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At parameter $1/2$, primal and dual bond percolation on the square lattice have the same probability distribution. Suppose an infinite percolation cluster exists. Large boxes meet it with probability tending to one. The square-root trick for positively associated events and quarter-turn symmetry imply that each side has an infinite exterior open arm with probability tending to one. The same holds for dual arms. A union bound then gives positive probability of primal top and bottom arms together with dual left and right arms.

Use a primal box $[-r,r{]}^2$ and a dual box with sides at $\pm (r+1/2)$. Open only primal edges whose endpoints both lie in the primal box, preserving all outward and exterior arm edges. The two primal arms are joined inside. Planarity prevents the two alternating dual arms from joining: a proposed dual connection together with the inner boundary arc separates one primal infinite arm from infinity. Thus the modified configuration has two infinite dual clusters. Its positive probability by finite modification of Bernoulli percolation contradicts uniqueness of the infinite percolation cluster. Consequently the percolation probability at $1/2$ is zero. This is often called Zhang's argument.

Take the increasing event that a finite square torus graph contains some long horizontal or vertical rectangle crossing. Translations and a quarter turn make this a transitive increasing event on its bonds. A positive critical crossing bound and the Friedgut-Kalai sharp threshold theorem make its supercritical probability tend to one. A bounded family of shorter, thicker rectangles intercepts every possible long crossing. The square-root trick for positively associated events transfers the high probability to each fixed intercepting rectangle. Using every possible translate would lose the useful bounded exponent.

We use Zhang's argument, obtaining the critical nonpercolation conclusion from uniqueness rather than assuming it as part of the Harris-Kesten theorem. For independent bond percolation on ${\mathbb{Z}}^2$, apply part (ii) to independent site percolation on its line graph; open sites correspond to open bonds. The line graph is amenable and of finite type. Therefore an infinite bond cluster, if it exists, is almost surely unique. The same holds for the dual bond percolation model.

Assume for contradiction that $\theta (1/2)>0$. The planar dual graph is another square lattice, and its open bonds correspond to closed primal bonds, also with parameter $1/2$. Both models then have an infinite percolation cluster almost surely, by part (i). As a disc $D_R$ expands, the probability that it meets an infinite cluster tends to one, for each model. Choose $R$ avoiding all primal and dual vertices and all intersections of either graph with the four endpoints of the following arcs. Divide the boundary circle into four equal arcs $A_1,A_2,A_3,A_4$ in cyclic order. Let $L_i$ be the event of an infinite primal open ray whose last exit from $D_R$ is in $A_i$; define $L_i^{\ast }$ similarly for an infinite dual open ray. After that exit the entire ray stays outside the disc. The existence of a ray follows by finite branching in an infinite locally finite graph.

Each $L_i$ is increasing in the primal bond states, and their union has probability tending to one. Four-fold rotation symmetry makes their probabilities equal. The FKG inequality for their decreasing complements gives the square-root trick for positively associated eventsThe left side tends to zero, so $\mathbb{P}(L_i)\to 1$. Apply the same argument to the dual states to get $\mathbb{P}(L_i^{\ast })\to 1$. A union bound, requiring no independence between the models, now shows that for sufficiently large $R$ the alternating eventhas positive probability.

The alternating exterior-arm separation lemma says that the two primal rays on this event cannot be joined by a primal open path lying outside $D_R$. Indeed, such a joining path contains a simple crosscut of the exterior between the two primal exit points. Join that crosscut to a chord inside the disc. The Jordan curve theorem puts one of the alternating dual exit points on the bounded side. Its dual ray, which is restricted to the exterior of the disc and runs to infinity, would have to cross the open primal crosscut. A dual open bond cannot cross a primal open bond. This is impossible.

Let $F$ be the finite set of primal bonds whose drawn edges meet $D_R$, and let $E$ be the projection of the alternating-arm event onto the states outside $F$. Because $F$ is finite, this projection is a finite union of measurable events; $\mathbb{P}(E)>0$. Every configuration in $E$ admits an assignment on $F$ witnessing the original four arms. Now close all bonds of $F$. The infinite tails of the two primal rays beyond their first exterior vertices survive; any path joining the surviving tails would now avoid the disc. Appending their original exterior segments back to their exit points would give the forbidden exterior connection. Thus the modified configuration contains at least two infinite percolation clusters. By independence, or the finite-energy property of Bernoulli percolation, $\mathbb{P}(E\cap \{F\text{ all closed}\})=2^{-|F|}\mathbb{P}(E)>0$. This contradicts the uniqueness obtained from part (ii). Therefore

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.

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 each of the $2d$ coordinate faces $F$ of $B(N)$, define the increasing event $A_F(m,N)$ that an open path in a graph lying in $B(N)$ joins $B(m)$ to $F$. Their union is exactly the event of connection from $B(m)$ to $\partial B(N)$ within $B(N)$. The bond percolation law and the centered boxes are invariant under coordinate permutations and sign changes, so all $2d$ events have the same probability.

Apply the square-root trick for positively associated events from part (a), with $2d$ in place of the number of events. If ${\delta }_m=1-{\mathbb{P}}_p(D_m)$, part (b.i) gives, for every $N=m+n$,In particular this holds for the left and right faces, and it proves the stronger statement in which each connecting path in a graph stays inside the outer box.