Quantitative Hyperbolic Geometry Toolkit

For completeness, the realization is geodesic as asserted in the definition. For two interior-edge points, any finite edge route either stays on their common edge, when there is one, or first reaches one of the at most two endpoints of the first edge and finally leaves one of the endpoints of the last edge. Between those vertices its length is at least their word distance. Conversely each of these at most four endpoint routes is attained by a shortest word, with the specified initial and final partial edges. Include the direct same-edge interval as another candidate. The minimum of this finite list is attained and positive for distinct points; it defines the path metric. Concatenation gives the triangle inequality. A minimizing route, parametrized by length, is isometric, since a shorter route between two of its points would shorten it. At vertices the same argument gives exactly word distance. Loops are covered by the two ends of their interval before identification; coincident endpoints cause no problem.

Put $a_n=\mathrm{\mid }{g}^n\mathrm{\mid }$ and $a_0=0$. Then $0\le a_{n+m}\le a_n+a_m$, so the nonempty set $\{a_k\mathrm{/}k:k\ge 1\}$ is bounded below by zero. Its infimum $t$ exists and satisfies $0\le t\le a_1$.

Fix $\eta >0$. By the defining property of the infimum there is a positive integer $k$ such that $a_k\mathrm{/}k<t+\eta \mathrm{/}2$. For each $n\ge 1$ write $n=qk+r$, $0\le r<k$. Repeated subadditivity gives $a_n\le q{a}_k+a_r$. With $C={\max }_{0\le r<k}{a}_r$, it follows that $t\le a_n\mathrm{/}n\le a_k\mathrm{/}k+C\mathrm{/}n$. Here $q\mathrm{/}n\le 1\mathrm{/}k$ and $a_k\ge 0$ justify the last inequality.

For all sufficiently large $n$, $C\mathrm{/}n<\eta \mathrm{/}2$, giving $t\le a_n\mathrm{/}n<t+\eta$. The requisite large integers exist in a real complete ordered field: if the natural numbers had a finite supremum $u$, some natural $m>u-1$ would give $m+1>u$, a contradiction. Thus $a_n\mathrm{/}n\to t$. This also treats $C=0$ and $k=1$ directly. If $g=e$, every $a_n$ and $t$ is zero; if $g$ has finite order $m$, $a_m=0$ gives $t=0$. Only one witness $k$ for a given tolerance and finitely many endpoint routes were used, so no AC is needed.

Subdivide at the at most three interior-edge points among $a,b,c$. This preserves connectivity, edge lengths and absence of cycles: contracting the finitely many new degree-two vertices in a cycle would produce a cycle in the original graph. Remove closed subwalks from any finite path until a simple path remains. Two distinct simple paths with the same endpoints would have a first divergence and a first subsequent reunion, and the intervening arcs form a cycle. Hence there is exactly one finite simple path between any two of the specified points.

Every finite edge route reduces to that simple path by deleting closed subwalks, so its length is at least the simple-path length. Consequently the latter attains the path distance and, parametrized by length, is isometric on every subinterval. It is also the only geodesic image: if a path leaves this simple path through an edge, it must return through the same attaching point (otherwise the finite routes to the two attaching points form a cycle). A geodesic cannot revisit a point. An interior point of any edge of the simple path separates its endpoints, so every continuous joining path must traverse that edge; thus no alternative geodesic can skip part of the simple path.

The intersection of the finite paths $[a,b]$ and $[a,c]$ is an initial path $[a,m]$. Indeed if a point belongs to both, their subpaths from $a$ to that point coincide by uniqueness. Since the two paths are finite unions of closed edge intervals, this common initial path has a last point $m$. After subdivision at $m$, their tails $[m,b]$ and $[m,c]$ meet only at $m$; their concatenation is the simple path from $b$ to $c$. Hence the three pairwise paths have precisely the three legs from $m$, and distances are sums of leg lengths.

Each side is a union of two legs, each also on another side; hence every side lies in the union of the other two. This gives $0$-slimness. If points coincide or $m$ is a vertex of the triangle, the corresponding legs have length zero and the same union identities hold. Only finitely many paths and subdivisions were needed, including in an infinite graph.

Put $\alpha =\min \{(x\mathrm{\mid }y{)}_o,(y\mathrm{\mid }z{)}_o\}$. For any point $v\in [x,y]$, the inequalities $d(o,x)\le d(o,v)+d(v,x)$ and $d(o,y)\le d(o,v)+d(v,y)$ add to give $d(o,v)\ge (x\mathrm{\mid }y{)}_o$. The same argument applies to $[y,z]$.

If $\alpha \le \delta$, nonnegativity gives the claimed lower bound. Otherwise fix $0\le t<\alpha -\delta$ and $0<h<\alpha -\delta -t$. Choose points $x_t,y_t,z_t$ at distance $t$ from $o$ on three chosen radial sides. These points exist since each radial length is at least $\alpha$. Slimness and the infimum convention supply a point within distance less than $\delta +h$ of $x_t$ on $[o,y]\cup [x,y]$. It cannot lie on $[x,y]$, whose points have distance at least $\alpha >t+\delta +h$ from $o$. Let it be $v\in [o,y]$. Then $\mathrm{\mid }d(o,v)-t\mathrm{\mid }<\delta +h$, so $d(x_t,y_t)<2(\delta +h)$. Applying the same reasoning to $y_t$ in triangle $(o,y,z)$ gives $d(y_t,z_t)<2(\delta +h)$.

Thus $d(x_t,z_t)<4(\delta +h)$ and the joining route through them gives $d(x,z)\le d(o,x)-t+4(\delta +h)+d(o,z)-t$. Expanding the product yields $(x\mathrm{\mid }z{)}_o\ge t-2(\delta +h)$. This is true for every sufficiently small positive $h$, so $(x\mathrm{\mid }z{)}_o\ge t-2\delta$. Otherwise a sufficiently small $h$ would contradict the strict gap.

Letting $t$ approach $\alpha -\delta$ from below in the same elementary real inequality gives $(x\mathrm{\mid }z{)}_o\ge \alpha -3\delta$. When $\delta =0$ and $\alpha >0$, the same positive $h$ and $t<\alpha$ argument applies; when both vanish step 2.1 applies. All side selections are finite and all near-point witnesses are used one at a time, so no AC or closest-point attainment was used.

For an ordered quadruple $(o,x,y,z)$ write $A=d(o,y)+d(x,z)$, $B=d(o,z)+d(x,y)$ and $C=d(o,x)+d(y,z)$, and put $R=d(o,x)+d(o,y)+d(o,z)$. Then $2(x\mathrm{\mid }z{)}_o=R-A$, $2(x\mathrm{\mid }y{)}_o=R-B$, and $2(y\mathrm{\mid }z{)}_o=R-C$. Consequently the product inequality for this ordered quadruple is exactly $A\le \max \{B,C\}+2\kappa$, since $\min \{R-B,R-C\}=R-\max \{B,C\}$.

Suppose the product condition holds for all ordered quadruples. Permuting $x,y,z$ in step 1.1 gives the three inequalities bounding each of $A,B,C$ by the maximum of the other two plus $2\kappa$. Apply the inequality with the largest sum on its left: the maximum on its right is the second-largest, including ties. This proves the four-point condition.

Conversely suppose the four-point condition holds. For every ordered quadruple, if $A$ is largest it is at most the second-largest plus $2\kappa$, while if it is not largest it is already at most $\max \{B,C\}$. Thus $A\le \max \{B,C\}+2\kappa$ in either case. Step 1.1 recovers the product inequality at the arbitrary basepoint $o$. The computations remain valid when points coincide or $\kappa =0$.

Write $\mathrm{\ell }=d(a,b)$, $t=d(a,p)$ and $\alpha =(b\mathrm{\mid }c{)}_a$. Suppose first that $t\le \alpha$. Since $\alpha \le d(a,c)$ there is $q\in [a,c]$ with $d(a,q)=t$. Products along a radial geodesic give $(p\mathrm{\mid }b{)}_a=t$ and $(c\mathrm{\mid }q{)}_a=t$. Applying the product inequality first through $b$, then through $c$, gives $(p\mathrm{\mid }c{)}_a\ge t-\kappa$ and $(p\mathrm{\mid }q{)}_a\ge t-2\kappa$. Hence $d(p,q)=2t-2(p\mathrm{\mid }q{)}_a\le 4\kappa$.

If $t\ge \alpha$, use $b$ as basepoint. Indeed $(a\mathrm{\mid }c{)}_b=\mathrm{\ell }-\alpha$ by expansion, and $d(b,p)=\mathrm{\ell }-t\le \mathrm{\ell }-\alpha$. The argument of step 1.1 with $a$ and $b$ interchanged gives a point on $[b,c]$ at distance at most $4\kappa$ from $p$. At $t=\alpha$ either construction works.

Thus every point of $[a,b]$ is within $4\kappa$ of the other two sides; relabeling vertices proves this for all sides and all specified triangles. Zero side lengths require only $t=0$, and when $\kappa =0$ the produced point equals $p$. There were only finitely many segment choices.

Fix $y_i\in D(x_i,x_1)$ for $i=0,2$, and write $E_i=d(x_i,y_i)$, $\eta =d(y_0,y_2)$. Let $u$ be the midpoint of a chosen segment from $y_0$ to $y_2$. Triangle inequalities and the halfspace assumptions give $d(u,x_i)\le E_i+\eta \mathrm{/}2$ and $d(u,x_1)\ge E_i-\eta \mathrm{/}2$ for each $i=0,2$.

For every $h>0$, slimness of triangle $(x_0,u,x_2)$ at $x_1$ supplies $v$ on $[x_0,u]$ or $[x_2,u]$ with $d(v,x_1)<\delta +h$. Suppose it is on $[x_i,u]$. Then $d(u,v)>E_i-\eta \mathrm{/}2-\delta -h$, while $d(x_i,u)\le E_i+\eta \mathrm{/}2$, so $d(x_i,v)<\eta +\delta +h$. Therefore $3\delta =d(x_i,x_1)<\eta +2\delta +2h$. Since this holds for every $h>0$, $\eta \ge \delta$.

Taking the infimum over $y_0,y_2$ gives the first bound; both sets are nonempty since they contain $x_0,x_2$, respectively. They are disjoint because $\delta >0$. Thus any $z\in D(x_2,x_1)$ does not belong to $D(x_0,x_1)$, which is precisely the asserted strict inequality.

If $y\in D(x_1,x_0)$, then $3\delta \le d(x_0,y)+d(y,x_1)\le 2d(x_0,y)$, proving the endpoint bound by taking an infimum. Interchanging $x_0,x_2$ gives the other bound. No closest point to a halfspace or bisector has been selected, and all segment selections are finite.

First let $\delta >0$ and fix $[s,t]\subseteq I$. Put $L=t-s$, $n=\lfloor L\mathrm{/}(3\delta )\rfloor$, $\sigma =L-3n\delta$, and $x_i=q(s+3i\delta )$ for $0\le i\le n$. Consecutive triples lie on an isometrically parametrized subpath of length $6\delta$. For $0\le i<n$ put $H_i=\{z:d(z,x_{i+1})\le d(z,x_i)\}$ and $B_i=\{z:d(z,x_{i+1})=d(z,x_i)\}$. For $i<n-1$, F2 gives $H_{i+1}\subset \{z:d(z,x_{i+1})<d(z,x_i)\}\subset H_i$ and $\mathrm{dist}(B_i,B_{i+1})\ge \delta$.

Now suppose $\delta =0$. A geodesic segment is closed: if $p$ has distance zero from the image of $\gamma :[0,\mathrm{\ell }]\to X$, set $r=d(\gamma (0),p)$. Arbitrarily close image points $\gamma (u)$ satisfy $\mathrm{\mid }u-r\mathrm{\mid }\le d(\gamma (u),p)$, so $r\in [0,\mathrm{\ell }]$ and $d(\gamma (r),p)=0$. A zero-slim degenerate triangle consisting of two segments between the same endpoints and a constant side consequently forces their images to agree; radial parameters then agree too. For any triangle $(a,b,c)$ let $\alpha =(b\mathrm{\mid }c{)}_a$. The points of $[a,b],[a,c]$ at radius $\alpha$ coincide: apply F3 twice, through $b$ and $c$, to see their product at $a$ is at least $\alpha$ and hence their distance zero. Call the common point $m$. The equality $d(b,c)=d(a,b)+d(a,c)-2\alpha =d(b,m)+d(m,c)$ shows that the concatenated tails form the unique geodesic $[b,c]$. Uniqueness prevents any further common tail. Thus every triangle is a tripod.

Suppose $n\ge 2$. A geodesic $\gamma :[0,d(x_0,x_n)]\to X$ from $x_0$ to $x_n$ starts outside every $H_i$ and ends inside every $H_i$ by nesting. For each $i$ its first entry time $t_i$ into $H_i$ exists and lies in $B_i$: the inverse image of $H_i$ is nonempty and closed, so its infimum belongs to it, and continuity of the distance difference forces equality at first entry. This closed-infimum assertion follows by taking real parameters approaching the infimum; continuity passes the nonpositive inequality to their limit. Nesting gives $t_i\le t_{i+1}$. F2 now gives $t_0\ge 3\delta \mathrm{/}2$, $t_{i+1}-t_i\ge \delta$, and $d(x_0,x_n)-t_{n-1}\ge 3\delta \mathrm{/}2$. Adding yields $d(x_0,x_n)\ge (n+2)\delta \ge n\delta$. For $n=0,1$, the bound $d(x_0,x_n)\ge n\delta$ follows directly from locality.

The final subarc has length $\sigma <3\delta$. Thus $d(q(s),q(t))\ge n\delta -\sigma =L\mathrm{/}3-4\sigma \mathrm{/}3\ge L\mathrm{/}3-4\delta$. Arc length gives $d(q(s),q(t))\le L\le 3L+4\delta$. Since $s,t$ were arbitrary, this is the required quasi-geodesic inequality. Increasing the locality radius preserves its $6\delta$ hypothesis.

Subdivide any compact parameter interval into finitely many equal positive pieces of length less than $k\mathrm{/}2$; a zero-length interval already is geodesic. Each piece is geodesic. Suppose the path through the first $j$ pieces is geodesic and append the next piece at $b$. In the tripod formed by the old starting point, $b$, and the new endpoint, failure of their concatenation to be geodesic would mean both segments from $b$ initially follow the same positive-length leg. Choose $h>0$ smaller than that leg and both adjacent mesh lengths. The points at parameter distances $h$ before and after the join would coincide, although their parameter distance is $2h<k$. This contradicts locality. Finite induction proves the compact restriction geodesic, hence the whole interval map is geodesic. Only finitely many auxiliary segments have been chosen for each restriction; AC is not used.

If $\epsilon =0$, set $p=q$: the upper inequality makes it $\lambda$-Lipschitz and continuous, and all approximation errors vanish. If $\epsilon >0$ and the endpoint distance is less than $2\epsilon$, the lower inequality gives $b-a<3\lambda \epsilon$, and the upper inequality gives $d(q(t),q(a))\le \lambda (b-a)+\epsilon <3{\lambda }^2\epsilon +\epsilon$. This includes a one-point domain. Henceforth assume positive $\epsilon$ and separated endpoints. Write $h=\epsilon \mathrm{/}\lambda$. The upper endpoint inequality implies $b-a\ge h>0$.

If $h\le b-a\le 2h$, use a single marked interval $[a,b]$. If $b-a>2h$, put $N=\lfloor (b-a)\mathrm{/}h\rfloor -2\ge 0$, mark $a,a+h,\dots ,a+Nh$, and then $(a+Nh+b)\mathrm{/}2,b$. The last two gaps lie in $[h,3h\mathrm{/}2)$; earlier gaps equal $h$. In particular all gaps are between $h$ and $2h$. Choose one geodesic for each consecutive pair of marked images, and interpolate it at constant speed to define $p$. At a zero image distance use the constant map. Endpoints of adjacent pieces agree. On a marked interval of length $H$, its speed is at most $(\lambda H+\epsilon )\mathrm{/}H\le 2\lambda$. Splitting a parameter interval at its finitely many marks proves the global $2\lambda$-Lipschitz bound and hence continuity.

For any parameter $t$, the closer endpoint $v$ of its marked interval satisfies $\mathrm{\mid }t-v\mathrm{\mid }\le h$ (including the single-interval case). Since $p(v)=q(v)$, the quasi-geodesic upper bound gives $d(q(t),p(v))\le \lambda h+\epsilon =2\epsilon$, while Lipschitz control gives $d(p(t),q(v))\le 2\lambda h=2\epsilon$. These prove the two Hausdorff inclusions individually. Adding the same two bounds gives $d(q(t),p(t))\le 4\epsilon$.

Within one marked interval of length $H$, constant speed gives $d(p(s),p(t))\le (\lambda +\epsilon \mathrm{/}H)\mathrm{\mid }s-t\mathrm{\mid }\le \lambda \mathrm{\mid }s-t\mathrm{\mid }+\epsilon$. In distinct intervals, join $p(s)$ to its right marked endpoint, then to the left marked endpoint for $t$, then to $p(t)$. The middle pair are original $q$ values. Adding their three upper estimates gives $d(p(s),p(t))\le \lambda (t-s)+3\epsilon \le \lambda (t-s)+4\epsilon$ for $s<t$.

For the lower bound, the single-interval case satisfies $t-s\le 2h$, so $(t-s)\mathrm{/}\lambda -4\epsilon \le 0$. In the longer construction, parameters in the same or adjacent marked intervals satisfy $t-s\le 3h$, again making that lower bound nonpositive. For separated intervals write $s\in [u,u+h]$ and $t\in [v,w]$, $u+h<v$, $w-v\le 3h\mathrm{/}2$. The first interval has length exactly $h$: the two exceptional intervals are the final adjacent ones, so neither can be the first of a separated pair. Put $A=(s-u)\mathrm{/}h\in [0,1]$ and $B=(t-v)\mathrm{/}h\in [0,3\mathrm{/}2]$.

In the separated case select marked endpoints $U,V$ as follows. If $A\le 3\mathrm{/}5$ take $U=u$, otherwise take $U=u+h$; if $B\le 3\mathrm{/}5$ take $V=v$, otherwise take $V=w$. The four cases give respectively the following upper bounds for $E=(\mathrm{\mid }s-U\mathrm{\mid }+\mathrm{\mid }t-V\mathrm{\mid })\mathrm{/}h$ and for $J=\max \{0,(t-s)-(V-U)\}\mathrm{/}h$: $(E,J)\le (6\mathrm{/}5,3\mathrm{/}5),(1,1),(3\mathrm{/}2,0),(13\mathrm{/}10,2\mathrm{/}5),$ in the order $(A\le 3\mathrm{/}5,B\le 3\mathrm{/}5)$, $(A>3\mathrm{/}5,B\le 3\mathrm{/}5)$, $(A\le 3\mathrm{/}5,B>3\mathrm{/}5)$, $(A>3\mathrm{/}5,B>3\mathrm{/}5)$. For example the last case has $u+h-s\le 2h\mathrm{/}5$ and $w-t\le 9h\mathrm{/}10$, while $(t-s)-(w-u-h)\le u+h-s\le 2h\mathrm{/}5$. In every case $1+2E+J\le 4$. Since $U<V$, the original lower inequality and the $2\lambda$ Lipschitz bound give $d(p(s),p(t))\ge (V-U)\mathrm{/}\lambda -\epsilon -2\lambda hE\ge (t-s)\mathrm{/}\lambda -\epsilon (1+2E+J\mathrm{/}{\lambda }^2)\ge (t-s)\mathrm{/}\lambda -4\epsilon$.

Combining the lower cases, upper estimate and approximation estimates proves all assertions. Equality $s=t$ is immediate and reversed parameter order follows by symmetry. The construction uses finitely many chosen geodesics, and requires no continuity of the original map.

We first justify projections to any specified geodesic segment $H=\gamma ([0,\mathrm{\ell }])$. The function $f(t)=d(x,\gamma (t))$ is $1$-Lipschitz. Let $m=\inf f([0,\mathrm{\ell }])$. Bisect the interval, keeping the left half if its infimum is $m$, and otherwise the right half, whose infimum must be $m$ since the original interval is their union. Iteration gives nested closed intervals of length $\mathrm{\ell }\mathrm{/}{2}^n$, each with infimum $m$. Their left endpoints have a supremum $r$ belonging to all the intervals, by completeness. Lipschitz control gives $0\le f(r)-m\le \mathrm{\ell }\mathrm{/}{2}^n$, hence $f(r)=m$. This includes $\mathrm{\ell }=0$. Moreover, if $p$ is a closest point of $x$ in $H$ and $x^{\mathrm{\prime }}$ lies on $[p,x]$, then $p$ is still closest to $x^{\mathrm{\prime }}$: for $z\in H$, $d(x^{\mathrm{\prime }},z)\ge d(x,z)-d(x,x^{\mathrm{\prime }})\ge d(x,p)-d(x,x^{\mathrm{\prime }})=d(x^{\mathrm{\prime }},p)$.

For a closest point $p$ of $x$ on a segment $H$ and $z\in H$, put $t=(x\mathrm{\mid }z{)}_p$ and choose $u\in [p,z]$ at radius $t$. Write $A=d(x,p)$ and $P=d(p,z)$; from the definition of $t$ we have $A+P=d(x,z)+2t$. Applying the product inequality at basepoint $x$ with bridge $u$ gives $(p\mathrm{\mid }z{)}_x\ge \min \{(p\mathrm{\mid }u{)}_x,(u\mathrm{\mid }z{)}_x\}-\kappa$, and the products expand to $(p\mathrm{\mid }z{)}_x=A-t$ and $(p\mathrm{\mid }u{)}_x=(u\mathrm{\mid }z{)}_x=(A+d(x,u)-t)\mathrm{/}2$, since $d(p,u)=t$. Hence $d(x,u)\le d(x,p)-t+2\kappa$. Minimality of $p$ forces $t\le 2\kappa$, or equivalently $d(x,p)+d(p,z)\le d(x,z)+4\kappa$. For projections $p,r$ of $x,y$ to the same segment, put $A=d(x,p)$, $B=d(y,r)$ and $s=d(p,r)$. The just-proved estimate gives $d(x,r)+d(y,p)\ge A+B+2s-8\kappa$. By the four-point inequality this sum is at most $\max \{A+B,d(x,y)+s\}+2\kappa$. If $s>5\kappa$, the first entry of the maximum cannot suffice, and consequently $s\le d(x,y)-A-B+10\kappa$. Thus always $s\le \max \{5\kappa ,d(x,y)-A-B+10\kappa \}$.

Here is the halving estimate. Suppose projections $p,r$ of $x,y$ lie on a segment $H$, $d(x,y)\le 10\rho +c$, $c\ge 0$, and $A,B\ge M+5\rho +c\mathrm{/}2$, with $M\ge 15\rho \mathrm{/}2$. Step 1.2 gives $s=d(p,r)\le 5\kappa \le 5\rho$. Let $x^{\mathrm{\prime }},y^{\mathrm{\prime }}$ be the radius-$M$ points on $[p,x],[r,y]$. Step 1.1 preserves their projections. At basepoint $p$, all three products $(x^{\mathrm{\prime }}\mathrm{\mid }x{)}_p,(x\mathrm{\mid }y{)}_p,(y\mathrm{\mid }{y}^{\mathrm{\prime }}{)}_p$ are at least $M$: the first is exactly $M$; the second follows from $d(p,y)\ge B$ and $A+B-d(x,y)\ge 2M$; the third follows from $d(p,y^{\mathrm{\prime }})\ge M$, $d(p,y)\ge B$ and $d(y,y^{\mathrm{\prime }})=B-M$. Two product inequalities give $(x^{\mathrm{\prime }}\mathrm{\mid }{y}^{\mathrm{\prime }}{)}_p\ge M-2\rho$. The same argument at $r$ gives the analogous bound. Writing $v=d(x^{\mathrm{\prime }},y^{\mathrm{\prime }})$, we obtain $d(p,y^{\mathrm{\prime }})\ge M-4\rho +v$ and $d(r,x^{\mathrm{\prime }})\ge M-4\rho +v$. Four-point control now yields $2M-10\rho +2v\le \max \{s+v,2M\}$. Since $s\le 5\rho \le 2M-10\rho$, if the first entry is maximal this forces $v=0$; otherwise it gives $v\le 5\rho$. In either case $d(x^{\mathrm{\prime }},y^{\mathrm{\prime }})\le 5\rho$.

For the original $Y$, choose its quasiconvexity segment $H=[p_a,p_b]$, and choose closest points $r_a,r_b$ of $q(a),q(b)$ on $H$ using step 1.1. For any $h\in H$ and any $e>0$, there exists $y\in Y$ with $d(h,y)<K+e$. Consequently $D\le d(q(t),y)\le d(q(t),h)+K+e$. Letting $e$ tend to zero and taking the infimum over $h$ shows $d(q(t),H)\ge D-K$. At $h=r_a$, minimality of $p_a$ in $Y$ similarly gives $d(q(a),p_a)\le d(q(a),r_a)+K$. Step 1.2 on $H$, with $x=q(a),p=r_a,z=p_a$, then gives $d(r_a,p_a)\le K+4\kappa \le K+4\rho$. The same estimate holds at $b$.

By finite induction, a chain $z_0,\dots ,z_{2^j}$ with projections $r_i$ to $H$, adjacent gaps at most $10\rho +c$, and projection distances at least $5\rho j+15\rho \mathrm{/}2+c\mathrm{/}2$, satisfies $d(r_0,r_{2^j})\le 5\kappa$. For $j=0$ this is step 1.2. For $j+1$, move each point along its projection segment to radius $M=5\rho j+15\rho \mathrm{/}2$. Step 2.1 makes adjacent new gaps at most $5\rho$. Discard odd-indexed points: adjacent retained gaps are at most $10\rho$, their projections are unchanged by step 1.1, and all their distances to $H$ equal $M$. The induction hypothesis with $c=0$ applies to this chain of $2^j$ gaps and proves the claim. This is an induction over finite chains, not a selection over all path parameters.