When a group acts on a metric space isometrically, transitively, and faithfully, is it a closed subspace of the isometry group? Suppose you have a group $G$ acting on $ (M,d)$ a compact metric space by isometries (meaning $d(gx,gy) = d(x,y)$ for all $x,y \in M$ and  all $g \in G$), transitively and faithfully. 
You can define a metric on $G$ thinking about it as a topological subgroup of the isometries on $M$ ($\operatorname{Isom}(M)$) with the sup metric, being more specific, if you take $g,h \in G$, let the distance between them be $$ \rho (g,h) = \displaystyle\sup_{x \in M} d(gx,hx) .$$ I think it's true that $G$ is a closed subspace of $\operatorname{Isom}(M)$, is that true? How can I prove it?  
 A: By the closed map lemma (http://en.wikipedia.org/wiki/Open_map), you can answer by the affirmative in the case where the group $G$ itself is compact. This is because metric spaces are always Hausdorff, so the inclusion map $i : G \to \mathrm{Isom}(M)$ is continuous, hence a closed map (by the closed map lemma ; note that it is not that hard to prove, you should give it a try). Since you were acting on a compact space $M$, this probably explains why you believe the result is true ; I had similar intuition, but all my examples involved a compact group $G$. 
I couldn't solve the problem in full generality.
Hope that helps,
A: This is already false for the sphere $S^3$ with the standard metric. The group G is isomorphic to $SU(2) \times R$.
Edit: Here is a simplified version of the above example, which assumes that you only know linear algebra (and general topology).


*

*Verify that the unitary group $U(2)< GL(2, {\mathbb C})$ acts isometrically on ${\mathbb C}^2={\mathbb R}^4$ (equipped with the standard metric) and acts faithfully on the unit sphere $S^3\subset {\mathbb C}^2$. In particular, this action is isometric with respect to the restriction of the standard metric to $S^3$ (you can also use the induced Riemannian metric -  this does not matter). 

*Verify that the subgroup $SU(2)= U(2)\cap SL(2, {\mathbb C})$ acts transitively on $S^3$. Bonus: The action is also simply transitive. 

*Verify that the group of scalar unitary 2-by-2 matrices $Z(U(2))\cong S^1$ contains a subgroup $H$ isomorphic to the infinite cyclic group ${\mathbb Z}$. Bonus: $Z(U(2))$ is the center of $U(2)$. 

*Verify that $H$ is not closed in $S^1$. Bonus: $H$ is dense in $S^1$. If this is too hard, replace the infinite cyclic subgroup of $S^1$ with the subgroup of $S^1$ consisting of rotations by angles which are rational multiples of $\pi$.  

*Take the subgroup $G< U(2)$ generated by $H$ and $SU(2)$. This subgroup acts  faithfully, transitively and isometrically on $S^3$, but $G$ is not closed in the full isometry group of $S^3$. Bonus A: $G$ is isomorphic to $H\times SU(2)$ (provided that $H\cong {\mathbb Z}$ of course). Bonus B: The full isometry group of $S^3$ is the orthogonal group $O(4)$. Bonus C: The topology on $O(4)$, defined via the sup-metric, is the same as the matrix topology. 
