= Solution
Order configurations coordinatewise. The <FKG lattice condition> on the masses of a finite Boolean lattice is
$$
P(\omega\wedge\eta)P(\omega\vee\eta)\geq P(\omega)P(\eta)\quad\text{for every }\omega,\eta.
$$
The <FKG inequality> states that under this condition, increasing real-valued functions $f,g$ satisfy $\mathbb E[fg]\geq\mathbb E[f]\mathbb E[g]$. In particular <increasing events> satisfy $P(A\cap B)\geq P(A)P(B)$. For a <product measure> with coordinate weights $w_e(0),w_e(1)$, the pair $(\min(\omega_e,\eta_e),\max(\omega_e,\eta_e))$ has the same two entries as $(\omega_e,\eta_e)$. Multiplying over coordinates gives equality in the lattice condition, including degenerate Bernoulli parameters without division by zero.
For independent <bond percolation> on the nearest-neighbor <cubic lattice>, write $\theta_d(p)=\mathbb P_p(|C(0)|=\infty)$. Define the <percolation critical probability> and the <connective constant> by
$$
\boxed{p_c(d)=\inf\{p\in[0,1]:\theta_d(p)>0\},\qquad\mu(d)=\lim_{n\to\infty}c_n(d)^{1/n}},
$$
where $c_n(d)$ counts rooted $n$-step <self-avoiding walks>. <Translation invariance> makes the root irrelevant, and part (i) proves existence of the latter limit.
First work on the <square lattice>, whose dual is another translated <square lattice>. Put $q=1-p$ and assume $q\mu(2)<1$. If the open cluster at the origin is finite, its exterior <edge> boundary contains a closed dual simple circuit surrounding the origin. To see this planar fact, surround its finitely many vertices by their unit square cells and follow the exterior boundary: its crossing primal <edges> are closed. Resolving repeated boundary vertices into simple circuits leaves a circuit separating the origin from infinity.
Let $a_n$ count such dual circuits of length $n$. Every one crosses the positive horizontal ray at distance at most $n$, because it surrounds the origin and its horizontal span is at most its length. Choosing a ray-crossing <edge>, an orientation and all but the last <edge> encodes it by one of at most $Cn$ rooted length-$n-1$ <self-avoiding walks>. Thus $a_n\leq Cn c_{n-1}(2)$. Choose $b>\mu(2)$ with $qb<1$. The root-count limit gives $c_{n-1}(2)\leq C_b b^n$, so
$$
\mathbb P_p(\text{some enclosing closed dual circuit of length at least }N)
\leq CC_b\sum_{n\geq N}n(qb)^n\longrightarrow0.
$$
This summable tail alone need not make the probability of every enclosing circuit less than one. Choose a large $R$ and let $A_R$ be the absence of enclosing closed dual circuits of length at least $R$, with $P(A_R)>1/2$. Let $B_R$ require every primal <edge> inside $[-R,R]^2$ to be open. It has positive probability. Both events are increasing. The finite-measure <Harris-FKG inequality> extends to $A_R$ by decreasing limits over finitely many circuit exclusions, so $P(A_R\cap B_R)\geq P(A_R)P(B_R)>0$.
On $B_R$ the origin is connected to every <graph vertex> of that box. Any closed dual circuit enclosing the origin must then enclose the whole box, hence have length at least $R$. On $A_R\cap B_R$ no such circuit exists, so the origin cluster is infinite. This proves the <connective-constant Peierls bound>, $p_c(2)\leq1-1/\mu(2)$. Finally the lattice in dimension $d\geq2$ contains a coordinate copy of the <square lattice>, whose <edge> law is unchanged. Percolation in that subgraph implies percolation in the full <graph>, and hence
$$
\boxed{p_c(d)\leq p_c(2)\leq1-\frac1{\mu(2)}}.
$$
Back to article page
