Let $k$ be a global field and $k_v$ denote its completion at a place $v$.
Let $Z$ be the closure of $SL_2(k)$ in $SL_2(\mathbb{A})$; $Z$ is a subgroup. Assume $Z$ contains $SL_2(\mathcal{O}_v)$ for each place $v$. Why does that mean $Z$ contains the infinite product $\prod_v SL_2(\mathcal{O}_v)$? I think it has to do with $Z$ being topologically closed and the topology adele groups but I think I'm missing the simple topological reason.
Below is the full argument given here http://math.stanford.edu/~conrad/248BPage/handouts/strongapprox.pdf and and I get all the other parts.
This is the definition of $SL_2(\Bbb{A}_k)$, its topology:
The elements $(a_v)\in \prod_v SL_2(k_v)$ such that $a_v\in SL_2(O_v)$ for all but finitely many $v$.
$x^{(n)} \to x$ if all the $x^{(n)}_{v_j}$ are in $SL_2(O_{v_j})$ for $j$ large enough and $x^{(n)}_v\to x_v$ (in the norm of $k_v^{2\times 2}$) for all $v$.