Proper action in "Metric spaces of non-positive curvature"

Question

(This question is related to Proper action and compactness)

In (I.8.2) of "Metric spaces of non-positive curvature", properly action is defined as: for each $x$, there exists a $r$ such that $\{\gamma | \gamma B(x,r) \cap B(x,r)\}$ is finite. (I think this does not implies that $\{\gamma | \gamma B(x,r) \cap B(x,r)\}$ is finite for any $r$).

Then I find a problem about the proper action on Page 233, which is highlighted:

From the proof of proposition 6.10 (3) on page 233,
we know that for all $n$ sufficently large $\gamma_n B(x, r+a+1) \cap B(x, r)\ne \emptyset$. Then the original proof say this contradicts to the properness of the action (see I.8.3 (1)). But the definition of proper action is I.8.2 as above, and I.8.3(1) just explain the proper action in (I.8.2) is more restrictive than the another proper action, which is defined as: for every compact subset $K$, the set $\{\gamma | \gamma . K \cap K\}$ is finite. Why I.8.3(1) implies a contradiction?

(In fact in Proper action and compactness , after the answer by Seirios, there are comments by GGT, which ask the same question)

Answer 1

I think they simply forgot to assume properness of $X$ or they forgot their definition of a proper action. Or they wrote the proof using one definition of a proper action, later change their definition but forgot to correct the proposition. The right thing to do is to write to Martin Bridson and ask him to add an erratum for this proposition.

Here is a counterexample to their claim of discreteness of the set of translation lengths (proposition 6.10(3)). Let $H$ be an infinite dimensional separable Hilbert space with an orthonormal basis $({\mathbf e}_n), n\in {\mathbb N}$. Take the sequence $a_n=1-\frac{1}{2n}$ and consider the translations $$ T_n: {\mathbf x}\mapsto {\mathbf x} + a_n {\mathbf e}_n, {\mathbf x}\in H. $$ Let $G$ denote the (free abelian) subgroup of isometries of $H$ generated by the translations $T_n$, $n\in {\mathbb N}$. Then the group $G$ acts properly on $H$ (in the sense of Bridson-Haefliger which, incidentally, is equivalent to the standard notion of properness for isometric actions of discrete groups): For each point ${\mathbf x}\in H$ the set $$ \{g\in G| g B({\mathbf x}, \frac{1}{3}) \cap B({\mathbf x}, \frac{1}{3})\ne \emptyset\} $$ is finite.

Let $X\subset H$ denote the $G$-orbit of zero, equipped with the restriction of the standard distance function on $H$. Then $G$ acts on $X$ properly, cocompactly (since the action is transitive). But the set of translation lengths of the elements of $G$ is not a discrete subset of ${\mathbb R}$ since it contains the sequence $(a_n)$.

Here is another counterexample, of combinatorial origin: Take the countably infinite bouquet of circles $Z$; assign length $a_n= 1- \frac{1}{2n}$ to the $n$th circle. This yields a $G$-invariant path-metric on the universal covering space $X$ of $Z$, where $G$ is the group of covering transformations. The group $G$ acts discretely and transitively on the vertex set of $X$. However, the set of translation lengths includes the sequence $(a_n)$, hence, is nondiscrete.