Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 204 / 2

Fix $ϵ>0$ and choose a cylinder event $B$ with ${\mathbb{P}}_p(A△B)<ϵ$. Translate $B$ far enough that $B$ and $\varphi (B)$ depend on disjoint edge sets. They are independent, while automorphism invariance gives $\varphi (A)=A$. The symmetric-difference inclusion supplied in the question givesAlso $|{\mathbb{P}}_p(A)-{\mathbb{P}}_p(B)|<ϵ$, and independence gives ${\mathbb{P}}_p(B\cap \varphi (B))={\mathbb{P}}_p(B{)}^2$. Letting $ϵ\downarrow 0$ yieldsso ${\mathbb{P}}_p(A)\in \{0,1\}$.

The number $N_{\infty }$ is invariant under lattice automorphisms, so part a makes it almost surely equal to one constant in $\{0,1,2,\dots ,\infty \}$. For $p>p_c$ the constant is nonzero.

Suppose it were a finite $k\ge 2$. As boxes increase to ${\mathbb{Z}}^d$, with positive probability one box meets all $k$ infinite clusters. On that event, open a finite collection of edges inside the box joining those clusters. The finite-energy property of Bernoulli percolation gives the modified event positive probability, but it has fewer than $k$ infinite clusters. This contradicts almost-sure constancy. Hence the only possibilities are

The open descendants of any vertex form a Galton-Watson process with offspring distribution $\mathrm{Bin}(k,p)$. If $kp\le 1$, it dies out almost surely, so $N_{\infty }=0$.

If $kp>1$, let $\theta >0$ be its survival probability. At level $n$, each vertex begins an independent descendant subtree; the event that its edge to its parent is closed while its descendant open cluster is infinite has probability $(1-p)\theta >0$. Thus the number of such vertices at level $n$ is binomial with $k^n$ trials and a fixed positive success probability. For every fixed $M$, the probability of at least $M$ successes tends to one. Their infinite clusters are separated by their closed parent edges, so $\mathbb{P}(N_{\infty }\ge M)=1$ for every $M$, and hence $N_{\infty }=\infty$ almost surely.