X$, we consider that $X$ is the root of $T_X$ as well as a leaf of $T_X$. Thus, if $X = \{x\}$, then we have $L(T_X) = \{\{x\}\}$.

The following proposition claims that the ballean of every finite ultrametric space $(X,d)$ coincides with the vertex set of $T_X$.

\[p2.8\] Let $(X,d)$ be a finite, nonempty ultrametric space with representing tree $T_X$. Then the following statements hold:

1.  For every node $v\in V(T_X)$ there is the unique $B \in \mathbf{B}_X$, $B = \{x_1, \ldots, x_n\}$, such that $L(T_v) = \{\{x_1\}, \ldots, \{x_n\}\}$. This ball $B$ is singular if and only if $v \in L(T_X)$.

2.  For every $B \in \mathbf{B}_X$, $B = \{x_1, \ldots, x_n\}$, there is the unique $v \in V(T_X)$ such that $L(T_v) = \{\{x_1\}, \ldots, \{x_n\}\}$. This vertex $v$ is a leaf of $T$ if and only if $B$ is singular.

This proposition is a simple modification of the corresponding result from [@Pet2013PoI] (see also Theorem 2.5 in [@Dov2019pNUAA]).

The proofs of Theorem \[t2.7\], Theorem \[t1.7\] and Proposition \[p2.8\] are based on the next basic theorem.

\[t2.9\] Let $(X, d)$ be a finite ultrametric space and let $x_1$ and $x_2$ be two distinct points of $X$. If $P$ 