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If an infinite percolation cluster exists almost surely, the probability that a large square box intersects one tends to one. The four rotationally symmetric exterior-side arm events are increasing. By the Harris-FKG inequality, their common probability is at least $1-(1-P(D_N){)}^{1/4}$, where $D_N$ is their union. Thus a pair of specified opposite sides both has an infinite exterior arm with probability at least $1-2(1-P(D_N){)}^{1/4}$, tending to one. The rays start at a boundary graph vertex and immediately leave the graph vertex box.

For nearest-neighbour bond percolation with $0<p<1$ on the strip $T_k=\mathbb{Z}\times \{-k,\dots ,k\}$, put $q_k(n)={\mathbb{P}}_p((0,0)\leftrightarrow (n,0)\text{ in }{T}_k)$. The Harris-FKG inequality and translation invariance imply $q_k(n+m)\ge q_k(n)q_k(m)$. Thus $-\log q_k(n)$ is a subadditive sequence, and the Fekete lemma givesThe direct horizontal graph path gives $0\le f_k(p)\le -\log p$. Enlarging the strip increases every percolation two-point connection probability, so $f_k(p)$ is nonincreasing in $k$ and has a nonnegative limit of a sequence.

The FKG inequality gives positive association under a log-supermodular probability measure on a finite distributive lattice.

The Russo-Seymour-Welsh theorem and the Harris-FKG inequality imply $\pi (2r)\ge c\pi (r)$ for a scale-independent constant $c>0$.

In bond percolation on the square lattice, every nearest-neighbour edge independently has state one, meaning open, with probability $p$, and state zero, meaning closed, with probability $1-p$. Thus the configuration law is the product measure ${\mathbb{P}}_p={⨂}_{e\in E}\mathrm{Bernoulli}(p)$. An open path in bond percolation uses only open edges. An increasing event $A$ satisfies $\omega \in A,{\omega }^{\prime }\ge \omega \Rightarrow {\omega }^{\prime }\in A$: opening further edges cannot destroy the event.

The Harris-FKG inequality for two increasing events isFor disjoint occurrence of increasing events, write $A\square B$ when disjoint sets of open edges separately witness $A$ and $B$, so each event is forced by its own witness regardless of the other edges. The Van den Berg-Kesten inequality statesFor events given by cylinder sets these are finite-coordinate assertions. Connection events have finite open-path witnesses; truncating to finite boxes and taking increasing limits gives the same disjoint-occurrence bound for them. Sharing a vertex between the two witness paths is permitted; sharing an edge is not.

For $1\le m<n$, take a simple open path from the origin to the radius-$n$ boundary and let $x$ be its first vertex on the radius-$m$ boundary. Its initial segment witnesses $0\leftrightarrow x$. If $z$ is its final vertex, then $\Vert z-x{\Vert }_{\infty }\ge n-m$. Consequently the remaining segment reaches $x+\partial {\Lambda }_{n-m}$, and its initial portion until that visit witnesses $x\leftrightarrow x+\partial {\Lambda }_{n-m}$. The two segments use disjoint edges. HenceThe union bound, Van den Berg-Kesten inequality and translation invariance of the product measure now give the weighted BK boundary-splitting estimateFor $m=n$ the same bound follows directly from the union bound, with $g_0=1$. With the conventional $\partial {\Lambda }_0=\{0\}$, $m=0$ gives equality.

Write $C(0)$ for the percolation cluster of the origin. By summing its membership indicators, the percolation susceptibility isIf it is finite, then $a_m\to 0$. For $0<p<1$, choose $m$ with $0<a_m=a<1$. Iterating the boundary-splitting estimate with this fixed $m$ givesFor $k\ge 2m$, $\lfloor k/m\rfloor \ge k/(2m)$, so $g_k\le \exp [-(-\log a)k/(2m)]$. To handle all smaller positive radii without an unspecified prefactor, observe that $g_k\le g_1=1-(1-p{)}^4<1$. ThereforeFor $1\le k<2m$, the second chosen rate gives $e^{-\gamma k}\ge g_1$, and for larger $k$ the first rate applies. At $k=0$ both sides equal one. The endpoint $p=0$ has $g_k=0$ for every $k\ge 1$, so any positive rate works; $p=1$ is excluded by finite susceptibility. This proves exponential one-arm decay from finite susceptibility with the required unit prefactor.

Write $a=P(A_1)=P(A_2)$. The complements are decreasing events. The Harris-FKG inequality for a product measure givesThe same correlation inequality for two decreasing events follows from that for their increasing complements, since their covariance is unchanged. ThusBoth $1-a$ and the square root are nonnegative, so rearranging gives the square-root bound for increasing events:No independence between the events themselves is asserted; independence is the coordinate property of the underlying product measure.

Let $E_T,E_B,E_L,E_R$ be the increasing events that an infinite simple open ray in a graph starts at the indicated side and immediately leaves the box, with all later graph vertices outside it. A corner exit is assigned according to the direction of its outgoing edge. The four events have the same probability $a_N$ by rotational symmetry. Their union is $D_N$ from the preceding step.

The Harris-FKG inequality applies to these exterior-ray events by approximation with finite connections to larger boxes and passage to decreasing limits. Positive association of the four decreasing events $E_s^c$, by iterating the Harris-FKG inequality, givesEquivalently, one may apply the preceding square-root bound for increasing events first to the two opposite-side unions and then to the two individual sides of one pair. ThereforeA union bound now showsChoose $N$ such that $1-\mathbb{P}(D_N)<(1/8{)}^4$. ThenThe infinite exterior clusters touched by these two sides need not be the same. The same argument works for dual boxes and any sufficiently large size; this permits the matching primal/dual contours used in the full proof.

Order configurations coordinatewise. The FKG lattice condition on the masses of a finite Boolean lattice isThe FKG inequality states that under this condition, increasing real-valued functions $f,g$ satisfy $\mathbb{E}[fg]\ge \mathbb{E}[f]\mathbb{E}[g]$. In particular increasing events satisfy $P(A\cap B)\ge P(A)P(B)$. 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 bywhere $c_n(d)$ 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 $a_n\le Cn{c}_{n-1}(2)$. Choose $b>\mu (2)$ with $qb<1$. The root-count limit gives $c_{n-1}(2)\le C_b{b}^n$, soThis 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 $P(A_R\cap B_R)\ge P(A_R)P(B_R)>0$.

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, $p_c(2)\le 1-1/\mu (2)$. Finally the lattice in dimension $d\ge 2$ 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

Use the coordinatewise partial order: $\omega \le \eta$ means $\omega (e)\le \eta (e)$ for every $e\in E$. An increasing event $A$ is upward closed: $\omega \in A$ and $\omega \le \eta$ imply $\eta \in A$.

Under the independent Bernoulli distribution product measure ${\mathbb{P}}_p$ on these coordinates, the Harris-FKG inequality statesfor increasing events $A,B$. For $K\subseteq E$, let $[\omega {]}_K$ be the cylinder of configurations agreeing with $\omega$ on $K$. The disjoint occurrence of increasing events $A\square B$ consists of configurations for which there exist disjoint $K,L$ with $[\omega {]}_K\subseteq A$ and $[\omega {]}_L\subseteq B$. For increasing events, witnesses can be taken to prescribe only open coordinates. The BK inequality isPositive association rewards simultaneous occurrence; the disjoint-witness requirement in the BK inequality gives a bound in the opposite direction. The inequalities also apply to different independent Bernoulli parameters in different coordinates.

Regard the boxes as sets of lattice graph vertices. Write ${\tau }_p(x,y)={\mathbb{P}}_p(x\leftrightarrow y)$, and put $g_0=1$. On the event defining $g_n$, choose an open self-avoiding walk from the origin to its first visit to $\partial {\Lambda }_n$. Let $y$ be its first visit to $\partial {\Lambda }_m$. Its initial segment witnesses $0\leftrightarrow y$. Its remaining segment ends at a point $z$ withFor $m<n$, truncate this remaining segment on its first visit to $y+\partial {\Lambda }_{n-m}$. The two segments use disjoint edges, so they are disjoint witnesses. For $m=n$, the second event is the sure event with empty witness. Thus the union bound, the BK inequality and translation invariance giveAn unrestricted connection event can depend on infinitely many edges. Apply the finite-coordinate BK inequality first to connections confined to growing boxes and take increasing limits; each occurrence has a finite path witness. This justifies its use here. The estimate is the weighted BK boundary-splitting estimate.

By summing cluster indicators and using Tonelli theorem, the percolation susceptibility isIf $\chi (p)<\infty$, then $S_m\to 0$. At $p=0$ all $g_k$ vanish, so any positive exponential rate works. Otherwise choose $m\ge 1$ with $a=S_m\in (0,1)$. Iterating the weighted BK boundary-splitting estimate gives, for $k=jm+r$ with $0\le r<m$,For $k\ge m$, $\lfloor k/m\rfloor \ge k/(2m)$, so $g_k\le \exp [-(-\log a)k/(2m)]$. To include the finitely many smaller indices without a prefactor, note that $p<1$ under finite susceptibility andTherefore one may takeFor $k<m$ this follows from $\mu (p)k\le -\log g_1$, and for $k\ge m$ it follows from the iterated estimate. This proves exponential one-arm decay from finite susceptibility with exactly the requested unit prefactor.

The purpose of the RSW lemma is to turn local crossing information into control at every fixed shape and scale. Work with independent bond percolation on the square lattice. Let $H(a,b)$ be the event of an open left-to-right graph path in $[0,a]\times [0,b]\cap {\mathbb{Z}}^2$, and write $h_p(a,b)={\mathbb{P}}_p(H(a,b))$.

A useful precise uniform version of the RSW lemma is the following uniform RSW crossing estimate. If for a fixed parameter $p$ and some $\delta >0$,then for every fixed aspect ratio $\rho >1$ there is $c(\delta ,\rho )>0$, independent of $n$, such thatIn particular this bounds length $2n$ in terms of the uniform square bound at length $n$. The constants need not be sharp. The hypotheses used in the proof are planarity, translation and reflection symmetries, the Harris-FKG inequality and independence of unexplored edges. At a self-dual parameter the same argument for dual crossings gives an upper bound strictly below one as well. Neither an exact value of $p_c$ nor the existence of an infinite cluster is an assumption of this estimate.

Here is the gluing picture behind the RSW lemma. For increasing crossing or attachment events, the Harris-FKG inequality provides a lower bound on their joint probability. If an event of probability at least $\delta$ is a union of two reflection-related increasing alternatives $A_1,A_2$, the square-root trick for positively associated events givesIndeed their decreasing complements are positively associated too, so ${\mathbb{P}}_p(A_1^c\cap A_2^c)\ge {\mathbb{P}}_p(A_1^c{)}^2$. This prevents a square crossing from concentrating all its useful attachment locations on just one side.

The nontrivial first gluing step enlarges a square to aspect ratio $3/2$. Explore an extremal square crossing, revealing the edges on the explored side but leaving its other side unexamined. In that unexplored region the conditional law is still independent bond percolation. Compare possible attachments to the crossing with their reflected alternatives. Reflection symmetry and the preceding square-root estimate give a positive bound, depending only on $\delta$, for the required attachment after averaging over the explored crossing. Carry out the reflected construction at the other end and use the Harris-FKG inequality to combine the increasing attachment events with the original crossing. Planarity ensures that the relevant transverse paths actually meet. This RSW reflection extension lemma yields a lower bound $u(\delta )>0$ for the $3n/2$-by-$n$ crossing. The exploration is important: a reflection compares the laws of fresh configurations; reflecting the picture of an open path does not make the reflected edges open.

From that first extension, longer rectangles are obtained by a genuinely transverse gluing. Two $3n/2$-by-$n$ rectangles shifted by $n/2$ overlap in an $n$-by-$n$ square. Require a horizontal crossing in each long rectangle and a vertical crossing of the overlap square. Each horizontal crossing crosses that overlap from left to right, and therefore meets its vertical crossing. Their union crosses the $2n$-by-$n$ rectangle. The Harris-FKG inequality gives a lower bound $u(\delta {)}^2\delta$. Repeat a bounded number of times for any fixed $\rho$. Integer rounding uses neighboring lattice rectangles and bounded additional gluing steps; finitely many smallest scales can be absorbed into the constant. The constants deteriorate with $\rho$ but not with $n$. Merely requiring horizontal crossings in adjacent squares would not suffice, since their endpoints need not coincide; the overlap crossing solves that problem.

I apply the RSW lemma to the exact threshold for bond percolation on the square lattice. The planar duality for rectangle crossings says that an open horizontal crossing and a closed dual vertical crossing are complementary. At $p=1/2$, the dual edge states have the same law as the primal ones. For the balanced lattice rectangle with side lengths $n+1,n$, the rotated dual crossing rectangle has those same side lengths: its transverse lengths before rotation are $n,n+1$. Thus the exact self-dual rectangle crossing probability is $1/2$. Restricting a crossing of this rectangle to its first visit to the shorter vertical side givesThe one-unit balance avoids assuming an exact one-half probability for every finite vertex-square convention. The RSW lemma now gives scale-independent positive bounds for all fixed-aspect-ratio primal and closed dual rectangle crossings.

Arrange four appropriately overlapping long rectangles around a square ring. Closed dual crossings along the four sides, with transverse overlap crossings if needed, join to a closed dual graph cycle surrounding the inner square. The Harris-FKG inequality, applied to the closed dual states, and the RSW lemma give a constant $b>0$ for this circuit event, uniformly over ring size. Choose disjoint rings with radii increasing, for example, by a factor of four. Their circuit events depend on disjoint edge sets, so they are independent. An open graph path from the origin to infinity would have to avoid every one of these dual barriers. Its probability is at most $(1-b{)}^j$ after $j$ rings and therefore zero. HenceThis is the independent annular barriers for percolation argument. It is worth separating it from the other inequality: the absence of an infinite cluster at one parameter alone does not prove that every larger parameter percolates.

For the reverse inequality use sharpness of the percolation transition: below $p_c$, independent bond percolation on the cubic lattice has exponentially decaying connection probabilities. One can see why this is the relevant general ingredient through the finite-set criterion for percolation sharpness. For a finite set $S$ containing the origin putIf ${\phi }_p(S)<1$, split a long open self-avoiding walk at its first exit from $S$. Its internal connection, exit edge and subsequent connection have disjoint witnesses. The BK inequality gives a contraction by ${\phi }_p(S)$ each time distance decreases by the diameter of $S$ plus a fixed step. Iteration proves exponential decay of subcritical percolation at such $p$.

To identify this finite-set threshold with $p_c$, the Margulis–Russo formula expresses the derivative of $g_n(p)$ as the sum of pivotal-edge probabilities. Explore the cluster attached to the box boundary, and let $S$ be its complement. On failure of the origin-to-boundary event, $0\in S$; all edges from $S$ to the boundary cluster are closed, while the internal edges of $S$ remain fresh. A boundary edge is pivotal precisely when its endpoint in $S$ is connected to the origin inside $S$. Removing the factor $1-p$ for a closed pivotal edge yields the general differential inequalityFor $p>{\tilde{p}}_c$ every finite-set quantity on the right is at least one. Integrating from any $p_0\in ({\tilde{p}}_c,p)$ gives a positive lower bound for $g_n(p)$ independent of $n$, and taking $n\to \infty$ gives $\theta (p)>0$. Combined with the contraction below ${\tilde{p}}_c$, this proves $p_c={\tilde{p}}_c$ and the stated sharpness conclusion. This outline supplies the extra threshold argument rather than assuming the desired critical value.

If $p_c(2)>1/2$, sharpness would give $g_n(1/2)\le C{e}^{-cn}$. But a square crossing starting somewhere on the left side entails an open connection from that starting graph vertex to distance $n$. A union bound over the $n+1$ possible starting graph vertices givescontradicting exponential decay. Therefore $p_c(2)\le 1/2$, and the two directions establishThis is the Harris-Kesten theorem. The overall mechanism is that self-duality supplies a square crossing, the RSW lemma transports it between shapes and creates barriers at every scale, and sharpness converts the finite-scale crossing information into the exact threshold.

Order configurations coordinatewise: $\omega \le {\omega }^{\prime }$ means ${\omega }_e\le {\omega }_e^{\prime }$ for every edge. An increasing event $A$ is a measurable set such that $\omega \in A$ and $\omega \le {\omega }^{\prime }$ imply ${\omega }^{\prime }\in A$. Opening additional edges cannot destroy an increasing event. The Harris-FKG inequality, the product measure version of the FKG inequality, statesfor two increasing events. Equivalently, bounded coordinatewise increasing random variables $f,g$ satisfy ${\mathbb{E}}_p(fg)\ge {\mathbb{E}}_{p}f{\mathbb{E}}_{p}g$. The event inequality also holds for two decreasing events, by taking complements.

For disjoint occurrence of increasing events, a finite open edge set $K$ witnesses $A$ in $\omega$ if every configuration agreeing with $\omega$ on $K$ belongs to $A$. Since $A$ is an increasing event and all edges of $K$ are open, this says that the configuration with exactly $K$ open already forces $A$. Define $A\square B$ to consist of configurations admitting disjoint finite witnesses $K,L$ for $A,B$. For events depending on finitely many edges, this is the usual disjoint-occurrence definition; it also applies to finite-connection events on the infinite graph. The Van den Berg-Kesten inequality statesOn a countable graph, its finite-witness version follows by taking increasing unions over finite edge sets. The Harris-FKG inequality concerns simultaneous occurrence, whereas the Van den Berg-Kesten inequality requires separate certificates using disjoint coordinates.

The union bound gives $g_k\le {\sum }_{x\in \partial {\Lambda }_k}{\mathbb{P}}_p(0\leftrightarrow x)$. Hence some boundary graph vertex has percolation two-point connection probability at least $g_k/(8k)$. The rotations and reflections of the square lattice let us choose such a graph vertex as $x=(k,j)$, $-k\le j\le k$. Reflection in the vertical line through $x$ sends $0$ to $e_{2k}=(2k,0)$ and fixes $x$. ThereforeThe Harris-FKG inequality applied to these two increasing events proves the reflection lower bound for two-point percolation:Every graph path from $0$ to $e_{2k}$ reaches $\partial {\Lambda }_{2k}$, so $h_{2k}\le g_{2k}$. Taking roots givesBoth outside expressions have limit of a sequence $\gamma$, proving the even case. For $p>0$, the Harris-FKG inequality with the last horizontal edge gives $h_{2k+1}\ge p{h}_{2k}$, and also $h_{2k+1}\le g_{2k+1}$. The lower bound has rootThe upper bound has the same limit of a sequence. At $p=0$ all positive-distance connection probabilities vanish. Thus ${\lim }_{n\to \infty }{h}_n^{1/n}=\gamma$ for every $p\in [0,1]$.

Write $H(R)$ and $V(R)$ for the horizontal and vertical open-crossing increasing events of a rectangle in the rotated square lattice. At parameter $1/2$, planar duality for rectangle crossings says that failure of $H$ is exactly a closed dual vertical crossing. Quarter-turn symmetry and the boundary convention of the rotated rectangles identify the two alternatives on a square. Thus its crossing probability is $1/2$; a rectangle obtained by a harmless one-step boundary adjustment has crossing probability at least $1/2$. This is the finite-domain exact self-dual rectangle crossing probability, not an assumption about infinite percolation clusters.

We first prove a RSW reflection extension lemma. In coordinate units adapted to the rotated drawing, take $S=[0,n{]}^2$, $T=[-n,n{]}^2$, and $R=[-n,2n]\times [-n,n]$. Reveal the rightmost open vertical crossing $\Gamma$ of $S$, when one exists, using only its edges and the region to its right. Reflect $\Gamma$ across the horizontal axis. The part $D$ of $T$ to the left of these two paths remains unexposed and has the independent parameter-$1/2$ law. A horizontal crossing of $T$ must reach their union. Its probability is at least $1/2$. Reflection symmetry and the union bound give a probability at least $1/4$ of reaching the upper path $\Gamma$ through $D$. Conditional opening of the edges on $\Gamma$ can only help. Since $V(S)$ has probability at least $1/2$, the increasing event $B$ that a vertical crossing of $S$ is connected inside $T$ to its left side has probability at least $1/8$.

Reflect this construction in the vertical midline of $S$ to obtain an increasing event $B^{\prime }$ of the same probability, connecting a vertical crossing of $S$ to the right side of $[0,2n]\times [-n,n]$. On $H(S)\cap B\cap B^{\prime }$, the horizontal crossing of $S$ meets both vertical crossings, so all three graph paths join and give $H(R)$. The Harris-FKG inequality now yieldsThe exploration only conditions the exposed side of the extremal graph path; reflecting a drawn path does not assert that its reflected edges are open. This distinction is what makes the extension argument valid.

To pass from aspect ratio $3/2$ to $3$, put four rectangles of width $3h/2$ and height $h$ next to each other, with successive left boundaries separated by $h/2$. Adjacent rectangles overlap in a square of side $h$. Require horizontal crossings of the four rectangles and vertical crossings of the three overlap squares. Every consecutive pair of horizontal crossings meets the vertical crossing in its overlap; together they form a horizontal crossing of width $3h$. The Harris-FKG inequality gives a strictly positive bound, for these compatible scales, ofThe construction works on the rotated square lattice with the boundary sites used in the source drawing. Integer roundings do not create a scale restriction: for an odd height use the largest smaller compatible even height, and use eight horizontal rectangles and seven overlap squares, whose union has auxiliary aspect ratio $5$, so that its width exceeds the desired $3m$ width for all sufficiently large $m$. Truncate its horizontal crossing on first reaching the desired right side. The number of extra rectangles and overlap-square crossings is bounded independently of $m$, and every factor is bounded below by the same positive constants. The finitely many smallest rectangles each have positive crossing probability, since a specified finite open graph path suffices. Therefore a constant $c>0$ works at every integer scale. Taking $\tau =c/2$ ensures the requested strict inequality:This proves the required instance of the Russo-Seymour-Welsh theorem through reflection, exploration and gluing. The parameter in this calculation is $1/2$.

First obtain a polynomial critical one-arm upper bound from part (a). Fix a sufficiently large integer radius $r$. Use rotated-coordinate boxes $B_r=[-r,r{]}^2$ for this geometric construction, in the axes of part (a). Surround $B_r$ by a square ring inside $B_{3r}$. It can be assembled from the four stripsThe first two require horizontal dual crossings; the last two require vertical dual crossings. Their corner overlaps are squares, so the adjacent long crossings intersect there and their union contains a dual graph cycle surrounding the inner box. Translate the strips by the half-lattice displacement of the planar dual graph and adjust the endpoints by one lattice unit if needed. All aspect ratios remain bounded. Part (a) and the same overlap gluing give a uniform lower bound $c_1>0$ for each dual crossing at $1/2$. Applying the Harris-FKG inequality to these decreasing events in the primal configuration gives a closed dual circuit with probability at least $\delta =c_1^4>0$. We may replace $\delta$ by a smaller number in $(0,1)$.

Use the independent annular barriers for percolation at radii $r_j=r_{0}10^j$. Their supporting edge sets are disjoint, so the circuit events are independent. In the original unrotated coordinates, $B_{3r}$ lies inside ${\Lambda }_{5r}$ and the ring still separates the origin from infinity. These fixed geometric comparisons let us bound the original $g_N$. An open graph path from $0$ to distance $N$ must avoid every such closed dual circuit inside ${\Lambda }_N$. Thus, for some constants $C<\infty$ and $\alpha >0$,where the first bound is used only when at least one of the indicated rings fits; increasing $C$ handles smaller $N$.

Now let $\epsilon =p-1/2>0$ and use the monotone coupling of Bernoulli percolation on edges. The event $\{0\leftrightarrow \partial {\Lambda }_N\}$ only needs edges whose endpoints lie in ${\Lambda }_N$, because a first boundary hit gives an internal graph path. There are exactly $4N(2N+1)$ such edges. Each changes state between parameters $1/2$ and $p$ with probability $\epsilon$. The union bound therefore gives the finite-box comparison for percolation parametersFor sufficiently small $\epsilon$, choose $N=\lfloor {\epsilon }^{-1/(\alpha +2)}\rfloor$, so that $N$ is at least half the unrounded value. Both terms then have order ${\epsilon }^{\alpha /(\alpha +2)}$. In particular,for a finite $A$. Increase $A$ to cover the remaining compact range of parameters by $\theta (p)\le 1$. This near-critical percolation power upper bound requires only a positive crossing constant, not the exact value of a critical exponent. It also implies $\theta (1/2)=0$ through the critical one-arm probability bound.

The strips are nested. Every connection permitted inside $T_k$ is also permitted inside $T_{k+1}$, so $q_k(n)\le q_{k+1}(n)$ for every fixed separation. ConsequentlyThe preceding nonnegative bound makes this a decreasing sequence bounded below. By monotone convergence of real sequences,There is no assertion that this limit of a sequence is strictly positive: positivity at each fixed width need not survive an increasing-width limit of a sequence.

One can also identify the limit of a sequence. Let $q(n)$ be the percolation two-point connection probability in the whole square lattice. Every finite connecting graph path has a bounded vertical extent, so $q_k(n)\uparrow q(n)$. Commuting infima gives the strip approximation to the planar connection decay rate:The same positive-association argument identifies the final infimum with the whole-plane normalized logarithmic limit of a sequence. This is an interchange of infima justified by monotonicity at fixed $n$. For example, when $p>1/2$, the threshold proved in question 2 and uniqueness give an infinite percolation cluster with root probability $\theta (p)>0$. The Harris-FKG inequality makes the probability that both endpoints belong to it at least $\theta (p{)}^2$, so $q(n)\ge \theta (p{)}^2$. The whole-plane rate, and hence ${\lim }_k{f}_k(p)$, is then zero, despite the strict positivity of every finite-strip rate.

Put $q_k(n)={\mathbb{P}}_p((0,0)\leftrightarrow (n,0)\text{ inside }{T}_k)$, with $q_k(0)=1$. The connection event is increasing in the open edges. Apply the Harris-FKG inequality to connection from $(0,0)$ to $(n,0)$ and connection from $(n,0)$ to $(n+m,0)$, both constrained to the strip. Their intersection implies connection between the outer endpoints, while horizontal translation invariance makes the second event's probability $q_k(m)$. HenceThese events may depend on infinitely many strip edges. To justify the inequality directly, first restrict each connecting graph path to a finite box, apply Harris-FKG inequality in the finite independent edge family, and then let the box grow. Every connection has a finite witness graph path, so continuity from below gives the stated inequality.

The direct horizontal graph path supplies $q_k(n)\ge p^n>0$, and probabilities are at most one. Thus $a_k(n)=-\log q_k(n)$ is a finite nonnegative subadditive sequence with $a_k(n)\le -n\log p$. The Fekete lemma now gives a finite connection decay rate in a percolation strip:Each individual ratio is at least its infimum, so $-\log q_k(n)\ge n{f}_k(p)$ and thereforeAt $n=0$ this reads $1\le 1$. Only nonnegative separations are intended by this notation; horizontal reflection supplies the analogous bound with $|n|$ for negative separations.

We use planar duality for rectangle crossings, Harris-FKG inequality and the permitted exponential decay of subcritical percolation. A critical-point detail matters: uniqueness stated only for $p>p_c$ cannot by itself be applied at $p=p_c$. We prove the Burton-Keane theorem by the boundary counting proof of percolation uniqueness, establishing at most one infinite percolation cluster at every parameter and avoiding that gap.

For $0<p<1$, the number $N$ of infinite percolation clusters is almost surely constant by translation ergodicity of Bernoulli percolation. It cannot be a finite constant greater than one. A sufficiently large box meets two different infinite percolation clusters with positive probability; opening its finitely many interior edges joins them without creating a new infinite percolation cluster, decreasing $N$. Finite modification of Bernoulli percolation gives positive probability to this modification, contradicting constancy.

Nor can $N=\infty$. A finite box then meets three different infinite percolation clusters with positive probability. Select one infinite exterior branch from each. In the box and a finite collar retain just an embedded three-armed tree joining those branches and close all other incident edges there. Its branching graph vertex separates three infinite components when removed, so is a trifurcation vertex in percolation. All selections concern finitely many possible entrances and edge patterns; finite-energy property of Bernoulli percolation gives positive probability for at least one such pattern. Translation invariance then gives a common positive trifurcation probability $q$.

Here is the counting contradiction in detail. For a finite box $W$, contract each open component outside $W$ that touches $W$ to a boundary terminal. In each cluster, the resulting incidence graph is finite and connected. Every trifurcation in $W$ separates at least three sets of terminals, since each infinite branch must leave $W$. A minimal subtree joining all terminals must therefore contain that graph vertex with degree of a vertex at least three. Its leaves are terminals. The tree identity ${\sum }_v(\mathrm{deg}(v)-2)=-2$ bounds the number of those branching graph vertices by the terminal count. Different terminals are represented by different exterior graph neighbours of $W$. Thus the trifurcation boundary-counting lemma givesFor $W=[-r,r{]}^2\cap {\mathbb{Z}}^2$, the two sizes are $(2r+1{)}^2$ and $8r+4$. Letting $r\to \infty$ contradicts $q>0$. Therefore $N\in \{0,1\}$ almost surely, with the same conclusion for the translated planar dual graph.

Now prove absence of percolation at $p=1/2$ by Zhang's argument. Suppose instead that $\theta (1/2)>0$. Almost surely an infinite primal cluster exists, and self-duality gives an infinite dual cluster as well. An infinite connected subgraph of this locally finite lattice contains a graph ray, by the König infinity lemma. Expanding square boxes meet the infinite percolation clusters with probability tending to one.

For $B_r=[-r,r{]}^2\cap {\mathbb{Z}}^2$, let $A_N,A_E,A_S,A_W$ mean that an open graph ray takes an outward edge through the indicated side and subsequently uses graph vertices outside $B_r$. The union occurs whenever the box meets an infinite percolation cluster: take the last exit of an infinite graph ray from the finite box. Conversely an exterior arm supplies an infinite percolation cluster meeting its boundary. Quarter-turn symmetry makes the four probabilities equal. Their complements are decreasing events, so the square-root trick for positively associated events givesFor the planar dual graph use the box with boundary coordinates $\pm (r+1/2)$ and define its exterior side-arm events in the same way. This box also has quarter-turn symmetry and meets the dual infinite percolation cluster with probability tending to one. Thus each dual side has an infinite exterior arm with probability tending to one. Primal outward edges cross this dual-box contour on the corresponding sides; their remaining graph vertices stay outside it. A union bound makes the simultaneous event of primal north/south arms and dual east/west arms have positive probability for a sufficiently large box.

The four arms alternate around the contour. Open all edges with both endpoints in $B_r$, leaving all outward and exterior edges intact. This joins the two primal entrance points and preserves all four exterior arms. Finite-energy property of Bernoulli percolation keeps the event's probability positive. The primal north and south arms are now connected through the box. The two dual arms must lie in different infinite dual clusters. To see the separation, a hypothetical finite dual graph path joining the east and west arms cannot enter the dual-box interior: its boundary-crossing and interior edges cross primal edges that were just opened. Erase its loops and cut it at successive hits of the contour. A resulting exterior dual crosscut, together with the corresponding contour arc, encloses one of the two intervening primal entrance points. The infinite primal graph ray from that point would have to cross a dual-open edge, which is impossible. This is the planar separation used in the alternating arms argument at the self-dual percolation parameter. It contradicts uniqueness of the infinite dual cluster. HenceFinally suppose $p_c(2)>1/2$. At $p=1/2$ the permitted exponential decay of subcritical percolation would supply $C,c>0$ with ${\mathbb{P}}_{1/2}(v\leftrightarrow \partial B_n(v))\le C{e}^{-cn}$. Use the rectangle with graph vertices $\{0,\dots ,n\}\times \{0,\dots ,n-1\}$, and let $H_n$ be its left-to-right open crossing. Planar duality for rectangle crossings identifies its complement with a dual top-to-bottom crossing of a rectangle of width $n-1$ and height $n$. Rotation and translation give the same crossing law; side edges at the entrance and exit boundaries are irrelevant. Thus the exact self-dual rectangle crossing probability isBut a crossing has some starting graph vertex on its $n$-vertex left side connected to distance $n$. The union bound and exponential decay of subcritical percolation imply ${\mathbb{P}}_{1/2}(H_n)\le nC{e}^{-cn}\to 0$, a contradiction. ThereforeNo Russo-Seymour-Welsh theorem or continuity assertion for $\theta$ is being used, and critical uniqueness was proved rather than inferred from the supercritical hypothesis.

For bond percolation, the Harris-FKG inequality states that any two increasing events $A,B$ satisfyThe same inequality holds for two decreasing events, either by reversing the coordinate order or by taking complements in the two-event identity. Intersections of decreasing events are decreasing events, so induction givesWrite $a={\mathbb{P}}_p(A_1)=\cdots ={\mathbb{P}}_p(A_n)$ and $u={\mathbb{P}}_p({\bigcup }_j{A}_j)$. Then $1-u\ge (1-a{)}^n$. Taking the nonnegative $n$-th root and rearranging yields the square-root trick for positively associated events:No independence among the events is needed.

For $m\ge 1$, let $S_m$ be the set of vertices of $B(m)$ connected to $\partial B(m)$ by an open path in a graph using only edges of $B(m)$. Every infinite-cluster vertex of $B(m)$ belongs to $S_m$, by stopping an infinite open path in a graph at its first boundary visit. Hence the increasing eventhas probability at least $\theta (p)/2$ by part (a). Importantly, $E_m$ depends only on the edges with both endpoints in $B(m)$.

Open all $8m$ perimeter edges joining successive vertices of $\partial B(m)$. They form a cycle in a graph, and opening them joins every vertex of $S_m$ into one percolation cluster, containing the specified perimeter vertex $v_m=(m,0)$. The Harris-FKG inequality bounds the probability of this event together with $E_m$ below by $p^{8m}\theta (p)/2$.

Close all $8m+4$ edges in the edge boundary of $B(m)$. These edges are distinct from the internal edges already considered, so their states are independent of those events. The resulting percolation cluster of $v_m$ is finite, contained in $B(m)$, and has at least $|S_m|$ vertices. Therefore, by translating $v_m$ to the origin,For $n\ge 1$, choose $m=\lceil \sqrt{n/(2\theta (p))}\rceil$. Then the threshold is at least $n$, and $m\le (1+(2\theta (p){)}^{-1/2})\sqrt{n}$. Put $A=-\log p$ and $B=-\log (1-p)$; both are positive. The preceding lower bound is at least $e^{-c_1\sqrt{n}}$ with, for example,The construction modifies only order-$m$ boundary edges while trapping order-$m^2$ vertices, which explains the stretched exponential scale. Here $\mathbb{N}$ is interpreted as the positive integers; the proposed lower bound at $n=0$ would be false because $\theta (p)>0$.

The Harris-FKG inequality givesHere is an induction proof of the product-measure version, also called Harris' inequality. The result is immediate for one coordinate. For $n$ coordinates, defineThe sections of an increasing event are increasing events in the first $n-1$ coordinates, so the induction hypothesis givesMonotonicity gives $a_1\ge a_0$ and $b_1\ge b_0$. The difference between the right side andisThis closes the induction and proves positive association.