Let $G$ be a topological group (or Lie group), $\Gamma <G$ a lattice, $H<G$ a closed connected subgroup such that $H\cap \Gamma$ is a lattice in $H$. let $K=N_G(H)\cap\Gamma$.
Let $x_i\in \Gamma\backslash K$ such that $x_i K$ converges to some $xK$ for $x\in \Gamma$. I want to prove $x\in\Gamma\backslash K$ (in the quotient topology).
I'm new to topological groups, but I felt like it should be obvious, yet couldn't prove it. My attempts were direct, taking a neighborhood of $xK$, taking the inverse of it, and using discreteness. This failed badly. For that matter, my original desire was to only assume $x\in G$, but I couldn't even prove $x\in \Gamma$... in that case
Any kind of conversation would be good, as this is just me trying to learn more about topological groups.
P.S: Could we actually get (as well, or as a first step), that if $x_i \in \Gamma\backslash K$ converges to $x\in G$ then $x\in \Gamma$? even $x\in \Gamma \backslash K$?
This has nothing to do with lattices, all you need is discreteness of $\Gamma$ ($H$ can be taken an arbitrary subgroup of $G$). The point is that $K\subset \Gamma$ and $\Gamma$ is discrete. Hence, $x_iK\in \Gamma/K$, a discrete topological space. Hence, convergence $x_iK\to xK$ implies that $x_iK= xK$ for large $i$. Since $x_i\notin K$ for all $i$, it follows that $x\notin K$.
As for your P.S.: It is a standard exercise in the theory of discrete subgroups of topological groups that they are closed. Hence, $x_i\to x\in G, x_i\in \Gamma$, implies that $x\in \Gamma$.