Proposition 3

A graph is a chain if and only if its number of vertices is finite and non-zero and it is a tree with no ramification point.

Suppose that the graph $\Gamma$ is a chain $(x_0, \ldots, x_m)$ with the properties listed before the statement of Prop. 3. It is clear that any vertex of $\Gamma$ is joined to at most two vertices. If $i < j$ the path $(x_i, \ldots, x_j)$ extracted from the path $(x_0, \ldots, x_m)$ joins $x_i$ to $x_j$; thus, $\Gamma$ is connected. Finally, let $(x_{p_1}, \ldots, x_{p_n})$ be a circuit in $\Gamma$, and let $p_k$ be the smallest of the distinct integers $p_1, \ldots, p_n$. There exist distinct indices $i$ and $j$ such that $x_{p_k}$ is joined to $x_{p_i}$ and $x_{p_j}$: this follows from the definition of a circuit. Since $p_k < p_i$ and $p_k < p_j$, we necessarily have $p_i = p_j = p_k + 1$, which is a contradiction since $p_1, \ldots, p_n$ are distinct. Thus, there is no circuit in $\Gamma$.

Conversely, let $\Gamma$ be a tree with no ramification point and with a finite non-zero number of vertices, and let $(x_0, \ldots, x_m)$ be an injective path of maximal length in $\Gamma$. Denote by $T$ the set of vertices other than $x_0, \ldots, x_m$. A vertex $b \in T$ cannot be joined to any vertex $x_i$, for we would have either

a) $i = 0$, but then $(b, x_0, \ldots, x_m)$ would be an injective path of length $m + 1$ in $\Gamma$, or b) $i = m$, but then $(x_0, \ldots, x_m, b)$ would be an injective path of length $m + 1$ in $\Gamma$, or c) $0 < i < m$, but then $x_i$ would be joined to three distinct vertices $x_{i-1}$, $x_{i+1}$ and $b$.

Since $\Gamma$ is connected, $T$ is empty by Cor. 1 of Prop. 1. Moreover, if there were $i, j$ with $j - i > 1$ and $x_i, x_j$ joined, there would be a circuit $(x_i, x_{i+1}, \ldots, x_j)$ in $\Gamma$. Thus, $\Gamma$ is a chain. Q.E.D.