Group acting on locally finite tree Let $G$ be a group acting on a locally finite connected tree $T$ i.e. each vertex degree is finite. Let $G$ has compact open topology i.e. for each compact set $C$ and an open set $U$ of $T$, the sets $\{f\in G\colon f(C)\subseteq U\}$ form a subbasis of this topology (see Topology and Geometry-Bredon). Prove that $G$ is locally compact. Can we replace tree by arbitrary graph?
Edit: (T.B.) This question is closely related to (but not a duplicate of) Automorphism Group of a graph.
 A: First of all, you asked in a comment about the compact open topology. I recommend that you read the extraordinarily good exposition in Chapter XII in Dugundji's Topology, Allyn and Bacon, 1966. Most other books on general topology contain discussions of that as well, e.g. Munkres, §46. The actual question is discussed in the next part of the answer. I'd also recommend to read Pete's answer in the related question first before reading here. People acquainted with the compact open topology can safely scroll down to the displayed Conclusion and read from there on.

So let $X$ and $Y$ be arbitrary topological spaces. Consider the set of continuous functions $C(X,Y)$ equipped with the compact open topology, i.e., the topology generated by the subbasis
$$O_{K,U} = \{f \in C(X,Y)\,:\,f(K) \subset U\} \qquad K \subset X \text{ compact, } U \subset Y \text{ open.} $$
It is often helpful to think of open sets in a topological space as consisting of points that are near to each other. This topology is inspired by this. More precisely, let $f$ be a continuous function then $f$ maps the compact set $K$ to a compact set $f(K) \subset Y$. Fixing a "small" neighborhood $U$ of $f(K)$ we consider a function $g$ as near to $f$ if $g(K) \subset U$ as well. By taking $K$ "very large in $X$" and $U$ a "very small neighborhood" of $f(K)$ then $g$ will be quite close to $f$, intuitively. By cutting up our "large" $K$ into finitely many very small pieces $K_{1}, \ldots, K_n$ and taking tiny neighborhoods $U_i$ of $f(K_i)$ then a $g \in O_{K_1,U_1} \cap \cdots \cap O_{K_n,U_n}$ will be forced to be extremely close to $f$ on all of $K$.
This can be made precise e.g. by the following observation:

Let $(Y,d)$ be a metric space. Then a sequence $f_n$ converges to $f$ in the compact-open topology if and only if $f_n \to f$ uniformly on compact sets: for every compact $K \subset X$ we have $$d_{K}(f,f_n) = \sup_{x \in K}{\;d(f(x),f_n(x))}\;\xrightarrow{n \to \infty}\;0.$$

You can find the proof of this in Dugundji, Proposition 7.2, p. 268. However, I recommend that you prove this on your own to get a certain familiarity with the compact-open topology.
Note that we have a means of quantifying closeness if $Y$ is metric: two points are $\varepsilon$-close if their distance is less than $\varepsilon$ and two functions $f,g$ are $\varepsilon$-close (relative $K$) if $d_{K}(f,g) \lt \varepsilon$, that is all values of all $x \in K$ are $\varepsilon$-close.
Next, we have the Arzelà-Ascoli theorem (in one of its many incarnations). Recall that a metric space $(Y,d)$ is called proper if all closed balls are compact. Metric spaces with this condition are sometimes also called metric spaces with the Heine-Borel property (closed and bounded sets are compact).

Let $X$ be an arbitrary space and let $Y$ be a proper metric space. Let $\mathscr{F} \subset C(X,Y)$ be a family of functions. Then $\mathscr{F}$ is relatively compact in the compact-open topology if and only if 
  
  
*
  
*The family $\mathscr{F}$ is pointwise bounded: for all $x \in X$ the set $\{f(x)\,:\,f \in \mathscr{F}\}$ is bounded.
  
*The family $\mathscr{F}$ is equicontinuous.
  

To relate the present answer to the paper by Woess mentioned by Pete in his answer to a related question, you'll need the following simple observation whose proof I leave to you:

Let $f_n$ be an equicontinuous family of functions. Then $f_n \to f$ pointwise if and only if $f_n \to f$ uniformly on compact sets.

The upshot of this long-winded preliminary discussion is:

Conclusion Let $(X,d)$ be a proper metric space, i.e., closed balls are compact. Let $G = \operatorname{Isom}{(X)}$ be the group of isometries, equipped with the compact-open topology. Then $G$ is a locally compact topological group, the action $G \times X \to X$ is continuous and the action is proper (in particular stabilizers of points are compact, orbits are closed and $X/G$ is Hausdorff—even metrizable).

Proof. Rather immediate:


*

*Local compactness. Let $g \in G$, let $x_0 \in X$ be an arbitrary point and $\varepsilon \gt 0$. Put $K = \{x_0\}$ and $U = B_{\varepsilon}(gx_0)$ then $\mathscr{F} = O_{K,U}$ is relatively compact.  Indeed, if $h \in \mathscr{F}$ and $x \in X$ is arbitrary then $$d(gx_{0},hx) \leq d(gx_0, hx_0) + d(hx_0,hx) \leq \varepsilon +  d(x_0,x),$$
so $\mathscr{F}$ is pointwise bounded. Since isometries are $1$-Lipschitz, $\mathscr{F}$ is equicontinuous and it follows from Arzelà–Ascoli that $\mathscr{F}$ is relatively compact. But this means that every point in $G$ has a compact neighborhood.

*The group operations are continuous, hence $G$ is a topological group.

*Continuity of the action. Exercise (use that $g_n \to g$ means uniform convergence on compact sets and that closed balls  in $X$ are compact).

*The verification of properness of the action is straightforward and the rest follows from generalities on proper actions (see my answer on MO).

Let $\Gamma$ be a connected locally finite graph and equip it (or rather its geometric realization) with the length metric: edges are isometric to the unit interval and the distance between two points on $\Gamma$ is the infimum over all lengths of paths connecting them. This gives a metric on the graph inducing the correct (locally compact) topology on $\Gamma$ — note that closed balls are compact in this topology since $\Gamma$ is locally finite and connected. So $\Gamma$ is proper and we may apply the results from the first part of the answer.
It is plain that a graph automorphism maps paths to paths, so the automorphism group of $\Gamma$ can then be identified with a closed subgroup of the isometry group of $\Gamma$. We equip $\operatorname{Aut}(\Gamma)$ with the compact open topology. Combining the facts above we see:


*

*Every $g \in \operatorname{Aut}(\Gamma)$ has a compact neighborhood.  Indeed, take any point $x_{0} \in \Gamma$ and let $K= \{x_{0}\}$ and let $U = B_{\varepsilon}(g(x))$. Then $O_{K,U}$ is relatively compact by Arzelà-Ascoli: Indeed, $O_{K,U}$ is equicontinuous since all elements are isometries (hence $1$-Lipschitz), so condition 2. is satisfied; condition 1. is satisfied because for any point $x \in X$ and $h \in O_{K,U}$ we have $d(gx_{0},hx) \leq d(gx_0, hx_0) + d(hx_0,hx) \leq \varepsilon +  d(x_0,x).$

*If $G$ is a group acting on $\Gamma$ by graph automorphisms let $N$ be the normal subgroup consisting of the elements acting trivially, then $G/N$ is a group acting on the graph $\Gamma$. There are two problems in the statement you ask about:


*

*The compact open topology, as you define it endows $N$ with the trivial topology, in particular $G$ is not Hausdorff if $N$ is non-trivial. Since $N$ is not interesting anyway, I wouldn't take the pains and worry about it:  simply throw it away and consider $G/N$ instead of $G$. But then a further unpleasant surprise awaits us:

*The image of $G/N$ in $\operatorname{Aut}(\Gamma)$ need not be closed — but the compact open topology is the relative topology of $G/N$ as a subgroup of $\operatorname{Aut}(\Gamma)$.  However, one can show that a locally compact subgroup of a Hausdorff group is closed, see e.g. Corollary 5.11 on p.35 of Hewitt–Ross, Abstract Harmonic Analysis I, so we're in trouble and 
this is a serious problem: Since $\operatorname{Aut}(\Gamma)$ is only rarely discrete for homogeneous enough graphs (e.g. complete $n$-ary trees) there will be plenty of non-closed subgroups (take a countable dense subset $D$ of any closed subgroup and consider the subgroup $\langle D \rangle$ generated by it, for example). So you need a condition ensuring that $G/N$ is a closed subgroup of $\operatorname{Aut}(\Gamma)$. Off the top of my head I can't think of any such condition.


*Finally, if $G/N$ is locally compact (i.e., closed in $\operatorname{Aut}(\Gamma)$) and $G$ is equipped with the compact-open topology then $G$ is locally compact as an extension of the locally compact group $G/N$ by the (locally) compact group $N$ by Theorem 5.25 on p.39 of Hewitt-Ross, (provided that you check that the compact-open topology indeed defines a group topology on $G$, which I haven't done but I'm not too worried about).
I think that's the best I can do at the moment without further information on what you actually want to do.

Edit: As a further remark (as Pete notes in his answer) the above extends easily to graphs with finitely many connected components, but can fail with infinitely many connected components, as already the trivial graph on a countable set shows.
