to go further in the Berkovich formalism.

Spaces {#Chap2}
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Trees of spheres on an example
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In this section we forgot about rational maps and fix a set $X$ containing at least three elements. As we saw above, we are interested in understanding the diverging sequences in $\Mod_X$.

For simplicity, let us first suppose that $\card X=4$ (we already saw that for ${\card X=3}$ the space is compact). We set $X=\{a,b,c,d\}$. Consider a sequence $[i_n]$ without converging subsequences in $\Mod_X$. The natural reflex when we are working modulo Moebius transformation, is to post-compose $i_n$ by such a transformation $M_n$ in order to control the position of three points. We call $M_n\circ i_n$ a normalization of the marked sphere.

For example $M_n\circ i_n$ maps $a$ to $0$, $b$ to $1$ and $c$ to $\infty$. Using the compactness of $\S$, we can suppose (maybe after extracting a subsequence) that the sequence $M_n\circ i_n(d)$ converges. As $[i_n]$ diverges, we deduce that the limit is $0,1$ or $\infty$. Suppose for example that $\lim M_n\circ i_n(d)=\infty= M_n\circ i_n(c)$.

A naive way to think would be to define the limit of $[i_n]$ to be the application that maps $a$ to $0$, $b$ to $1$ and 