a sequence of nested compacta in $\Omega^{t_0}$. The intersection, $$S^\infty(t_0, a) = \bigcap_{n = 0}^\infty S^{n}(t_0, a)\,,$$ is not empty and compact. In fact, $S^\infty(t_0, a)$ is an abstract family of integral funnels. Indeed, it is easy to see that properties [**S1**]{}, [**S3**]{}, and [**S4**]{}, are satisfied. As the intersection of compact-valued maps into a Polish space, $S^\infty(t_0, \cdot)$ is measurable by [@A-B Lemma 18.4].

It turns out that each funnel $S^\infty(t_0, a)$ is a singleton. Indeed, if $u, v\in S^\infty(t_0, a)$, then, for every $\vp\in \Phi$, $$\int_0^\infty e^{-\lambda t} \vp(u(t_0 + t))\,dt = \int_0^\infty e^{-\lambda t} \vp(v(t_0 + t))\,dt\,,\quad \forall \lambda\in {{\mathbb{Q}}}_+\,.$$ By the uniqueness of the Laplace transform, $\vp(u(t_0 + t)) = \vp(v(t_0 + t))$ for all $t\ge 0$. Because this is true for every $\vp\in\Phi$ and the family $\Phi$ separates the points of $X$, we obtain $u(t_0 + t) = v(t_0 + t)$ for all $t\ge 0$, i.e., $u = v$ as paths.

For $u\in S^\infty(t_0, a)$, we will use the notation $u(t; t_0, a)$, where $t\ge t_0$. If $t_1 > t_0$, then $(\sigma_{t_1 - t_0} u)(t) = u(t; t_1, u(t_1; t_0, a))$ for $t\ge t_1$, by property [**S3**]{}. Thus, if $t_2 > t_1$, $(\sigma_{t_1 - t_0} 