Let $T$ be an $\R$-tree with a very small action of a free group $F_N$
which has dense orbits. Such a tree $T$ or its metric completion
$\bar T$ are not locally compact. However, if one adds the Gromov
boundary $\partial T$ to $\bar T$, then there is a coarser
\emph{observers' topology} on the union $\bar T \cup \partial T$, and it is
shown here that this union, provided with the observers' topology, is
a compact space $\Tobs$.
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To any $\R$-tree $T$ as above a \emph{dual lamination} $L^{2}(T)$ has
been associated in \cite{chl1-II}. Here we prove that, if two such
trees $T_{0}$ and $T_{1}$ have the same dual lamination $L^{2}(T_{0})
= L^{2}(T_{1})$, then with respect to the observers' topology the two
trees have homeomorphic compactifications: $\Tobszero = \Tobsone$.
Furthermore, if both $T_{0}$ and $T_{1}$, say with metrics $d_{0}$ and
$d_{1}$, respectively, are minimal, this homeomorphism restricts to an
$\FN$-equivariant bijection ${T_{0}} \to {T_{1}}$, so that on the
identified set $ {T_{0}} = {T_{1}}$ one obtains a well defined family
of metrics $\lambda d_{1}+(1-\lambda)d_{0}$. We show that for all
$\lambda \in[0,1]$ the resulting metric space $T_{\lambda}$ is an
$\R$-tree.