intersection of infinitely many closed subtrees may fail to be closed.

A *linear* subtree of $T$ is a subtree in which any three points are contained in a segment. It is an easy exercise to prove that a maximal linear subtree of $T$ is closed in $T$. Finally, any linear subtree $L\subset T$ is isometric to a convex subset of $\Lambda$ and any two isometries $L{\rightarrow}\Lambda$ differ by an isometry of $\Lambda$.

Isometries
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An isometry $g$ of a $\Lambda$–tree $T$ can be of one of the following exclusive types:

elliptic: $g$ has a fix point in $T$

inversion: $g$ has no fix point, but $g^2$ does

hyperbolic: otherwise.

In all cases, the set $A_g=\{x\in T\,|\, [g{^{-1}}x,x]\cap[x,g.x]=\{x\}$ is called the *characteristic set* of $g$.

If $g$ is elliptic, $A_g$ is the set of fix points of $g$ which is a closed subtree of $T$. Moreover, for all $x\in T$, the midpoint of $[x,g.x]$ exists and lies in $A_g$.

If $g$ is an inversion, then $A_g={\emptyset}$. Actually, for any $x\in T$, $d(x,g.x)\notin 2\Lambda$ so $[x,g.x]$ has no midpoint in $T$. In particular, if $2\Lambda=\Lambda$ (which occurs for instance if $\Lambda={{\mathbb{R}}}^n$), inversions don’t exist. Moreover, one can perform the analog of barycentric 