Proper and cocompact group action by isometries on a metric space

Question

I am reading the book "Metric spaces of Non-Positive Curvature" by Bridson and Haefilger and got stuck with the following :

Proposition II.6.10(2) Let $\Gamma$ be a group that acts properly by isometries on a metric space $X$. If the action is cocompact then every element of $\Gamma$ is semi simple.

In the proof I couldn't understand the highlighted line

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According to the book a group action is proper if for every $x\in X$ there exist $r>0$ such that the set $\{\gamma\in \Gamma|\gamma B(x,r)\cap B(x,r)\not=\phi\}$ is finite.(section II.8.2)

And an action is cocompact if there exists a compact set $K$ such that $X=\Gamma K$.(section II.8.1)

I have spent a lot of time trying to justify this argument. Although I was able to find justification for this assuming $X$ is a proper metric space , but couldn't do it for general case .

Any help shall be highly appreciated. Thanks in advance

Answer 1

I don't think there is any missing assumption. Indeed cocompactness is enough to take advantage of compactness.

Let me also clarify the properness assumption. A action of a discrete group on a metric space is topologically proper (resp. metrically proper) if for every compact (resp. bounded) subset $K$, the set of $g\in G$ such that $K\cap gK$ is finite. "Metrically proper" implies "topologically proper", and the book uses "proper" to mean "metrically proper".

The definition of proper in the Bridson-Haefliger book should be taken as "metrically proper". (Their definition is flawed and should not be used as the given result fails otherwise, see the comments.)

Write $h_n=\gamma_n\gamma\gamma_n^{-1}$. They justify that $d(h_ny_n,y_n)$ is bounded. It follows that $d(h_nx,x)$ is bounded for every $x$ (namely by $2\sup_n d(x,y_n)$, which is bounded since $y_n\in K$). Metric properness of the action precisely means that every sequence $(h_n)$ such that $d(h_nx,x)$ is bounded for some $x$, is bounded (i.e., lies in one finite subset).

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Not only properness of the space is not a missing assumption, but it is a consequence of the assumptions.

Proposition. Let $X$ be a metric space with a cocompact metrically proper action of a discrete group $G$. Then $X$ is a proper metric space.

[The action is not supposed isometric, not even continuous. For an isometric action properness is equivalent to $d(gx,x)\to\infty$ when $g\in\infty$, for some $x\in X$ and hence every $x$.]

Proof. Let $(x_n)$ be a bounded sequence in $X$. Write $x_n=g_nk_n$ with $g_n\in G$, $k_n\in K$. By metric properness of the action, $g_n$ remains in a finite subset. So, after extracting, we can suppose $g_n$ constant, say $g_n=g$, so $x_n=gk_n$. Hence for any limit point $k$ of $(k_n)$, $gk$ is a limit point.