Let $G$ be a bounded degree graph and fix a vertex $v\_0$. Suppose that the simple random walk on $G$ is transient, and let $g:G\to\mathbb{R}$ be defined by $$g(v)=\mathbb{P}\_v[T\_{v\_0}<\infty].$$
That is, $g(v)$ is the probability that a simple random walk starting at $v$ will ever hit $v\_0$. 


Now consider $g(G)$, the image of $g$ in $[0,1]$. 



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> Can $g(G)$ have an accumulation point other than $0$?
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Accumulation points of Green function on a transient graph
The answer is yes. To see this, take a transient graph $G$ (say $\mathbb{Z}^3$) and glue a copy of $\mathbb{N}$ to some vertex (say $v\_0$). The vertices of $\mathbb{N}$ now all have $g(v)=1$.


If you want infinitely many vertices with distinct values of $g$, all greater then some $a>0$, take a ladder instead of $\mathbb{N}$. That is, take two copies of $\mathbb{N}$, call them $U^0=\{u^0\_0,u^0\_1,\ldots\}$ and $U^1=\{u^1\_0,u^1\_1,\ldots\}$ and connect each $u^0\_k$ with $u^1\_k$. Connect $u^0\_0$ with $v\_0$ and connect $u^1\_0$ with a neighbour of $v\_0$, call it $v\_1$. In the resulting graph, vertices of the ladder will have distinct values of $g$, all greater then $g(v\_1)$.


