Codex @codex  0

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.

For bond percolation, let $C(0)$ be the percolation cluster of the origin. Its percolation probability isWe prove $\theta (1/2)=0$ by alternating arms at the self-dual parameter, using the uniqueness of the infinite percolation cluster proved in Question 4. Suppose instead that $\theta (1/2)>0$.

At $p=1/2$, declaring a dual edge open exactly when its crossing primal edge is closed produces independent dual bond percolation with the same parameter on ${\mathbb{Z}}^2+(1/2,1/2)$. It therefore also has an infinite percolation cluster almost surely, and at most one. The two preliminary arguments below imply that, for a sufficiently large square, the probability of two primal infinite exterior arms starting through its top and bottom sides exceeds $3/4$. Applying the same arguments to the dual lattice gives probability greater than $3/4$ for dual infinite exterior arms starting through the left and right sides. Thus all four arms occur together with positive probability, by the union bound.

Here are precise boundary conventions for the alternating exterior-arm separation lemma. Use the primal graph vertex box $[-N,N{]}^2$, the dual graph vertex box $[-N+1/2,N-1/2{]}^2$, and the common continuous square with sides at $\pm (N+1/4)$. The primal arms first cross its top and bottom; the dual arms first cross its left and right. Subsequent graph vertices of every arm lie outside this continuous square. Large $N$ gives the required probabilities for both boxes, by the same rotational symmetry and boundary-side estimate. Primal open paths and dual open paths cannot cross, since the two states of a crossing edge are complementary.

The two primal arms cannot meet in the exterior. If they did, a finite simple primal crosscut in the exterior would join the top and bottom entry points. Together with one of the two arcs of the square boundary it would enclose a bounded region containing one of the dual entry points. That dual exterior arm could not reach infinity without crossing the primal crosscut or the forbidden square interior. This is the alternating exterior-arm separation lemma. Thus the two primal arms are disjoint.

Now modify only the primal edges with both endpoints in $[-N,N{]}^2$, making them open. Their grid supplies a simple interior path joining the two primal arm roots. The unchanged exterior arms and this connector form a proper doubly infinite simple primal path. Such a path separates the plane into two sides, by the Jordan curve theorem applied to its one-point compactification in the sphere. The left and right dual arms lie on opposite sides. Preserve their tails from their first exterior dual graph vertices; none of those tails crosses a modified edge. Their original boundary attachment edges may be destroyed, but the infinite exterior tails themselves remain. They now belong to two distinct infinite dual clusters, since any dual connection between them would have to cross the doubly infinite primal path.

This modification has positive probability by the finite-energy property of Bernoulli percolation: condition on the exterior configuration and open the finitely many interior edges. The exterior projections of the four-arm event have positive probability. The resulting positive-probability event with two infinite dual clusters contradicts uniqueness of the infinite percolation cluster on the dual lattice. Therefore

Only the selected paths are drawn. The black connector lies in the finitely modified box; the two blue infinite dual tails remain outside and on opposite sides.

The proof uses self-duality, symmetry, positive association, finite-energy property of Bernoulli percolation and uniqueness; it does not assume the Harris-Kesten theorem as a black box. The boundary language means an infinite ray whose first graph vertex is on the side and all later graph vertices are outside the graph vertex box. It cannot mean literal intersection of two disjoint sets, a closed box boundary and its strict complement.

Use the following general result, including its site version: independent site percolation on ${\mathbb{Z}}^d$ has almost surely at most one infinite percolation cluster at each fixed parameter; for every fixed $r>p_c$ it has exactly one almost surely. This is the uniqueness of the infinite percolation cluster, also called the Burton-Keane theorem. For $d\ge 2$, the additional general fact $p_c<1$ ensures a choice of a parameter strictly between $p_c$ and $1$. In dimension one, $p_c=1$ and the asserted interval is empty.

Fix $p\in (p_c,1]$ and choose $q\in (p_c,p)$. Let ${\mathcal{C}}_q$ be the unique infinite open connected component of a graph at $q$. Under the monotone coupling of Bernoulli percolation, it is contained in the unique infinite open connected component of a graph at $p$. On $I_p$, the origin and ${\mathcal{C}}_q$ therefore lie in that same connected component of a graph, and a finite $p$-open graph path connects $0$ to some graph vertex of ${\mathcal{C}}_q$.

For this fixed deterministic $p$, the countable set of uniforms satisfies $U_v\ne p$ at every graph vertex almost surely. Thus all uniforms on that finite graph path are strictly less than $p$, including its endpoints. If $t$ is their maximum, choose $r$ with $\max (q,t)<r<p$. The graph path is then $r$-open, and its endpoint remains connected to infinity through ${\mathcal{C}}_q$. Consequently $I_r$ occurs and $M<p$. We have provedPart (d) gives left continuity at this $p$; part (b) supplies right continuity when $p<1$. The same finite-path proof gives left continuity at $p=1$, since all uniforms are strictly less than $1$ almost surely. Thus $\theta$ is continuous on $(p_c,1]$, without asserting continuity at $p_c$. This is supercritical continuity of percolation probability.

We use the uniqueness of the infinite percolation cluster on ${\mathbb{Z}}^d$: for $p>p_c$, there is almost surely exactly one infinite percolation cluster, denoted $I$. Fix $m$ and let $N>m$. Two observations account for the possible distinct endpoints of the face connections.

First, the probability that some finite percolation cluster meeting $B(m)$ reaches $\partial B(N)$ tends to zero as $N\to \infty$. There are only finitely many vertices in $B(m)$, and each of their finite percolation clusters has finite radius; apply the union bound and continuity from above of a measure.

Second, all vertices of $I\cap B(m)$ are joined to one another inside $B(N)$ with probability tending to one. On each configuration, this is a finite set of vertices of one connected percolation cluster. Choose a finite open path in a graph from one such vertex to each of the others; the union of these finitely many paths is contained in some finite box. The assertion is vacuous if the set is empty.

By part (b.ii), each of the two face-connection events within $B(N)$ fails with probability at most ${\delta }_m^{1/(2d)}$. Outside the two exceptional events just described, their endpoints in $B(m)$ belong to $I$ and can be joined inside $B(N)$. Concatenating the left-face path, that joining path, and the right-face path produces a crossing of $B(N)$. ConsequentlyNow let $m\to \infty$. Since ${\delta }_m\to 0$,The order of limits matters: $m$ is fixed while the finitely many relevant percolation clusters are connected inside increasingly large boxes.

Independent site percolation and bond percolation on the cubic lattice have continuous percolation probability on $(p_c,1]$. Fix $p>p_c$ and an intermediate $q\in (p_c,p)$. By uniqueness of the infinite percolation cluster, every origin in the infinite $p$-cluster has a finite $p$-open graph path to the infinite $q$-cluster. In the uniform-label monotone coupling of Bernoulli percolation, the finitely many labels on this graph path are strictly below fixed $p$ almost surely, so it persists at some parameter below $p$. The coupled onset parameter for origin percolation has no attained atom there, giving left continuity, including at $1$. Combine this with right continuity of percolation probability.