Suppose we have a metric space $V$, a group $G$ and an action $\cdot: G \times V \rightarrow V$.
What assumptions must I make so that the following is true?
Claim: For each $x, y \in V$, if there exists $(g_n)_{n \in \mathbb{N}} \subset G$, such that $g_n \cdot x \rightarrow y$ (i.e. $d(g_n \cdot x, y) \rightarrow 0)$, then $\text{Orb}(x) = \text{Orb}(y)$.
I do have available that for each $g \in G$, the map $x \mapsto g \cdot x$ is an isometry. So this is in common the wikipedia page on geometric group actions https://en.wikipedia.org/wiki/Geometric_group_action, however, I don't think the groups I'm working with are finitely generated.
I'm not very well versed in algebra or topology, but this popped up in something I was working on.
In my particular context, I am working on a normed vector space $V$ so that $\text{Orb}(0) = \{0\}$, I'm not sure if this helps. It seems that $G$ needs to be some sort of topological group, but one of the vector spaces/groups I am working with, $V = L^1(X, \mathcal{A}, \mu)$ and $G = \text{Aut}(X, \mathcal{A}, \mu)$ (i.e. measurable bijections that preserve pushforward measure), doesnt seem to have an inherent topology nor is finitely generated.
Best