Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 209 / 1

It suffices to use the two-dimensional coordinate half-plane inside $H$. A Peierls argument shows that when $p$ is sufficiently close to one, the probability that a fixed open vertex is separated from distance $n$ by a closed dual contour is summable over contour lengths: the number of length-$m$ contours is at most exponential in $m$, whereas each is closed with probability $(1-p{)}^m$. Hence the origin has positive probability of belonging to an infinite open cluster. The event that some infinite cluster exists is invariant and therefore has probability zero or one because the boundary translation is an ergodic transformation; its positive probability makes it almost sure. The same cluster is also an infinite cluster of $H$ in every $d\ge 2$.

Translation by $e=(0,1,0,\dots ,0)$ preserves $H$, so $(\omega (x+e):x\in H)$ has the same independent Bernoulli-$p$ law as $\omega$. This shift is ergodic, and the number $N$ of infinite clusters is shift-invariant. Thus $N=k_0$ almost surely for some deterministic $k_0\in \mathbb{N}\cup \{\infty \}$.

Suppose $2\le k_0<\infty$. With positive probability a finite box meets all $k_0$ infinite clusters. By the finite-energy property of Bernoulli percolation, forcing finitely many sites in the box open has positive conditional probability and joins those clusters without affecting infinity outside the box. The resulting configuration has fewer than $k_0$ infinite clusters on an event of positive probability, contradicting the almost-sure constancy. Hence $k_0\in \{0,1,\infty \}$.

Use independent percolation configurations in $H$ and in its reflected copy $H^{\prime }=(-\infty ,-2]\times {\mathbb{Z}}^{d-1}$. Require event $A$ in both copies and close the intervening site $(-1,0,\dots ,0)$. This has probability $(1-p)Q^2$. Because either distinguished infinite cluster meets its boundary hyperplane only at its distinguished origin, the closed intervening site prevents it from leaving its own half-space. The resulting whole-space configuration therefore has two distinct infinite clusters. Whole-space supercritical Bernoulli percolation has an almost surely unique infinite cluster, so $(1-p)Q^2=0$ and hence $Q=0$.

Fix the finite set $S=\{x_1,\dots ,x_n\}\subset B$. If an infinite half-space cluster met $B$ exactly in $S$ with positive probability, take independent reflected occurrences in $H$ and $H^{\prime }$ and force the finitely many intervening sites adjacent to $S$ closed. The finite-energy property of Bernoulli percolation gives this combined event positive probability, while it creates two distinct whole-space infinite clusters, contradicting uniqueness. Thus the probability is zero for every finite $S$. Since $B$ has only countably many finite subsets, almost surely every infinite cluster that meets $B$ meets it infinitely often.

By part (c2), $C\cap B_{n_0+1}$ is infinite almost surely. For each $x$ in this intersection, its neighbor $x-e_1\in B_{n_0}$ is open independently with probability $p$. The probability that all infinitely many such neighbors are closed is zero. Hence some open site of $B_{n_0}$ is adjacent to $C$, and the open cluster $C^{\prime }$ containing their union in $H_{n_0}$ is infinite, contains $C$, and intersects $B_{n_0}$.

If an infinite cluster $C$ avoids $B$, the set of first coordinates of its vertices has a least value $n_0\ge 1$. Then $C$, viewed in $H_{n_0}$, intersects $B_{n_0}$. Repeated application of part (d1) embeds it in an infinite cluster of $H_{n_0-1}$ meeting $B_{n_0-1}$, eventually producing an infinite cluster of $H$ meeting $B$. Because each enlarged cluster contains $C$, already the first step contradicts the minimality of $n_0$. Thus such a cluster has probability zero.