= Solution
\b[Independence or another ergodicity hypothesis is necessary.] Read literally for an arbitrary translation-invariant <site percolation> measure, the assertion is false: the mixture assigning <probability> $1/2$ to the all-open configuration and $1/2$ to the all-closed configuration is translation-invariant, but the invariant event of an <infinite percolation cluster> has <probability> $1/2$. We prove the zero–one law for the invariant independent law used in part (ii); the proof also works for the stated finite-range independent alternative.
Let $\Gamma$ be the group of translations, or the specified group of <graph automorphisms>, and let $A$ be invariant under it, up to null sets. An event determined by the states in a finite vertex set is a <cylinder set>. For every $\varepsilon>0$, approximation in the product <sigma-algebra> provides a <cylinder set> $B$, supported on a finite set $F$, with $\mathbb P(A\mathbin\triangle B)<\varepsilon$. Choose $g\in\Gamma$ so that $F$ and $gF$ are disjoint. Then $B$ and $gB$ are <independent events>, have equal <probabilities>, and $\mathbb P(A\mathbin\triangle gB)<\varepsilon$. Hence
$$
\bigl|\mathbb P(A)-\mathbb P(B\cap gB)\bigr|<2\varepsilon,\qquad \mathbb P(B\cap gB)=\mathbb P(B)^2,
$$
and
$$
|\mathbb P(A)-\mathbb P(A)^2|\le 2\varepsilon+2|\mathbb P(B)-\mathbb P(A)|<4\varepsilon.
$$
Letting $\varepsilon\downarrow0$ proves \b[the zero–one law] $\boxed{\mathbb P(A)\in\{0,1\}}$. This is <translation ergodicity of Bernoulli percolation>.
For a <quasi-transitive graph>, the required displacement exists: put $F$ in a ball about a vertex $v$; its orbit is infinite, so an automorphism can move $v$, and therefore that whole ball, beyond any prescribed finite neighborhood. A $k$-independent <site percolation> law means that the <sigma-algebras> generated by vertex sets at <graph distance> greater than $k$ are independent. Move $F$ far enough that $\operatorname{dist}(F,gF)>k$ and the same proof applies. Thus finite-range independence does give the zero–one law, although it will not give uniqueness.
Back to article page
