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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.

If $b_n$ counts the $n$-step self-avoiding walks from a fixed vertex of a transitive lattice, the connective constant is the exponential growth rateSubmultiplicativity of $b_n$ and the Fekete lemma applied to $\log b_n$ prove existence of this limit.

This is the Fekete lemma, with its possible extended-real limit made explicit. Put $L={\inf }_{m\ge 1}{x}_m/m$. Since $x_n\le n{x}_1$, $L$ cannot be $+\infty$; it may be $-\infty$.

Fix $m\ge 1$ and write $n=qm+r$, with $0\le r<m$. Repeated subadditivity giveswhere for this calculation set $x_0=0$. The finitely many possible remainder terms are bounded. Since $q/n\to 1/m$,This holds for every $m$, whereas $x_n/n\ge L$ by the definition of an infimum. If $L$ is finite, the upper and lower bounds agree. If $L=-\infty$, choosing $m$ with arbitrarily negative $x_m/m$ shows that $x_n/n\to -\infty$. ThusA finite real limit is not guaranteed: $x_n=-n^2$ is a subadditive sequence and has $x_n/n=-n\to -\infty$. If the limit is intended to be real rather than extended real, one must add a linear lower bound.

Split a self-avoiding walk of length $m+n$ after its first $m$ steps. Its prefix is one of the $c_m$ walks, and its translated suffix is a self-avoiding walk of length $n$ from the origin. The pair determines the original walk, although not every pair can be concatenated without intersection. ThereforeAll $c_n$ are positive, so $x_n=\log c_n$ is a subadditive sequence. The Fekete lemma gives existence of ${\lim }_n{n}^{-1}\log c_n$ and consequently of the connective constant.

For a lower bound, all words of length $n$ using only north and east steps produce distinct self-avoiding walks: the sum of the two coordinates increases at each step. Thus $c_n\ge 2^n$. For an upper bound, there are four choices for the first step and at most three thereafter, since a self-avoiding walk cannot immediately reverse its previous edge. Hence $c_n\le 43^{n-1}$. Taking $n$th roots provesThe elementary upper bound counts all non-backtracking walks, including some that self-intersect, so it is an upper estimate rather than an equality.

A self-avoiding walk is a finite sequence of graph vertices $v_0,\dots ,v_n$ in which successive vertices are adjacent and no graph vertex is repeated; its length is $n$, the number of edges. Include the zero-length walk.

Split a length-$m+n$ walk from $v$ into its first $m$ edges and remaining $n$ edges. After fixing the first segment, the number of possible suffixes is at most ${\sigma }_n$, since forgetting the prohibition on revisiting its earlier vertices can only increase that number. Thus ${\sigma }_{m+n}(v)\le {\sigma }_m(v){\sigma }_n$, and taking the supremum givesEvery finite-radius ball is finite because degrees are bounded. An infinite connected graph has vertices arbitrarily far from each root, so shortest paths give ${\sigma }_n(v)\ge 1$. Also ${\sigma }_n\le \Delta (\Delta -1{)}^{n-1}$ for $n\ge 1$: after the first edge, an immediate reversal is forbidden. In particular $\Delta \ge 2$ and these suprema are finite.

Apply the Fekete lemma: a real subadditive sequence $a_{m+n}\le a_m+a_n$ satisfies $\lim a_n/n={\inf }_{n\ge 1}{a}_n/n$, allowing negative infinity. Here $a_n=\log {\sigma }_n\ge 0$, so the limit is finite. The uniform connective constant of a bounded-degree graph is therefore

Give each nearest-neighbor edge of the cubic lattice an independent Bernoulli distribution state, open with probability $p$. The resulting product measure is denoted ${\mathbb{P}}_p$. In this bond percolation model, $C(0)$ is the percolation cluster of the origin, andThe uniform-label monotone coupling of Bernoulli percolation shows that ${\theta }_d$ is increasing.

Let $c_n(d)$ count $n$-step self-avoiding walks starting at the origin, with $c_0(d)=1$. Splitting a walk after $n$ steps, and discarding the avoidance constraint between the two pieces, gives $c_{n+m}(d)\le c_n(d)c_m(d)$. The Fekete lemma therefore gives the connective constantThe locally finite graph structure means that an infinite percolation cluster at the origin supplies an open self-avoiding walk of every length. Each specified walk has $n$ distinct edges and is open with probability $p^n$. The union bound givesFor $p\mu (d)<1$ the right side tends to zero. Hence the connective-constant lower bound for percolation is $p_c(d)\ge \mu (d{)}^{-1}$.

For the upper bound first work on the square lattice. A finite open percolation cluster has an outer boundary containing a simple closed graph cycle of dual edges, all crossing closed primal edges. Such a dual bond percolation circuit separates that cluster from infinity. Write $N_{\ell }$ for the number of simple dual circuits of length $\ell$ surrounding the origin. Each circuit meets the positive horizontal ray at distance at most $\ell$: its bounding box contains the origin and its diameter is bounded by its length. Choose such an intersection as an anchor and orient the circuit. Removing its last edge leaves a rooted self-avoiding walk of length $\ell -1$ in the translated square lattice. Consequently, for an absolute constant $K$,The exact constant and this possible overcount do not matter. If $(1-p)\mu (2)<1$, the root test givesA summable circuit count alone need not give a total sum below one. To use its tail correctly, let $B_R=[-R,R{]}^2\cap {\mathbb{Z}}^2$ and condition every edge internal to $B_R$ to be open. This finite event has positive probability. A closed dual circuit surrounding all of $B_R$ crosses no internal edge of $B_R$, and so its closed-edge probability remains $(1-p{)}^{\ell }$ under this conditioning. Its length tends to infinity with $R$. Choose $R$ so large that the union bound for all such circuits is below one. With positive conditional probability, none occurs.

On that event, $C(0)$ contains $B_R$ and cannot be finite: a finite cluster containing $B_R$ would have an enclosing closed dual circuit. Thus ${\theta }_2(p)>0$ whenever $p>1-\mu (2{)}^{-1}$. This is the connective-constant Peierls bound, proved by excluding short circuits through the open-box conditioning. An embedded coordinate plane in the cubic lattice has exactly the same bond percolation law as the square lattice, so $p_c(d)\le p_c(2)$. Together,There are $2d$ choices for the first step of a self-avoiding walk and at most $2d-1$ thereafter, because immediate reversal is forbidden. Hence $c_n(d)\le 2d(2d-1{)}^{n-1}$ and $\mu (d)\le 2d-1$. In particular $\mu (2)\le 3$. Substituting with the correct directions of the inequalities gives

Use integer graph vertices in the boxes: ${\Lambda }_n=[-n,n{]}^2\cap {\mathbb{Z}}^2$, and set ${\Lambda }_{-1}=\varnothing$. Thus $\partial {\Lambda }_0=\{0\}$, $g_0=1$, and $|\partial {\Lambda }_m|=8m$ for $m\ge 1$.

If $0\leftrightarrow \partial {\Lambda }_{m+n}$, erase loops from an open graph path to obtain a self-avoiding walk. Let $x$ be its first graph vertex on $\partial {\Lambda }_m$. Split at $x$. The prefix certifies $A_x=\{0\leftrightarrow x\}$, while a segment of the suffix certifies $B_x=\{x\leftrightarrow x+\partial {\Lambda }_n\}$: the final graph vertex is at supremum norm distance at least $n$ from $x$, so the suffix first hits that translated boundary. The two certificates use disjoint edges. Therefore the BK boundary-splitting estimate, the union bound, the van den Berg-Kesten inequality, and translation invariance giveThe cases $m=0$ or $n=0$ follow directly from $g_0=1$, so this proves the bound including the endpoints.

Here is an explicit way to remove the polynomial boundary factor. For $p>0$ put $b_n=32n^2{g}_n$, $n\ge 1$. By symmetry in $m,n$, assume $m\le n$. ThenConsequently $a_n=\log b_n$ is a subadditive sequence. The Fekete lemma states that any real subadditive sequence satisfies $\lim a_n/n={\inf }_{n\ge 1}{a}_n/n$, possibly $-\infty$. In this case the direct horizontal open graph path gives $g_n\ge p^n$, so this limit of a sequence is bounded below by $\log p$. Since $g_n\le 1$, its upper bound is zero. Finally,has the same limit of a sequence. Thus the percolation one-arm decay rate exists andFor $p=0$, every $g_n$ with $n\ge 1$ is zero and $\gamma =0$. No exponential-decay theorem or assumption that $p$ is below the percolation critical probability is needed.

Let $c_n$ count $n$-step self-avoiding walks on the square lattice from one fixed graph vertex. Splitting a self-avoiding walk after $m$ steps and dropping the avoidance constraint on its suffix gives $c_{m+n}\le c_m{c}_n$. The Fekete lemma therefore gives the connective constant $\mu =\lim c_n^{1/n}$. The planar dual graph of the square lattice is a translated copy of it, so its connective constant is also $\mu$.

Put $q=1-p$ and assume $q\mu <1$. Choose $a>\mu$ with $qa<1$. The definition of the connective constant provides a finite $K$ such that $c_j\le K{a}^j$ for every $j\ge 0$. A simple dual graph cycle of length $\ell$ surrounding $0$ has a graph vertex in a box of radius $\ell +1$: its diameter is at most $\ell$, and its coordinate ranges straddle the origin. Choose such a graph vertex as the starting point, orient the graph cycle, and omit its closing edge. The remaining self-avoiding walk has length $\ell -1$. Consequently the number $N_{\ell }$ of these graph cycles obeysfor fixed finite constants. A specified graph cycle is open in dual bond percolation, equivalently all its crossed primal edges are closed, with probability $q^{\ell }$. HenceTo make this tail argument valid for every $q\mu <1$, rather than only extremely small $q$, use a finite modification of Bernoulli percolation. Condition all primal edges within ${\Lambda }_N$ to be open. This event has positive probability for $p>0$. If the resulting percolation cluster of $0$ is finite, its outer boundary contains a simple closed dual graph cycle surrounding the whole box. Such a graph cycle has length tending to infinity with $N$ and crosses no forced-open edge. Under the conditioning its remaining edge states retain the original independent law, so its probability is still $q^{\ell }$. Choose $N$ so that the above tail is less than $1/2$. The conditional probability that $0$ belongs to an infinite percolation cluster is then at least $1/2$, and thus $\theta (p)>0$.

This connective-constant Peierls bound proves that every $p>1-{\mu }^{-1}$ is above or at the onset of positive percolation probability. Taking the infimum givesThe case $p=1$ is immediate. This proof uses the Peierls argument, without assuming the exact Harris-Kesten theorem.

On a probability measure-preserving system, for a finite measurable partition $\xi =\{A_1,\dots ,A_r\}$, the entropy of a finite measurable partition isUse natural logarithms, so information entropy is measured in nats; another fixed logarithm base rescales all answers. The join of measurable partitions is their common refinement, and write ${\xi }_a^b={\bigvee }_{j=a}^b{T}^{-j}\xi$, with an empty join the trivial partition. The entropy rate of a measurable partition and Kolmogorov-Sinai entropy are respectivelyThe block entropies form a subadditive sequence, so the first limit exists by the Fekete lemma. For finite partitions the conditional entropy of finite measurable partitions is $H(\eta \mid \zeta )=H(\eta \vee \zeta )-H(\zeta )$.

Put $c_0=H(\xi )$ and $c_k=H(\xi \mid {\xi }_1^k)$ for $k\ge 1$. The chain rule for information entropy, applied from the last coordinate backwards, and measure preservation giveThe second equality uses invariance of the joint partition atom probabilities under the common pullback $T^{-j}$; invertibility is unnecessary. Since conditioning reduces entropy, $c_k$ decreases to a nonnegative limit $c$. The Cesaro convergence of a sequence of this convergent sequence has the same limit. ThereforeEquivalently $h_{\mu }(T,\xi )=H_{\mu }(\xi \mid {\mathcal{F}}_1)$, where ${\mathcal{F}}_1=\sigma ({\bigvee }_{j\ge 1}{T}^{-j}\xi )$. Here conditional entropy of a countable measurable partition conditioned on a sigma-algebra is computed using conditional partition atom probabilities; the Martingale convergence theorem gives continuity under increasing conditioning sigma-algebras. This is the infinite-future formula for partition entropy rate.

The Kolmogorov-Sinai generator theorem states that if a finite or countable measurable partition $\eta$ has finite entropy of a countable measurable partition and its iterates generate the whole completed sigma-algebra modulo null sets, then $h_{\mu }(T)=h_{\mu }(T,\eta )$. For an invertible system, generating means $\sigma ({\bigvee }_{j\in \mathbb{Z}}{T}^{-j}\eta )=\mathcal{B}$ modulo null sets. For a noninvertible system a one-sided generator, using $j\ge 0$, suffices. The two-sided and one-sided versions must not be confused.

For a Bernoulli shift with discrete symbol probabilities $(p_i)$, the coordinate-zero measurable partition has independent coordinate iterates. ThusFor a finite alphabet the coordinate partition is a generator of finite entropy of a finite measurable partition; on the two-sided sequence space use all integer coordinate iterates, and on the one-sided space use the nonnegative ones. The Kolmogorov-Sinai generator theorem proves the displayed answer in both cases. The same calculation applies to countably many symbols when their Shannon entropy is finite, using the countable finite-entropy version of the theorem. If the Shannon entropy is infinite, merge all but the first $r$ symbols into one cell. These finite coordinate partitions have entropy rate $-{\sum }_{i\le r}{p}_i\log p_i-p_{>r}\log p_{>r}\to \infty$, so the system entropy is infinite. Zero-probability symbols contribute zero. In particular a fair $r$-symbol shift has entropy $\log r$.

For the final assertion, let ${\mathcal{F}}_0=\sigma ({\bigvee }_{j\ge 0}{T}^{-j}\xi )$ and complete it modulo null sets. The approximation property forces ${\mathcal{F}}_0=\mathcal{B}$ modulo null sets. Indeed for each $A\in \mathcal{B}$, choose approximating sets from finite blocks with error tending to zero. Their indicators approach $1_A$ in $L^2$, and $L^2({\mathcal{F}}_0)$ is a closed vector subspace, so $A$ is ${\mathcal{F}}_0$-measurable modulo a null set.

Invertibility now gives ${\mathcal{F}}_1=T^{-1}{\mathcal{F}}_0=T^{-1}\mathcal{B}=\mathcal{B}$ modulo null sets. In particular the present partition $\xi$ is measurable with respect to its entire future, so $H(\xi \mid {\mathcal{F}}_1)=0$. The infinite-future formula gives $h_{\mu }(T,\xi )=0$. Since the given one-sided generator is also a two-sided generator, the Kolmogorov-Sinai generator theorem finishes the proof:This is finite one-sided generator of an invertible system forces zero entropy. Invertibility is essential: a fair binary one-sided Bernoulli shift has a finite one-sided generator and Kolmogorov-Sinai entropy $\log 2$.

Use the fact that infinite cyclic subgroups of hyperbolic groups are undistorted: if $z$ has infinite order in a hyperbolic group, then its stable word lengthis strictly positive. The limit exists by subadditivity and the Fekete lemma. It is invariant under conjugation, because conjugating $z^m$ changes its length by at most twice the conjugator's length. It also satisfies $\tau (z^k)=|k|\tau (z)$.

In the given Baumslag-Solitar group, $b$ has infinite order. To verify this without presuming the presentation's normal form, let $a$ act on the real line by $t\mapsto 3t$ and $b$ by $t\mapsto t+1$. These affine bijections satisfy $ab{a}^{-1}=b^3$, and every nonzero power of $b$ is a nonidentity translation. Thus the element $b$ in the presented group cannot have finite order.

If an injective group homomorphism to $G$ existed, its images $z$ of $b$ and $u$ of $a$ would satisfy $uz{u}^{-1}=z^3$, with $z$ still of infinite order. Consequentlyforcing $\tau (z)=0$, a contradiction. ThereforeEquivalently, the relation gives $a^{k}b{a}^{-k}=b^{3^k}$, so powers of $z$ with exponent $3^k$ would have length at most $2k|u|+|z|$. This contradicts the linear lower bound for an undistorted infinite cyclic subgroup. This alternative also makes the exponential compression obstruction explicit.

Let $\alpha ={\inf }_{k\ge 1}{x}_k/k$. The relevant version of the Fekete lemma is the extended-real Fekete lemma: $\alpha$ may be $-\infty$, but cannot be $+\infty$ because $x_1$ is real. We show that this is exactly the limit of a sequence.

Fix $k\ge 1$ and, for $n\ge k$, write $n=qk+r$, where $q\ge 1$ and $0\le r<k$. Iterating the defining inequality gives $x_{qk}\le q{x}_k$. If $r>0$, it also gives $x_n\le q{x}_k+x_r$; if $r=0$, there is no remainder term. Thus, with $C_k=\max (0,x_1,\dots ,x_{k-1})$ and $C_1=0$,As $n\to \infty$, $q/n\to 1/k$, so ${\mathrm{limsup}}_n{x}_n/n\le x_k/k$. This holds for every positive $k$. On the other hand every $x_n/n\ge \alpha$. If $\alpha$ is finite, these two bounds give the desired equality. If $\alpha =-\infty$, for every real $M$ choose $k$ with $x_k/k<M-1$; the same upper bound makes $x_n/n<M$ for all sufficiently large $n$. ThereforeThe index $k=0$ is excluded because its ratio is undefined, and $x_0$ is irrelevant since the defining inequality was required only at positive indices. If the printed word “limit of a sequence” is interpreted as a finite real limit of a sequence, an additional lower bound is necessary: $x_n=-n^2$ defines a subadditive sequence but $x_n/n=-n\to -\infty$. The extended-real statement is the correct unrestricted conclusion.

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.

PutFix $k\ge 1$ and write $n=qk+r$ with $0\le r<k$. Repeated subadditivity givessoThe finitely many values $x_0,\dots ,x_{k-1}$ are bounded, henceTaking the infimum over $k$ gives $\mathrm{limsup}{x}_n/n\le \alpha$, while the definition of $\alpha$ gives $x_n/n\ge \alpha$ for every $n$. ThereforeThis is Fekete lemma.

LetThe events $\{0\leftrightarrow e_n\}$ and $\{e_n\leftrightarrow e_{n+m}\}$ are increasing events of bond percolation. The FKG inequality and translation invariance giveThus $(a_n)$ is a subadditive sequence. Since every probability is positive for $p>0$, Fekete lemma applies and gives

Splitting any $(m+n)$-step self-avoiding walk after $m$ steps and translating its remaining segment to the origin injects it into an ordered pair of an $m$-step and an $n$-step self-avoiding walk. Hence $b_{m+n}\le b_m{b}_n$. The sequence $a_n=\log b_n$ is subadditive, so the Fekete lemma givesExponentiating proves existence of the connective constant $\kappa ={\lim }_n{b}_n^{1/n}$.

Let $\alpha ={\inf }_{r\ge 1}{a}_r/r$. The inequality $a_n/n\ge \alpha$ gives ${\mathrm{liminf}}_n{a}_n/n\ge \alpha$. Fix $r$ and write $n=qr+s$, where $0\le s<r$. Repeated subadditivity gives $a_n\le q{a}_r+a_s$ when $s>0$, with the evident omission when $s=0$. Since the finitely many remainders $a_s$ are bounded,Taking the infimum over $r$ proves the Fekete lemma:

The arithmetic--geometric mean inequality gives $x_n\ge 1$ from $n=2$ onward. For $x>1$,so the tail decreases to a limit $L\ge 1$. Passing to the recurrence gives $2L=L+L^{-1}$, hence $L=1$.

For the subadditive sequence, $0\le x_n/n\le x_1$, so it is bounded. The Fekete lemma can be proved directly here as follows. Let $\alpha ={\inf }_k{x}_k/k$. Fix $k$ and write $n=qk+r$, $0\le r<k$. ThenThus $\mathrm{limsup}{x}_n/n\le \alpha$, while the definition gives $\mathrm{liminf}{x}_n/n\ge \alpha$. Hence $x_n/n\to \alpha$.

For bond percolation on the square lattice, the one-arm probability $g_n={\mathbb{P}}_p(0\leftrightarrow \partial [-n,n{]}^2)$ has a root limit of a sequence $\gamma (p)\in [p,1]$. The BK boundary-splitting estimate implies that $32n^2{g}_n$ is submultiplicative for positive integers. Applying the Fekete lemma to its logarithm proves existence when $p>0$; at $p=0$ the rate is zero. The same rate is the root limit of the percolation two-point connection probability along a coordinate axis, by the reflection lower bound for two-point percolation.