If $G$ is a topological group acting effectively on a topological space $X$ and $H$ is a subgroup of $G$ then is it true that $\overline{H}x = \overline{Hx}$, where $\overline{H}$ is the closure of $H$ in $G$ and $Hx$ denotes the $H$ - orbit of $x$.
I thought one way would be easy - proving $\overline{Hx} \subseteq \overline{H}x$. However for that I need to show that $\overline{H}x$ is closed and I am unsure how to do that (or if it is even true). So I am not sure how to prove either one is contained in the other.
I am particularly trying this for the case when $G=\text{ Homeo}(X)$ with compact open topology, $X$ is locally connected, locally compact hausdorff space and $H$ is a subgroup of $G$.
a. Is $\overline{Hx} = \overline{H}x$ for any $G$ and $X$?
b. If not then is it true for my particular case?
This was a fun question. Unfortunately, what you hope for is false.
Let $X = S^1$. In this answer it will frequently be useful to consider $S^1$ as the quotient $[0,1]/(0 \sim 1)$. (Actually, the answer will be even easier if we just consider $X= [0,1]$, but I figured you might like a manifold counterexample.)
Let $f: S^1 \to S^1$ be the homeomorphism given by $f([x]) = [x^2]$, and let $H$ be the $\Bbb Z$ subgroup of $\text{Homeo}(S^1)$ generated by $f$. For any point $x \neq [0]$, $f^n(x) \to [0]$ as $n \to \infty$, so $\overline{Hx}$ includes $[0]$ (and indeed has that as its only accumulation point). However, $\overline{H}x$ cannot include $[0]$; because $H$ fixes $[0]$, so does $\overline{H}$, and in particular there is no $g \in \overline{H}$ with $g[0] = x$. So $\overline{Hx} \not\subset \overline Hx$. You can modify this example so that $f$ is smooth if you want to, or even 'linear' ($SL_2(\Bbb R)$ acts on $S^1$ by first acting on the plane and then projecting radially.)