Given a length space $X$. Suppose there is a group $G$ acting properly and cocompactly on $X$ by isometries, then $X$ is locally compact and complete. How can we prove this statement? Any idea will be appreciated.
A metric space $(X,d)$ is called a length space if for any two points $x,y\in X$, $$d(x,y) = \inf_{\text{c is a curve in $X$ joining $x$ and y}} l(c),$$ where $l(c)$ denotes the length of the curve $c$ associated to the metric $d$.
The action is proper means that for each point $x\in X$, there exists $r>0$ such that the set $\{g\in G\mid g\cdot B(x,r)\cap B(x,r)\neq \emptyset\}$ is finite. In particular, each compact subset $K$ of $X$ has an open neighborhood $U$, such that the set $\{g\in G\mid g\cdot U\cap U\neq \emptyset\}$ is finite. Some references call the second condition "metrically proper".
The action is cocompact if there exists a compact subset $K\subset X$ such that $G\cdot K = \cup_{g\in G}g\cdot K = X$.
I am not sure whether the condition of length space is necessary. The full statement says: Suppose there is a group $G$ acting properly and cocompactly on a length space $X$ by isometries, then $X$ is locally compact and complete. Hence, by Hopf-Rinow theorem, it is proper geodesic. I know the length space property is used in the second part, but I am not sure about the first part.
Actually, the assumption that $X$ is a length space (aka a path-metric space) is redundant. I will give a proof as a sequence of lemmata that you should be able to verify on your own:
Lemma 1. For any two points $x, y\in X$, there exists $r>0$ such that the subset $$ \{g\in G: g B(x,r)\cap B(y,r)\ne \emptyset\}\subset G $$ is finite. (Hint: argue by contradiction and use properness of the $G$-action on $X$ in the traditional sense.)
Lemma 2. For every $y\in X$ and a compact $C\subset X$, there exists $r=r(y,C)>0$ such that the subset $$ \{g\in G: g C\cap B(y,r)\ne \emptyset\}\subset G $$ is finite. (Hint: Use Lemma 1 and compactness of $C$.)
Lemma 3. Let $K\subset X$ be a compact such that $GK=X$. For $x\in X$ take $r=r(x, K)$. Then $B(x,r)$ is contained in a finite union of $G$-translates of $K$. In particular, $B(x,r)$ is relatively compact. In particular, $X$ is locally compact.
Lemma 4. For $K$ as above, there exists $r>0$ such that for every $x\in K$ the ball $B(x,r)$ is relatively compact. (Hint: Take $r$ to be a Lebesgue number of the cover of $K$ by the balls $B(x, r(x,K))$.)
Corollary. There exists $r>0$ such that for every $x\in X$, the ball $B(x,r)$ is relatively compact. (Hint: $GK=X$.)
Lemma 5. Let $(x_n)$ be a Cauchy sequence in $X$. Then there exists $N$ such that for all $n\ge N$, $x_n\in B(x_N, r)$, where $r$ is as above. Conclude that $(x_n)$ converges.