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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 $p$?

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    $\begingroup$ 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. $\endgroup$ Oct 23, 2019 at 12:53

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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 $d(c_N(t),c(t))<b$ and $d(\alpha_N(t),\alpha(t))<b$, 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 $X$.

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    $\begingroup$ 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? $\endgroup$
    – Square
    Oct 27, 2021 at 0:25
  • $\begingroup$ I agree with you @Square. This is not right. If you have figured a way around this please let me know. $\endgroup$ Feb 10 at 16:26

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