Loxodromic elements have north south boundary dynamics

Put $z_n=g^{n}o$, $w_n=g^{-n}o$ and $a_n=d(o,z_n)=d(o,w_n)$. By the quantitative conclusion in F1, choose $\tau >0$, an integer $s\ge 1$ and $C\ge 0$ such that $a_n\ge n\tau$ and, for every $m\ge n\ge s$, $(z_n\mathrm{\mid }{z}_m{)}_o\ge a_n-C,(w_n\mathrm{\mid }{w}_m{)}_o\ge a_n-C\mathrm{.}$ Thus each pole's canonical sequence has product with its corresponding $n$th orbit point at least $a_n-C$ on its tail.

For any Gromov sequence $(u_j)$ and fixed $v\in X$, put $b_v(u)={\mathrm{lim\hspace{0.17em}inf}}_j(u_j\mathrm{\mid }v{)}_o$. It is finite in $[0,d(o,v)]$. For all sufficiently large $i,j$, the Gromov property gives $(u_i\mathrm{\mid }{u}_j{)}_o>d(o,v)+\kappa$. Since $(u_j\mathrm{\mid }v{)}_o\le d(o,v)$, F3 yields $(u_i\mathrm{\mid }v{)}_o\ge (u_j\mathrm{\mid }v{)}_o-\kappa$. Interchanging $i,j$ bounds every difference on that tail by $\kappa$. Taking tail infima and suprema therefore gives ${\mathrm{lim\hspace{0.17em}sup}}_j(u_j\mathrm{\mid }v{)}_o\le b_v(u)+\kappa$. This argument uses only bounded real sequences.

For representing sequences $u$ and $v$ and an interior point $x$, apply F3 with bridge $x$. For every $\epsilon >0$, all sufficiently late products $(u_i\mathrm{\mid }x{)}_o$ and $(v_j\mathrm{\mid }x{)}_o$ are at least $b_x(u)-\epsilon$ and $b_x(v)-\epsilon$. Thus their joint liminf satisfies $P(u,v)\ge \min \{b_x(u),b_x(v)\}-\kappa$ after letting $\epsilon$ decrease to zero. This is a joint tail estimate; neither boundary class nor the bridge point is being selected simultaneously for a family.

Choose $R,T\ge 0$ with $U_R(p_{-})\subseteq U_{-}$ and $U_T(p_{+})\subseteq U_{+}$, using F2 and increasing the two thresholds to zero if necessary. For any $\xi \mathrm{\notin }{U}_{-}$ we have $B(\xi ,p_{-})\le R$. Fix any representing sequence $u$ of this one class. For $n\ge s$ with $a_n-C>R+\kappa$, step 1.1 gives $b_{w_n}(w)\ge a_n-C>R+\kappa$ for the canonical negative orbit $w$. By step 2.1 and F2, $R\ge P(u,w)\ge \min \{b_{w_n}(u),b_{w_n}(w)\}-\kappa$. The second entry exceeds $R+\kappa$, hence $b_{w_n}(u)\le R+\kappa$. Step 1.2 now gives ${\mathrm{lim\hspace{0.17em}sup}}_j(u_j\mathrm{\mid }{w}_n{)}_o\le R+2\kappa$. This bound is uniform in every class outside $U_{-}$, although its individual sequence tail cutoff need not be uniform.

The product formula and isometry identities give exactly $(g^n{u}_j\mathrm{\mid }{z}_n{)}_o=a_n-(u_j\mathrm{\mid }{w}_n{)}_o$. Taking liminf and using step 3.1 yields $b_{z_n}(g^{n}u)\ge a_n-R-2\kappa$. The image sequence represents $g^n\xi$ by F1. Step 1.1 gives $b_{z_n}(z)\ge a_n-C$ for the canonical positive orbit. Applying step 2.1 with bridge $z_n$ therefore gives $B(g^n\xi ,p_{+})\ge P(g^{n}u,z)\ge a_n-\max \{R+2\kappa ,C\}-\kappa \mathrm{.}$ All expressions involving an interior point are finite, so no subtraction of infinite boundary products has occurred.

Choose $N_{+}\ge s$ so large that $N_{+}\tau >C+R+\kappa$ and $N_{+}\tau >T+\max \{R+2\kappa ,C\}+\kappa$. Then for every $n\ge N_{+}$ we have both the hypothesis of step 3.1 and $B(g^n\xi ,p_{+})>T$ in step 4.1, for every $\xi \mathrm{\notin }{U}_{-}$. Hence $g^n(\mathrm{\partial }X\setminus U_{-})\subseteq U_T(p_{+})\subseteq U_{+}$. The integer is chosen from constants independent of $\xi$; there is no appeal to pointwise convergence to claim uniformity.

The inverse isometry is loxodromic, with poles interchanged, the same sequence of lengths $a_n$, and the same orbit constants in step 1.1. Apply step 5.1 to $g^{-1}$, now using the excluded neighbourhood $U_{+}$ and the target neighbourhood $U_{-}$. This supplies $N_{-}$ such that $g^{-n}(\mathrm{\partial }X\setminus U_{+})\subseteq U_{-}$ for all $n\ge N_{-}$. Taking $N=\max (N_{+},N_{-})$ gives both inclusions for the same tail.

F4 and the boundary/isometry conventions in F1 specialize the result to the given group. If one excluded complement is empty its inclusion is automatic; the uniform proof also covers singleton complements, including a boundary consisting only of the two poles. The constants permit $\kappa =0$, and $\tau >0$ is the only quantity whose positivity is needed to choose large integers. Each argument fixes at most one representative of one given class; the final bound is independent of that representative and class. Only finitely many threshold witnesses are used. Thus neither AC, proper-ray selection, compactness, the general quasi-isometry extension theorem nor Morse stability is used.