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Independent bond percolation on ${\mathbb{Z}}^2$ has at most one infinite percolation cluster at every parameter, including a possible critical parameter. The number of infinite percolation clusters is constant almost surely by translation ergodicity of Bernoulli percolation. A finite constant larger than one is impossible: a box meeting two clusters can be made entirely open, joining them and decreasing that number with positive probability by finite modification of Bernoulli percolation.

If infinitely many clusters existed, a box would meet three with positive probability. Preserve one infinite exterior arm from each and replace the finitely many interior and boundary edges by a three-armed tree joining them, closing the remaining edges. Its branch graph vertex becomes a trifurcation vertex in percolation. Translation invariance therefore gives a positive density $q$. But the trifurcation boundary-counting lemma gives $q|W|\le |{\partial }^{+}W|$ for every large box, contradicting vanishing boundary-to-volume ratio. This proves uniqueness without assuming absence of an infinite percolation cluster at criticality.

An infinite percolation cluster is an open connected component of a graph with infinitely many graph vertices. On a countable translation-invariant lattice, a zero root percolation probability implies that no infinite percolation cluster exists almost surely, by the countable union over all graph vertices. A positive root probability implies almost-sure existence by translation ergodicity of Bernoulli percolation; uniqueness is an additional theorem.

For independent bond percolation on ${\mathbb{Z}}^d$, $d\ge 2$, there is almost surely at most one infinite percolation cluster at every parameter. This is the Burton-Keane theorem. It does not require an assumption about the behavior at the critical probability. If the probability that the origin lies in an infinite percolation cluster is positive, translation ergodicity of Bernoulli percolation also gives almost-sure existence and hence exactly one.

The two structural ingredients are translation invariance and finite-energy property of Bernoulli percolation. For $0<p<1$, every prescribed configuration of finitely many bonds has positive conditional probability given all the other bonds. In particular, opening or closing a fixed finite set of bonds transforms events of positive probability into events of positive probability. The product measure is ergodic under translations. One way to see this is to approximate a translation-invariant event by a finite-cylinder event, translate the cylinder sufficiently far to make its coordinates disjoint, and use independence. Sending the approximation error to zero gives $\mathbb{P}(A)=\mathbb{P}(A{)}^2$.

Let $K$ be the number of infinite percolation clusters. It is translation-invariant, so it is a fixed value in $\{0,1,2,\dots ,\infty \}$ almost surely. First exclude a finite value $k\ge 2$. Some finite box meets at least two infinite percolation clusters with positive probability. Force every internal bond in that box to be open. This joins those clusters and can only merge other clusters; it cannot split an infinite percolation cluster. It also cannot create a new infinite percolation cluster out of finite ones, since only finitely many clusters touch the modified region. Thus the number of infinite percolation clusters decreases. finite-energy property of Bernoulli percolation makes the modified event have positive probability, contradicting its almost-sure value $k$.

Suppose next that $K=\infty$. Some finite box meets three different infinite percolation clusters with positive probability. Each such cluster has an infinite component outside the box: deletion of a finite set in a locally finite graph leaves only finitely many components adjacent to that set, so an infinite percolation cluster cannot have all of them finite. Choose one infinite exterior component from each of the three clusters, and one bond connecting it to the box.

Inside the box, connect the three selected exterior endpoints by a finite tree, preserving just their three selected entrance bonds. Take the minimal subtree connecting these endpoints; they are leaves, since their only retained connection into the box is their chosen entrance bond. This tree has a branch graph vertex of degree three. Close every other internal or boundary bond of the box. Its three exterior components remain distinct outside, since they belonged to different original clusters. The branch graph vertex now has exactly three open incident bonds, and deleting it separates its cluster into three infinite components. Such a graph vertex is a trifurcation vertex in percolation. There are finitely many choices for the endpoints and tree within the chosen region, so at least one fixed finite modification of Bernoulli percolation has positive probability. finite-energy property of Bernoulli percolation consequently gives positive probability of a trifurcation at some fixed graph vertex. Translation invariance makes this probability the same number $q>0$ at every graph vertex.

The decisive fact is that a finite box cannot contain a positive-volume density of these branching points. Here is the boundary count, including its combinatorial justification. For a finite graph vertex set $W$, let ${\partial }^{+}W$ be the exterior graph vertices adjacent to it. In each infinite percolation cluster meeting $W$, contract each of its components outside $W$ that is adjacent to $W$ to a terminal graph vertex. The quotient graph consisting of the cluster's graph vertices in $W$ and these terminals is finite and connected. A trifurcation graph vertex in $W$ separates its terminals into at least three nonempty groups, since each of its infinite branches must leave $W$.

Take a minimal subtree of the quotient graph connecting all the terminals. Every such trifurcation graph vertex must belong to that subtree and have degree at least three: otherwise the terminals in one of its separated groups could not connect to the other groups. All leaves of the minimal subtree are terminals. A cluster with only one terminal has no counted trifurcation. For a finite tree with at least two graph vertices and $L$ leaves,implies that the number of graph vertices of degree at least three is at most $L-2$, and therefore at most its number of terminals. Distinct exterior components can be assigned distinct graph vertices of ${\partial }^{+}W$. Terminals from different clusters are disjoint as well. Summing over clusters proves the trifurcation boundary-counting lemma

Take $W=[-n,n{]}^d\cap {\mathbb{Z}}^d$. The expected left side is $q|W|$, of order $q{n}^d$, while the right side is $O_d(n^{d-1})$. Division by $|W|$ and passage to infinity force $q=0$, contradicting the positive probability obtained by finite modification of Bernoulli percolation. Thus infinitely many infinite percolation clusters are impossible. The surface-to-volume comparison is the geometric reason for uniqueness on the lattice; the same argument is tied to amenability on more general graphs.

At $p=0$ there is no infinite percolation cluster, and at $p=1$ the whole lattice is the unique infinite percolation cluster. This completes all parameter cases. Finally, if the origin's infinite-cluster probability is zero, the countable union over all graph vertices shows there are no infinite percolation clusters almost surely. If it is positive, the translation-invariant existence event has positive probability and hence probability one. Together with uniqueness this distinguishes absence from the case of a single infinite percolation cluster, without making an unsupported assertion that every graph vertex belongs to it.

Independence or another ergodicity hypothesis is necessary. Read literally for an arbitrary translation-invariant site percolation measure, the assertion is false: the mixture assigning probability $1/2$ to the all-open configuration and $1/2$ to the all-closed configuration is translation-invariant, but the invariant event of an infinite percolation cluster has probability $1/2$. We prove the zero–one law for the invariant independent law used in part (ii); the proof also works for the stated finite-range independent alternative.

Let $\Gamma$ be the group of translations, or the specified group of graph automorphisms, and let $A$ be invariant under it, up to null sets. An event determined by the states in a finite vertex set is a cylinder set. For every $\epsilon >0$, approximation in the product sigma-algebra provides a cylinder set $B$, supported on a finite set $F$, with $\mathbb{P}(A△B)<\epsilon$. Choose $g\in \Gamma$ so that $F$ and $gF$ are disjoint. Then $B$ and $gB$ are independent events, have equal probabilities, and $\mathbb{P}(A△gB)<\epsilon$. HenceandLetting $\epsilon \downarrow 0$ proves the zero–one law $\boxed{\mathbb{P}(A)\in \{0,1\}}$. This is translation ergodicity of Bernoulli percolation.

For a quasi-transitive graph, the required displacement exists: put $F$ in a ball about a vertex $v$; its orbit is infinite, so an automorphism can move $v$, and therefore that whole ball, beyond any prescribed finite neighborhood. A $k$-independent site percolation law means that the sigma-algebras generated by vertex sets at graph distance greater than $k$ are independent. Move $F$ far enough that $\mathrm{dist}(F,gF)>k$ and the same proof applies. Thus finite-range independence does give the zero–one law, although it will not give uniqueness.

Assume $\theta (1/2)>0$. The event that some infinite percolation cluster exists is invariant under lattice translations. By translation ergodicity of Bernoulli percolation, its positive probability is therefore one.

Let $D_N$ be the event that an infinite percolation cluster meets the finite graph vertex box $B_N=[-N,N{]}^2$. These increasing events increase with $N$, and their union is the event that some infinite percolation cluster exists. Continuity from below of a measure givesAn infinite percolation cluster meeting $B_N$ contains an infinite simple open ray in a graph from a graph vertex of the box. This follows from local finiteness and the König infinity lemma. That ray visits the finite box only finitely many times; its last box graph vertex lies on the boundary, and the subsequent ray is outside. Thus $D_N$ is also the event that the box boundary intersects an infinite percolation cluster, and is the union of the four exterior-side arm events.

Consequently, for every $ϵ>0$, choose $N$ withFor $ϵ\ge 1$ any sufficiently large $N$ still satisfies the strict inequality. No uniqueness assertion is needed for this preliminary step.

The Burton-Keane theorem says that for independent bond percolation on the hypercubic lattice ${\mathbb{Z}}^d$, $d\ge 2$, at every fixed $p\in [0,1]$, there is almost surely at most one infinite percolation cluster. It does not assert that such a cluster exists, nor does it assert absence of percolation at the critical parameter. The cases $p=0$ and $p=1$ are immediate; assume $0<p<1$ for the proof.

Let $N_{\infty }$ be the number of infinite percolation clusters. This random variable is translation invariant. By translation ergodicity of Bernoulli percolation, it is almost surely a constant $k\in \{0,1,2,\dots ,\infty \}$. One way to see ergodicity is to approximate an invariant event by events on finite sets of edges, translate one approximation far enough to make the two approximations independent, and then let the approximation error vanish. Its probability equals its square, hence is zero or one.

The other probabilistic input is the finite-energy property of Bernoulli percolation. Conditionally on all edges outside a finite set, every prescribed inside configuration has positive probability, namely a product of powers of $p$ and $1-p$. Thus an event of positive probability can be followed by a specified finite modification and still gives a positive-probability image event. Outside infinite arms are unaffected.

First exclude a finite $k\ge 2$. Some finite box meets at least two infinite percolation clusters with positive probability. Open all edges in and incident to that box. These clusters merge. Opening finitely many edges cannot create an infinite percolation cluster solely out of finite clusters, since it joins only finitely many such clusters. Hence the number of infinite percolation clusters decreases with positive probability, contradicting its almost-sure constant value. This rules out every finite multiplicity greater than one.

Now suppose $k=\infty$. With positive probability a finite box meets at least three distinct infinite percolation clusters. Retain an exterior infinite ray from each of three of them. Replace the finitely many interior and boundary edges by a minimal tree joining their three attachment graph vertices, and close all unused edges there. The exterior rays belong to distinct original clusters, so have no outside connections to each other. The resulting joined cluster has a branch graph vertex $v$ with exactly three incident open edges, and deleting $v$ leaves three infinite components. If a terminal lies in the middle of the interior connecting path, its exterior ray supplies the third branch there. Such a graph vertex is a trifurcation vertex in percolation. A finite box has only finitely many candidate locations and modifications; therefore some fixed graph vertex has positive trifurcation probability. By translation invariance, that probability is a common number $\tau >0$ at every graph vertex.

The decisive deterministic input is the trifurcation boundary-counting lemma. Let $T$ be all these degree-three trifurcation vertices in percolation in a finite box $B_n$. In each infinite percolation cluster meeting $T$, delete the graph vertices of $T$, contract each remaining connected component of a graph adjacent to $T$ to a node, and retain the incidences with the deleted graph vertices, including direct edges between graph vertices of $T$. This is a finite tree: a cycle would connect two neighbours of an original trifurcation vertex in percolation without passing through it, contradicting its three-way separation. Every deleted graph vertex has degree three. Every leaf component is infinite, since a finite leaf would give a finite branch after removing its adjacent trifurcation vertex in percolation. The elementary tree degree identity therefore supplies at least $|T\cap C|+2$ infinite leaf components for a cluster containing $T$.

Each such infinite component reaches the outer graph vertex boundary ${\partial }^{+}{B}_n$, and different components reach different boundary graph vertices. Summing over the clusters yields, in particular,For a nearest-neighbour lattice box,Taking expectations and using the common trifurcation probability givesThis contradicts $\tau >0$ and excludes infinitely many infinite percolation clusters. Thus only $k=0$ or $k=1$ remains:The proof explains the role of the amenable boundary growth of lattice boxes: finite-energy property of Bernoulli percolation would make branching points have positive volume density, while their disjoint infinite branches can only be supplied through a boundary of lower order. It uses no planar duality and works in every indicated dimension.

Use the nearest-neighbour lattice graph with graph vertex set ${\mathbb{Z}}^d$: an unordered pair $\{x,y\}$ is an edge precisely when $\Vert x-y{\Vert }_1=1$. A bond configuration is $\omega \in \{0,1{\}}^{E_d}$; under the product measure ${\mathbb{P}}_p$, its edge coordinates are independent Bernoulli distribution with success probability $p$. Edges with coordinate one are open. The open cluster $C(0)$ is the connected component of a graph reachable from the origin through open edges.

Define the percolation probability and percolation critical probability byEquivalently, $p_c(d)=\sup \{p:{\theta }_d(p)=0\}$. The monotone coupling of Bernoulli percolation proves the equivalence: assign independent uniform distribution $U_e$ to the edges and open $e$ at parameter $p$ when $U_e\le p$. Connection events and ${\theta }_d$ then increase with $p$. For $d\ge 1$, ${\theta }_d(0)=0$ and ${\theta }_d(1)=1$, so the defining set is nonempty. The definition does not settle what happens at $p=p_c(d)$.

By translation invariance all roots have the same probability. Since the graph vertex set is countable, ${\theta }_d(p)=0$ implies that no graph vertex lies in an infinite percolation cluster almost surely. Conversely a positive root probability gives positive probability of an infinite percolation cluster, and translation ergodicity of Bernoulli percolation then makes existence an almost-sure event. In question 2(b), $\theta$ means ${\theta }_2$.

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.

Put $I=\{x:|C(x)|=\infty \}$, the union of all infinite percolation clusters. Since $p>p_c$, the percolation probability $\theta (p)$ is positive. The translation ergodicity of Bernoulli percolation makes $\{I\ne \varnothing \}$ a zero-one event; it has positive probability because ${\mathbb{P}}_p(0\in I)=\theta (p)>0$, hence it occurs almost surely.

Let $D_m=\{B(m)\cap I\ne \varnothing \}$. These events increase to $\{I\ne \varnothing \}$, so continuity from below of a measure gives ${\mathbb{P}}_p(D_m)\to 1$. On $D_m$, some vertex of $B(m)$ has an infinite open path in a graph. For every $N\ge m$, the segment up to its first visit to $\partial B(N)$ lies entirely in $B(N)$. Thus even with this restriction on the connecting path in a graph,This proves the requested uniformity and supplies the version needed for crossings later.