# Topology on the Gromov Boundary of a Hyperbolic Space

Let $X$ be a proper geodesic metric space that is $\delta$-hyperbolic.

Definition. We define the Gromov boundary $\partial X$ of $X$ as the set of all the geodesic rays $c:[0, \infty)\to X$, where we regard two geodesic rays $c$ and $c'$ as the same if the Hausdorff distance between then is finite. We denote the equivalence class of a geodesic ray $c$ by $c(\infty)$.

Now we have a set $\bar X:=X\cup \partial X$. We want to define a topology on it. Before that we need to make a definition.

Definition. A generalized ray in $X$ is a geodesic $c:[0, T]\to X$, which we extend on $[0, \infty]$ by definition $c(t)=c(T)$ for all $t>T$ (So $c(\infty)=c(T)$ also).

Definition. We define a topology on $\bar X$ as follows. Fix a point $p\in X$. For a sequence of points $(x_n)$ in $\bar X$, and a point $x$ in $\bar X$, we write $x_n\to x$ if there is a sequence of generalized rays $(c_n)$ in $X$, with $c_n(0)=p$ and $c_n(\infty)=x_n$, such that every subsequence of $(c_n)$ has a subsequence which converges uniformly on compact sets to a generalized ray $c$ satisfying $c(\infty)=x$.

This defines a topology as follows: The closed sets are precisely those subsets $C$ of $\bar X$ for which the following holds: Whenever $(x_n)$ is a sequence in $C$ with $x_n\to x$ for a point $x\in \bar X$, then $x\in C$.

Now here is my question.

Question. Why this topology is independent of the choice of the point $p$?

• This is Proposition 3.7 (pg. 429) in Bridson-Haefliger: Metric Spaces of non-positive Curvature. There is a proof given there. Note that a generalized ray can either be $c:[0,T] \to X$ or $c:[0, \infty] \to X$. The case $c(t) = c(T)$ is a notation shortcut to treat rays that are only defined on $[0,T]$ as if they were defined everywhere. Oct 23, 2019 at 12:53

Let $$x_n\to x$$ with respect to $$p$$ in $$X$$, i.e, there is a sequence of generalized rays $$(c_n)$$ in $$X$$, with $$c_n(0)=p$$ and $$c_n(\infty)=x_n$$, such that every subsequence of $$(c_n)$$ has a subsequence which converges uniformly on compact sets to a generalized ray $$c$$ satisfying $$c(\infty)=x$$. We aim to show that $$x_n\to x$$ with respect to $$q$$ in $$X$$. To this end, note first that for all $$n\in X$$, we can find a generalized ray $$\alpha _{n}$$ such that $$\alpha _{n}(0)=q$$ and $$\alpha _{n}(\infty)=c_n(\infty)=x_n$$. Let $$(\alpha _{n_k})$$ be a subsequence of $$(\alpha _{n})$$, then $$(c _{n_k})$$ is a subsequence of $$(c _{n})$$, hence $$(c _{n_k})$$ contains a subsequence $$(c _{m})$$ which converges uniformly on compact sets to a generalized ray $$c$$ satisfying $$c(\infty)=x$$. Also $$(\alpha _{m})$$ is a subsequence of $$(\alpha _{n_k})$$ and by Arzelà-Ascoli lemma, $$(\alpha _{m})$$ contains a subsequence (without loss of generality we may assume that subsequence is $$(\alpha _{m})$$ itself) which converges uniformly on compact sets to a generalized ray $$\alpha$$. For any constant $$b>0$$, we can find a natural number $$N$$ large enough such that $$d(c_N(t),c(t)) and $$d(\alpha_N(t),\alpha(t)), for all $$t\geq 0$$. Also as $$\alpha _{N}(\infty)=c_N(\infty)=x_N$$, therefore we can find a positive number $$r>0$$ such that $$Im(c_N)\subseteq B_r(Im(\alpha_N))$$ and $$Im(\alpha_N)\subseteq B_r(Im(c_N))$$. Hence $$Im(c)\subseteq B_{r+2b}(Im(\alpha))$$ and $$Im(\alpha)\subseteq B_{r+2b}(Im(c))$$, so $$\alpha (\infty)=c(\infty)=x$$. In particular if $$C$$ is closed subset of $$\bar X$$ with respect to the base-point $$p$$ in $$X$$, then $$C$$ is closed subset of $$\bar X$$ with respect to the base-point $$q$$ in $$X$$.
• I don't understand the part where you get the constant $b$. If $d(c_N(t),c(t)) < b$ for all $t \geq 0$, wouldn't this mean $c_N(\infty) = c(\infty) = x$? How do you get this uniform convergence along the entire geodesic? Oct 27, 2021 at 0:25