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I was reading this book by Coorneart, Delzant, and Papadopoulos. I am stuck with this proposition in Chapter 1 (Proposition 1.5)

If $Y$ is a bounded $\delta$-hyperbolic subset of $X$, then $X$ is $\delta'$-hyperbolic with $\delta' = \delta + 6 \eta$ where $\eta = \sup_{x \in X} \operatorname{dist}(x,Y)$.

The authors have not given proof of the proposition. I have tried using the $\delta$-hyperbolicity condition of $Y$ but I am getting stuck. Please help.

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  • $\begingroup$ @AmirSagiv Thanks for editing the question $\endgroup$ Sep 23 '20 at 7:04
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Let $x,y,z,b \in X$ be given and assume that $\eta = \sup_{x\in X} d(x,Y) < \infty$. Fix $x' , y' , z' , b' \in Y$ such that $|x-x'|, |y-y'|$, etc. are all $\leq \eta$. (For a 'properly done' proof, use $|x-x'|<\eta + \varepsilon$ for some $\varepsilon>0$ and take $\varepsilon \to 0$ at the end).

Using the triangle inequality, one can show that \begin{align*} & (x'|y')_{b'} \leq (x|y)_b + 3\eta \\ & (x|z)_b \leq (x'|z')_{b'} + 3\eta \\ \text{and } & (y|z)_b \leq (y'|z')_{b'} + 3\eta . \end{align*} Then combining the above with the fact that Y is $\delta$-hyperbolic, we get \begin{align*} (x|y)_b & \geq (x'|y')_{b'} -3\eta \\ & \geq \min\{ (x'|z')_{b'} , (y'|z')_{b'} \} - \delta - 3\eta \\ & \geq \min\{ (x|z)_{b} -3\eta , (y|z)_{b} -3\eta \} - \delta - 3\eta \\ & = \min\{ (x|z)_{b}, (y|z)_{b} \} - \delta - 6\eta . \end{align*} So $X$ is $\delta'$-hyperbolic with $\delta' = \delta + 6\eta$.

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