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 cannot by itself be applied at . 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 , the number 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 . Finite modification of Bernoulli percolation gives positive probability to this modification, contradicting constancy.
Nor can . 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 .
Here is the counting contradiction in detail. For a finite box , contract each open component outside that touches to a boundary terminal. In each cluster, the resulting incidence graph is finite and connected. Every trifurcation in separates at least three sets of terminals, since each infinite branch must leave . 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 bounds the number of those branching graph vertices by the terminal count. Different terminals are represented by different exterior graph neighbours of . Thus the trifurcation boundary-counting lemma gives
For , the two sizes are and . Letting contradicts . Therefore almost surely, with the same conclusion for the translated planar dual graph.
Now prove absence of percolation at by Zhang's argument. Suppose instead that . 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 , let mean that an open graph ray takes an outward edge through the indicated side and subsequently uses graph vertices outside . 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 gives
For the planar dual graph use the box with boundary coordinates 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 , 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. Hence
Finally suppose . At the permitted exponential decay of subcritical percolation would supply with . Use the rectangle with graph vertices , and let 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 and height . 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 is
But a crossing has some starting graph vertex on its -vertex left side connected to distance . The union bound and exponential decay of subcritical percolation imply , a contradiction. Therefore
No Russo-Seymour-Welsh theorem or continuity assertion for is being used, and critical uniqueness was proved rather than inferred from the supercritical hypothesis.