t)} \overline{{M \backslash K}}.$$ See Section \[sec:notation\] for a description of these special Jacobi fields. In order to make our application of Lemma \[lemma:jacobigrowth\] transparent, rescale the time parameter by $\lambda = t - \sigma_r(t_0)$. Then $J_s( 0 ) = \dot\tau^{(t_0)}(s)$, and for $s \in (t_0, \infty)$, $t \in (\sigma_r(t_0), \infty)$ and $\lambda \in (0, \infty)$, Lemma \[lemma:jacobigrowth\] implies $$|J_s(\lambda)| \geq |J_s(0)| \cosh( a \lambda ).$$ Therefore we may write: $$1 \geq |\dot\sigma_Y(s)|_{g_Y(\sigma_r(s))} = |J_s(\sigma_r(s))| \geq |\dot\tau^{(t_0)}(s)| \cosh( a ( \sigma_r(s) - \sigma_r(t_0))).$$ Consequently using this estimate and the estimate on $\sigma_r$ from Lemma \[lemma:geodesicrgrowth\], we find that $$|\dot\tau^{(t_0)}(s)| \leq \frac{1}{\cosh( a( C s + B - \sigma_r(t_0)) )},$$ for constants, $C, B$. Upon integration of this expression we find $$\begin{aligned}
len( \tau^{(t_0)} ) &\leq& \int^{\infty}_{t_0}  \frac{1}{\cosh( a( C s + B - \sigma_r(t_0)) )}\\
                    &\leq& C(a) \tanh^{-1} (e^{ a( Cs + B - \sigma_r(t_0) ) } ) \big|^{s=\infty}_{s=t_0}\\
                    &\leq& C(a) \frac{\pi}{2}.\end{aligned}$$

Thus $len( \tau^{(t_0)})$ is bounded independent of $t_0$. This completes step 2.

The proof of the above proposition can be extended to yield a stronger result that will be useful in proving that the topology on ${M^{*}}$ is well defined. In the lemma below we consider two hypersurfaces 