I'm currently reading through a proof of "If $X$ and $X'$ are geodesic/length spaces that are quasi-isometric, then $X$ is hyperbolic iff $X'$ is."
In the proof, we let $\phi :X\rightarrow X'$ be a quasi-isometry. Near the middle of the proof, it is then stated that $\phi(X)$ is cobounded. I'm not sure why this is true. Can someone help me see why this is the case? I don't think it has anything to do with what was proved above it but I can elaborate if needed.
Coboundedness of $\phi(X)$ holds by definition.
There is a more general concept of a quasi-isometric embedding $\phi : X \to X'$, which means that there exist $K \ge 1$, $C \ge 0$ such that $$\frac{1}{K} d(\phi(x_1),\phi(x_2))-C \le d(x_1,x_2) \le K d(\phi(x_1,x_2))+C \quad\text{for all} \quad x_1,x_2 \in X $$ Notice, so far coboundedness is not required.
To say that $\phi$ is a quasi-isometry means that it is a cobounded quasi-isometric embedding, where coboundedness of $\phi$ says that $\exists D \ge 0$ such that $\forall x' \in X' \exists x \in X$ such that $d(f(x),x') \le D$.