Percolation threshold

Remarks

Since $G$ is connected, $p_c$ does not depend on the choice of $o$. By Kolmogorov's zero–one law, an infinite open cluster exists with probability $0$ when $p < p_c$ and with probability $1$ when $p > p_c$. Example: for the line graph $\Z$, $\theta(p) = 0$ for every $p < 1$, so $p_c(\Z) = 1$. Example: take the infinite tree in which every vertex has $d \ge 3$ neighbours. After the first step, the open cluster of $o$ grows as a branching process with mean offspring $(d-1)p$, so $p_c = 1/(d-1)$. Non-example: a finite graph does not satisfy the hypothesis; there every cluster is finite, so $\theta \equiv 0$ and the supremum carries no information. For the lattice $\Z^d$ with $d \ge 2$, one has $0 < p_c < 1$.