Identification of ends of Bruhat-Tits Tree I am trying to understand why a canonical identification exists between the "ends" or "rays" of the Bruhat-Tits tree defined on $\mathbb{Z}_p$-lattices (with metric invariant under the action of $PGL(2,\mathbb{Q}_p)$) and $\mathbb{P}^1(\mathbb{Q}_p)=\mathbb{Q}_p\cup\{\infty\}$. I have consulted several online references on the Bruhat-Tits tree, all of which more or less avoid constructing the actual identification.
Intuitively, I see that since any vertex in the tree has $p+1$ neighbors, a path that travels outwards without backtracking should correspond to a $p$-adic number - for instance, if $p=2$, you could label a path and end up with some binary string such as $0101010011...$, which can be converted into a $2$-adic by using the 0s and 1s as coefficients of a formal Laurent series. I am unsure, however, of what the end that corresponds to $\infty$ should look like, as well as how the exact identification between a path and a $p$-adic would work - given a binary string, for instance, how would we know which power of $p$ would be the starting point? I am also unsure of what the implications of there being $p+1$ choices for the first outgoing edge are.
Edit: After thinking further on the matter, I've realized that the $p+1$-th edge (i.e. the "extra" one at your starting point) can be used to denote elements of $\mathbb{Q}_p\setminus\mathbb{Z}_p$, by letting "extra edge followed by $n$ zeroes" denote multiplication of the subsequent binary sequence by $p^{-n-1}$. So any path that does not travel through the extra edge initially simply belongs to $\mathbb{Z}_p$, and I can identify $\infty$ with the path that goes through the extra edge and is followed by infinite 0s (since this one seems to be degenerate/lack interpretation via the above). Feedback on this identification and whether it is correct/can be made more precise would be appreciated.
 A: I am more or less sure that following those lines you'll get your result, after proving that the Bruhat-Tits graph is a tree, ie. that my sequence $(L_n)$ is determined by any subsequence $(L_{k_j})$.

*

*Let $A_n$ be the set of lattices $$A_n= \{ L\subset Z_p^2, [Z_p^2:L] = p^n\}$$
Write $$L = (a,b)Z_p+(c,d)Z_p$$ Wlog $r=\min(v(a),v(b))\le \min(v(c),v(d))$.   If $v(a)=r$, with $d' = d-\frac{c}ab$ then $$L = (a,b)Z_p+(0,d')Z_p= (a,b)Z_p+(0,p^{v(d')})Z_p=(a,b)Z_p+p^{v(d')}Z_p^2$$
Since $$[(a,b)Z_p+p^{v(d')}Z_p^2:p^{v(d')}Z_p^2]= p^{v(d')-r}$$ we get that $$p^n=[Z_p^2:(a,b)Z_p+p^{v(d')}Z_p^2]=p^{2 v(d')}/p^{v(d')-r}=p^{v(d')+r}$$
Thus in fact $L$ is determined by $(a,b)$. Obviously it works the same way when $v(b)=r$ so $$A_n = \{ (a,b)Z_p+p^e Z_p^2,(a,b)\in Z_p^2,  \min(v(a),v(b))\le n/2,e= n-\min(v(a),v(b))\}$$


*Then let's look at the sequences $(L_n), L_n\in A_n$ such that $L_{n+1}\subset L_n$ and $p^k L_n \ne L_{n+2k}$.
Write $$L_n = (a_n,b_n)Z_p+p^{e_n} Z_p^2, \qquad e_n\ge r_n = \min( v(a_n),v(b_n)), n=r_n+e_n$$
The $p^k L_n \ne L_{n+2k}$ condition imposes that $e_n-r_n\to \infty$. This is because once $n$ and $e_n-r_n$ are known then there are finitely many choices for $L_n$.
Thus in fact there is $(a,b)\in Z_p^2-(0,0)$ such that for all $n\ge N$ large enough $$L_n = (a,b) Z_p+p^{n-\min(v(a),v(b))} Z_p^2$$
We impose $p^k L_n \ne L_{n+2k}$ because we are going to say that two lattices are equivalent if $L=\lambda L'$, so our $(a,b)$ becomes an element of $(\Bbb{Q}_p^2-(0,0))/\Bbb{Q}_p^* =\Bbb{P^1(Q_p)}$.
