Is $SL(2, \mathbb{R})$ dense in saturated elementarily extensions of the reals? 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 $SL(2,A)$?
 A: Turning my comment into an answer:
No, it's not. In fact, every "new" element of $SL(2,A)$ will have a neighborhood missing all of $SL(2,\mathbb{R})$ - and this uses merely the non-Archimedeanness of $A$.
Suppose $X\in M_2(A)\setminus M_2(\mathbb{R})$ (so $X$ has at least one non-real entry). Let $\epsilon$ be a "small enough" infinitesimal - specifically, let $\epsilon$ be smaller than (the absolute value of) the difference between each finite non-real entry $x$ of $X$ and the real part of $x$. The ball of radius $\epsilon$ around $X$ contains no elements of $M_2(\mathbb{R})$.
Now just make sure the determinant of $X$ is $1$. For example, fixing an infinitesimal $\delta$ we could use $$\begin{bmatrix}\delta & 0\\0 & {1\over\delta}\end{bmatrix}$$ (we could use $\epsilon=\delta^2$ as our corresponding radius).
EDIT: as Adayah observes, this in fact shows that $SL(2,\mathbb{R})$ is nowhere dense in $SL(2,A)$. As a further point of interest it's not hard to show that $SL(2,\mathbb{R})$ is closed as well (see their answer).
A: 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 $c$.
