$ or $(\frac{\alpha}{2}, \infty)$ is a neighborhood of $\alpha$ disjoint from $\mathbb{R}$. Otherwise $\alpha = r + \varepsilon$ for some $r \in \mathbb{R}$ and infinitesimal $\varepsilon$, and then $\left( \alpha-\frac{|\varepsilon|}{2}, \alpha+\frac{|\varepsilon|}{2} \right)$ is a neighborhood of $\alpha$ disjoint from $\mathbb{R}$.
It follows that $M_{2 \times 2}(\mathbb{R})$ is closed in $M_{2 \times 2}(A)$, so
$$SL(2, \mathbb{R}) = M_{2 \times 2}(\mathbb{R}) \cap SL(2, A)$$
is closed in $SL(2, A)$.
To see that it has empty interior, take any $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL(2, \mathbb{R})$ and its neighborhood. If $ad \neq 0$, choosing sufficiently small $\delta, \varepsilon \in A \setminus \mathbb{R}$ such that $(a+\varepsilon)(d+\delta) = ad$, we get that $\begin{pmatrix} a+\varepsilon & b \\ c & d+\delta \end{pmatrix}$ is in the neighborhood but not in $SL(2, \mathbb{R})$. If $ad = 0$, then $bc = -1$ and we can apply the same method to $b$ and $c$.