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If some choice $\omega$ produced a graph with largest component of a graph smaller than $2n/3$, the balanced component cut result with packing parameter one would give an empty graph cut whose sides both have at least $n/3$ vertices. For any fixed such vertex set partition, a uniform edge of the complete graph crosses with probabilityTo avoid crossing in row $i$, at least one of its $k$ offered edges must be noncrossing. The row's probability is $1-q^k\le 1-(4/9{)}^k$. The rows consist of independent random variables, so at $m=cn$ a union bound over at most $2^n$ vertex set partitions givesFor example, the positive integermakes this bound at most $e^{-n}$. Therefore every choice has a component of order at least $2n/3$, with probability tending to one. The simultaneous giant for fixed random-edge choice estimate includes choices made after seeing the entire array; it does not assume a particular online selection rule.

Assume $0<\lambda =1-\epsilon <1$, as required for this nontrivial subcritical binomial random graph. Put $\delta =\lambda -1-\log \lambda >0$. A breadth-first exploration of a binomial random graph is dominated by a Galton-Watson process whose offspring counts are independent random variables with law $\mathrm{Bin}(n,\lambda /n)$. If its total progeny $T$ is at least $j$, the first $j-1$ offspring counts sum to at least $j-1$. Their moment-generating function is bounded by that of a Poisson distribution of mean $\lambda (j-1)$. The exponential Markov inequality, optimized at $t=-\log \lambda$, yieldsTaking $j=\lceil (1+\eta )\log n/\delta \rceil$ for any fixed $\eta >0$, a union bound over all starting vertices gives $\mathbb{P}(L_1\ge j)=o(1)$.

For the matching lower bound, let $X_j$ count tree components of order $j$. The tree-component expectation in the Erdős-Rényi model is exactlyFor $j=O(\log n)$, the Stirling formula givesChoose $j=\lfloor (1-\eta )\log n/\delta \rfloor$, with $0<\eta <1$. Then $\mathbb{E}{X}_j\to \infty$. Distinct overlapping vertex sets cannot both be graph components. For disjoint sets, the joint probability differs from the product only because the $j^2$ between-set edges were counted twice as absent. ThusIt follows that $\mathrm{var}{X}_j/(\mathbb{E}{X}_j{)}^2=o(1)$. The second moment method gives $X_j>0$ with high probability. Combining both bounds, and then taking arbitrarily small fixed $\eta$, proves the sharp subcritical largest-component scaleFinally the Taylor expansion of $-\log (1-\epsilon )$ gives $\delta ={\epsilon }^2/2+{\epsilon }^3/3+\cdots$. The endpoint $\lambda =0$ has no edges and is excluded from this logarithmic asymptotic.

The persistence of unsampled graph components prevents the mass from each size band from disappearing entirely before $t_c$. More precisely, a band $[k,Dk)$ containing at least $\delta n/4$ vertices retains at least $\beta n$ vertices in that same band after at most $n/k$ steps, with high probability, for a fixed $\beta >0$ depending only on $\delta ,D$. To see why, each initial graph component in that band has a fixed positive probability that none of the offered edge endpoints touches it during the interval. Such a graph component remains unchanged regardless of the selection rule. The McDiarmid inequality concentrates the total mass of these untouched graph components. The reusable lemma supplies the full uniform-in-starting-time argument, including the version that samples absent edges.

Choose finitely many disjoint bands $[D^j,D^{j+1})$, $j=0,\dots ,J$, with $(J+1)\beta >1$. Part (i) provides at least $\delta n/4$ vertices in band $j$ at its time $t_{D^j}$. Each interval from that time to $t_c$ has length at most $n/D^j$, so persistence leaves at least $\beta n$ vertices in every one of these disjoint bands at the common time $t_c$. A union bound over this fixed finite number of bands makes all the conclusions simultaneous. Their total exceeds $n$, a contradiction. No fixed two-choice Achlioptas rule satisfies the explosive percolation hypothesis.

This continuity of fixed-choice percolation is the fixed-choice obstruction proved by Riordan and Warnke; their original research paper also treats a broader class of rules. The argument here does not apply when the number of offered choices grows with $n$.

Use the standard Achlioptas process with two independent uniform candidate edges per step and a rule choosing one. The component-band vertex count $N_{[k,Dk)}$ counts vertices, not the number of graph components. Integer time rounding changes the arguments below by $O(1)$ steps.

First, at $t_k$ the eventual graph component of order at least $\delta n$ is assembled from graph components already present at $t_k$. Contract those initial graph components. During $s=t_c-t_k$ steps, at most $s$ new edges are added, so the eventual connected contracted graph contains at most $s+1$ initial graph components. Initial graph components of order less than $k$ can therefore contribute at mostvertices, for sufficiently large $n$ and fixed $k$, since $s=\delta n/(2k)+O(1)$. For $k=1$ their contribution is zero. Consequently $N_{\ge k}(G_{t_k})\ge \delta n/2$.

We now use forced merging of large components. If at a time $m$ the graph components of order at least $K$ contain at least $an$ vertices, then after at most $A(a)n/K+O(1)$ steps there is a graph component of order at least $an/3$, except on an event of probability exponentially small in $n/K$. One explicit choice is $A(a)=81(\log 2+2)/a^4$. The proof is a union bound over set partitions of the at most $n/K$ initial large graph components: if all final graph components were small, a balanced component cut would split their initial vertices into sets of size at least $an/3$, and both candidate edges cross that fixed cut with probability at least $a^4/81$ in every step. The rule cannot avoid such a forced crossing. The failure bound is $e^{-2n/K}$, and a union bound makes the estimate simultaneous over all starting times at most $3n$, for each fixed $K$.

Set $a=\delta /4$ and choose a fixed integer $D\ge 2$ with $D\ge 4A(a)/\delta$. If $N_{\ge Dk}(G_{t_k})\ge \delta n/4$, this lemma would produce a graph component of order at least $\delta n/12$ by timeThat contradicts the assumed $L_1(G_{t_c^{\prime }})=o(n)$. Thus $N_{\ge Dk}(G_{t_k})<\delta n/4$, and subtraction givesThe same $D$ works for every fixed $k$; it depends only on $\delta$, with the number of offered edges fixed at two.

Use the phase-distance convention for a Bohr set:where the norm is distance to the nearest integer. This is the Bohr set in phase-distance convention; the alternative condition $|e(rx/N)-1|\le \rho$ has a different radius and corresponding constants. Assume $d\ge 1$, as $1/d$ requires.

For each residue $x$, form $t_x=(rx/N{)}_{r\in R}$ in the $d$-dimensional torus. Around these $N$ points place translates of a box of side length $ϵ$ in each coordinate. For $0<ϵ\le 1$, their volumes sum to $N{ϵ}^d$. If $ϵ>N^{-1/d}$, two boxes overlap, and their distinct indices differ by a nonzero $q$ with every phase distance less than $ϵ$. Thus the Bohr set contains a nonzero element at the stated threshold. For larger radii the entire group is already present.

Letting the box width decrease to $N^{-1/d}$ and using the finite number of possible nonzero $q$, obtain a nonzero $q$ with $\Vert rq/N\Vert \le N^{-1/d}$ for every $r\in R$. By the triangle inequality, $jq$ lies in $B(R,ϵ)$ whenever $0\le j<ϵN^{1/d}$. Since $N$ is prime, these multiples are distinct up to length $N$. Therefore, in the usual radius range $0<ϵ\le 1/2$, the arithmetic progressionhas the requested length. For literally unrestricted $ϵ>0$, the correct lower bound is $\min (N,\lceil ϵN^{1/d}\rceil )$: a progression of distinct residues cannot exceed $N$. For example, $d=1$, $ϵ=2$ makes the uncapped printed bound impossible. If $R$ is empty, the Bohr set is the whole group and has a progression of length $N$; the displayed exponent is simply inapplicable.

Now transfer the dense integer set to a larger cyclic group. Choose a prime $8N<p<16N$, using Bertrand's postulate, and regard $A$ as a subset of $G=\mathbb{Z}/p\mathbb{Z}$. Its density $\alpha =|A|/p$ is at least $1/1600$. Normalize Fourier coefficients on a finite abelian group and convolution byThe Fourier inversion and Parseval identity formulas are $f(x)={\sum }_r\hat{f}(r)e(rx/p)$ and ${\sum }_r|\hat{f}(r){|}^2=\mathbb{E}|f{|}^2$; convolution turns into multiplication of Fourier coefficients on a finite abelian group.

For $f=1_A$, set $R=\{r:|\hat{f}(r)|\ge {\alpha }^{3/2}/2\}$. Then Parseval identity boundsAlso, with $\tilde{f}(x)=f(-x)$, the nonnegative convolution $Q=f\ast f\ast \tilde{f}\ast \tilde{f}$ is supported exactly on the modular set $2A-2A$, andThe frequencies outside $R$ contribute in absolute value at most $({\alpha }^3/4){\sum }_r|\hat{f}(r){|}^2={\alpha }^4/4$. For $x\in B(R,1/6)$, the real part of each phase from $R$ is at least $1/2$; moreover $0\in R$ and $|\hat{f}(0){|}^4={\alpha }^4$. Hence $Q(x)\ge {\alpha }^4/4>0$. We have proved the cyclic Bogolyubov lemma with explicit phase radius:The previous Bohr set argument gives a modular arithmetic progression of length at least $p^{1/D}/6$. The remaining step is lifting a short modular progression to an integer progression. Every one of its elements has a unique integer lift belonging to $2A-2A\subseteq [-2N,2N]$. Successive lifted differences lie in $[-4N,4N]$ and are congruent to the same residue. Since $p>8N$, at most one integer in this difference interval represents that residue, so every successive difference is the same. The lifts form a genuine integer arithmetic progression, not merely a modular one.

For sufficiently large $N$, the constants can be absorbed into half the exponent. ThusNo attempt to optimize this absolute constant is needed.

A counterexample comes from dense integer sets with only logarithmic-length progressions. To show that $A$ itself need not have such a progression, choose a random subset $B\subseteq [1,N]$ by independent inclusion with probability $p_0=0.02$. Its size has mean $0.02N$ and variance at most $0.02N$, so Chebyshev's inequality shows $\mathbb{P}(|B|<0.01N)\le 200/N$. There are at most $N^2$ increasing arithmetic progressions of any fixed length $L\ge 2$, and each belongs to $B$ with probability $p_0^L$. Taking $L=\lceil 4\log N/\log (1/p_0)\rceil$ makes the union bound at most $N^{-2}$. For large $N$, some realization has at least $0.01N$ elements and no length-$L$ arithmetic progression. Take a subset of exactly $0.01N$ elements. It retains that avoidance. Such a set has no progression longer than $O(\log N)$, and therefore no progression of length $N^c$ for any fixed $c>0$, eventually. Interpret the prescribed exact cardinality on values of $N$ where it is integral, or use its integer part.

A connection to a box boundary can be witnessed by a finite self-avoiding open path stopped at its first boundary hit. On the event of reaching radius $m+n$, split such a path at its first graph vertex $z\in \partial {\Lambda }_n$. Its prefix witnesses connection from zero to $z$ within ${\Lambda }_n$. Its remaining segment reaches a point $y$ with $\Vert y{\Vert }_{\infty }=m+n$, so $\Vert y-z{\Vert }_{\infty }\ge m$; truncate it at its first hit of $z+\partial {\Lambda }_m$. The two witnessing edge sets are disjoint, giving disjoint occurrence of increasing events.

Use the BK inequality: for increasing finite-coordinate events in a Bernoulli product measure, $P(A\square B)\le P(A)P(B)$, where the square means disjoint open-edge witnesses. For each fixed $z$, the first event has probability at most ${\beta }_n$, and the second has probability exactly ${\beta }_m$ by translation invariance. A union bound over $z$ givesThe two boxes can overlap, so replacing BK by an independence claim would be incorrect. Restricting each event to paths stopped at its local boundary makes all the relevant coordinate sets finite, as required by the stated inequality.

Put $x_n=\log {\beta }_n$ and ${\alpha }_n=\log |\partial {\Lambda }_n|$. Here $|\partial {\Lambda }_n|=(2n+1{)}^d-(2n-1{)}^d$, so ${\alpha }_n=O(\log n)=o(n)$. A fixed straight open path of length $n$ gives $p^n\le {\beta }_n\le 1$. Part (a) therefore gives the almost-subadditive percolation decay rateIn particular the limit is finite for the prescribed $0<p<1$.

For every site, $P(R(x)\ge r)={\beta }_r$. This event depends only on edges with both endpoints in $x+{\Lambda }_r$: any connecting path can be stopped on its first hit of that boundary. Since $\lambda >0$, ${\beta }_r\to 0$, so all clusters have finite radius almost surely, by a countable union over sites. Set the radius of an isolated site to zero.

Write $a_n=(d/\lambda )\log n$. It suffices to take $0<ϵ<1$, since larger error intervals contain one such interval. For any small $\delta >0$, the decay-rate limit gives, for all large $r$,For the upper tail use $r_{+}=\lceil (1+ϵ)a_n\rceil$ and a union bound:provided $\delta <\lambda ϵ/(1+ϵ)$.

For the lower tail let $r_{-}=\lfloor (1-ϵ)a_n\rfloor +1$. Choose sites in ${\Lambda }_{n-r_{-}}$ spaced by $2r_{-}+1$ in each coordinate. Their radius-$r_{-}$ boxes are vertex-disjoint, so their local connection events are independent. There are $N_n\ge c(n/r_{-}{)}^d$ such sites for large $n$. If $M_n\le (1-ϵ)a_n$, none of these events occurs. Thuswhereif $\delta <\lambda ϵ/(1-ϵ)$. Choose one $\delta$ satisfying both restrictions. The integer choices give the required strict inequalities. Thus the maximum cluster radius under exponential one-arm decay has convergence in probability:The logarithmic box-packing loss does not change the leading constant $d/\lambda$.

Use the following normalization for the Lasso estimator, leaving the intercept unpenalized:A different convention for the factor in front of squared error rescales the regularization parameter. With this convention, centering of the columns gives $\hat{a}=\overline{Y}$. Put $Y_c=Y-\overline{Y}{1}_n$ and ${ϵ}_c=ϵ-\overline{ϵ}{1}_n$. Then $Y_c=X\beta +{ϵ}_c$ and $X^T{ϵ}_c=X^Tϵ$. The Lasso minimizes $\Vert Y_c-Xb{\Vert }_2^2/(2n)+\lambda \Vert b{\Vert }_1$. A minimum exists since $\lambda >0$ makes this continuous objective coercive; the bound below holds for every minimizer.

We first prove the needed sharp two-sided Gaussian tail bound. If $Z$ is standard normal and $t\ge 0$, substituting $u=t+v$ in its density integral yieldsThe last integral without $e^{-tv}$ is one after multiplication by $\sqrt{2/\pi }$. In particular this proves the two-sided bound with no extra factor of two.

Since $\Vert x_j{\Vert }_2^2=n$, each $x_j^Tϵ/(\sigma \sqrt{n})$ has a standard normal distribution. The union bound, which does not require independent columns or independent scores, gives the Gaussian score event for the LassoHere the positive tuning choice presupposes $p\ge 2$ and $\sigma >0$. When the displayed lower bound is negative it is simply a valid, vacuous lower bound. For $p=1$ the prescribed parameter is zero and the probability lower bound is also zero, so no positive-parameter assertion is supplied by that prescription.

Let $\delta ={\hat{\beta }}_{\lambda }^L-\beta$ and $Q=\Vert X\delta {\Vert }_2^2/n$. Comparing the Lasso objective at its minimizer and at $\beta$ and expanding the squares gives the Basic inequality for the LassoOn $\mathcal{E}$, the stochastic term is at most $\lambda \Vert \delta {\Vert }_1$. Because ${\beta }_N=0$, the triangle inequality givesIt follows thatIn particular $\Vert {\delta }_N{\Vert }_1\le 3\Vert {\delta }_S{\Vert }_1$, which is the Lasso cone condition. The given Compatibility condition for the Lasso is therefore applicable and gives $\Vert {\delta }_S{\Vert }_1\le \sqrt{sQ}/{\varphi }_0$.

Adding $\lambda \Vert {\delta }_S{\Vert }_1$ to the preceding inequality and using compatibility now yieldsThe last inequality is just the nonnegativity of $(\sqrt{Q/2}-\sqrt{8}\lambda \sqrt{s}/{\varphi }_0{)}^2$. Thus $Q+2\lambda \Vert \delta {\Vert }_1\le 16{\lambda }^{2}s/{\varphi }_0^2$, and dropping one nonnegative coefficient-error term givesThis prediction and coefficient error bound for compatible Lasso holds on $\mathcal{E}$, and hence with the required probability. If $s=0$, the earlier basic inequality directly forces $\delta =0$ on $\mathcal{E}$, so the zero right side is also correct.

Let $I_0$ be the indices of the true null hypotheses, let $m_0=|I_0|$, and let $V$ count false rejections. The familywise error rate isThe Bonferroni correction rejects a null exactly when its p-value is at most $\alpha /m$. The union bound givesThis controls the familywise error rate for every configuration of true and false nulls, with arbitrary dependence between their p-values. Marginal super-uniformity, $\mathbb{P}(P_i\le t)\le t$, would also suffice.

For the Holm step-down procedure, inspect the ordered p-values in ascending order and stop at the first failed comparison. If the first comparison fails, reject none; this is the convention $k=0$ when the defining set is empty. Ties can be ordered by any fixed rule.

If $m_0=0$ there can be no false rejection. Otherwise let $j$ be the rank of the first true null. Since at most $m-m_0$ false nulls precede it, $j\le m-m_0+1$ and thus $m-j+1\ge m_0$. If any true null is rejected, the step-down rule must have rejected this first true null, which requiresUsing the union bound on the true-null p-values provesThis first true null argument for Holm control also requires no independence. Holm controls the familywise error rate at level $\alpha$ under arbitrary dependence.

Now let $R$ count all rejections. The false discovery rate is the expected proportion of rejections that are false, with zero assigned when nothing is rejected:It differs from the familywise error rate: several false rejections can still represent a small proportion of a large collection of discoveries.

The BH procedure is a step-up rule. SetIf $K=0$, reject nothing; otherwise reject all hypotheses with $P_i\le \alpha K/m$. Exactly $K$ hypotheses are rejected: if more than $K$ p-values were below that threshold, the next ordered value would also satisfy its own larger threshold, contradicting maximality. Thus $R=K$. Unlike the step-down rule, a failed early comparison does not make this procedure stop.

For the proof, assume that each true-null p-value is uniform on $[0,1]$ and independent of the entire vector of the other p-values. Joint independence of all p-values is a sufficient condition; the false-null marginal distributions can be arbitrary. Mere uniformity of the true-null marginals without a dependence condition is insufficient for this argument or for general unmodified BH control.

Fix a true null $i$. Replace its p-value by zero and let $R^{(i)}$ be the number of rejections made by the Benjamini-Hochberg procedure on the modified vector. This variable depends only on the other p-values, and $R^{(i)}\ge 1$. The Benjamini-Hochberg leave-one-out identity isTo prove it, suppose $i$ is rejected with $R=r$. Decreasing its p-value to zero leaves every ordered value above rank $r$ unchanged, since it was already among the first $r$ values. Those higher ranks still fail their thresholds, while rank $r$ still succeeds; hence $R^{(i)}=r$. Conversely, suppose $R^{(i)}=r$ and $P_i\le \alpha r/m$. Restoring $P_i$ still leaves at least $r$ values at most $\alpha r/m$, so $R\ge r$. Increasing one p-value cannot increase the maximal successful rank, so $R\le R^{(i)}=r$. This gives equality and rejection of $i$.

Independence and uniformity now giveSumming over the $m_0$ true nulls proves the exact false discovery rate under independent null p-values:If the independent true-null p-values are only super-uniform, the same calculation gives the inequality instead of equality. The distinction between the two types of error control and the dependence conditions is essential: Bonferroni and Holm have the preceding guarantees without independence, while the BH proof here explicitly uses it.

A probability space consists of a sample space $\Omega$, a sigma-algebra $\mathcal{F}$ of events, and a probability measure $\mathbb{P}$ satisfying:

For arbitrary events, Boole's inequality is $\mathbb{P}({\bigcup }_j{A}_j)\le {\sum }_j\mathbb{P}(A_j)$. To prove it, replace each $A_j$ by $D_j=A_j\setminus {\bigcup }_{i<j}{A}_i$. The $D_j$ are disjoint, have the same union, and satisfy $D_j\subseteq A_j$. Countable additivity and monotonicity give $\mathbb{P}({\bigcup }_j{A}_j)={\sum }_j\mathbb{P}(D_j)\le {\sum }_j\mathbb{P}(A_j)$. Finite unions are included by taking subsequent events empty.

Let $H_i$ be the $i$th heads event. For every $m$, the event of infinitely many heads lies inside ${\bigcup }_{i\ge m}{H}_i$. By the union bound, its probability is at most ${\sum }_{i\ge m}{p}_i$, which tends to zero. Thus the probability of infinitely many heads is $0$. This is the First Borel-Cantelli lemma, and requires no independence of the coin tosses.

For the dice experiment there are $4$ ordered outcomes giving a sum of $5$ and $6$ giving $7$, out of $36$. Each trial therefore has $p_5=1/9$, $p_7=1/6$ and irrelevant-outcome probability $13/18$. By independence between trials, summing over the number of initial irrelevant outcomes givesThe geometric series also shows that one of the two relevant sums eventually appears almost surely. Equivalently, first-step analysis gives $p=p_5+(13/18)p$.

Color each integer independently and uniformly with one of $k$ colors. For a translate $S+t$, let $E_t$ be the event that some color is missing. The union bound givesJoin two events in the dependency graph of events when their translates overlap. Nonneighbors depend on disjoint sets of independent colors, giving the required joint independence. Overlap occurs only if $u-t\in S-S$, so the maximum degree is at most $s(s-1)$.

With natural logarithms, $s\ge 6k\log k$ and $k\ge 20$ implyThe middle estimate uses that $s^2{e}^{-s/k}$ is decreasing for $s\ge 2k$, and the next uses $\log k\le \sqrt{k}$. Therefore the Lovász local lemma supplies a coloring avoiding every bad event in any specified finite family of translates.

To obtain one coloring for all translates, use the compactness extension of the Lovász local lemma. At level $N$, consider colorings of $[-N,N]\cap \mathbb{Z}$ satisfying every constraint whose translate is contained in that interval. There are finitely many constraints, and the preceding argument makes the level nonempty. Restrictions connect these colorings into a finitely branching tree. The König infinity lemma gives an infinite branch, hence a coloring of all integers. Every translate eventually lies in a level of this branch, so it contains every color.

Thus there exists a $k$-coloring of $\mathbb{Z}$ in which every translate of $S$ contains all $k$ colors. This is a polychromatic coloring of integer translates; the compactness step establishes existence of one simultaneous coloring.

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 the coordinatewise partial order: $\omega \le \eta$ means $\omega (e)\le \eta (e)$ for every $e\in E$. An increasing event $A$ is upward closed: $\omega \in A$ and $\omega \le \eta$ imply $\eta \in A$.

Under the independent Bernoulli distribution product measure ${\mathbb{P}}_p$ on these coordinates, the Harris-FKG inequality statesfor increasing events $A,B$. For $K\subseteq E$, let $[\omega {]}_K$ be the cylinder of configurations agreeing with $\omega$ on $K$. The disjoint occurrence of increasing events $A\square B$ consists of configurations for which there exist disjoint $K,L$ with $[\omega {]}_K\subseteq A$ and $[\omega {]}_L\subseteq B$. For increasing events, witnesses can be taken to prescribe only open coordinates. The BK inequality isPositive association rewards simultaneous occurrence; the disjoint-witness requirement in the BK inequality gives a bound in the opposite direction. The inequalities also apply to different independent Bernoulli parameters in different coordinates.

Regard the boxes as sets of lattice graph vertices. Write ${\tau }_p(x,y)={\mathbb{P}}_p(x\leftrightarrow y)$, and put $g_0=1$. On the event defining $g_n$, choose an open self-avoiding walk from the origin to its first visit to $\partial {\Lambda }_n$. Let $y$ be its first visit to $\partial {\Lambda }_m$. Its initial segment witnesses $0\leftrightarrow y$. Its remaining segment ends at a point $z$ withFor $m<n$, truncate this remaining segment on its first visit to $y+\partial {\Lambda }_{n-m}$. The two segments use disjoint edges, so they are disjoint witnesses. For $m=n$, the second event is the sure event with empty witness. Thus the union bound, the BK inequality and translation invariance giveAn unrestricted connection event can depend on infinitely many edges. Apply the finite-coordinate BK inequality first to connections confined to growing boxes and take increasing limits; each occurrence has a finite path witness. This justifies its use here. The estimate is the weighted BK boundary-splitting estimate.

By summing cluster indicators and using Tonelli theorem, the percolation susceptibility isIf $\chi (p)<\infty$, then $S_m\to 0$. At $p=0$ all $g_k$ vanish, so any positive exponential rate works. Otherwise choose $m\ge 1$ with $a=S_m\in (0,1)$. Iterating the weighted BK boundary-splitting estimate gives, for $k=jm+r$ with $0\le r<m$,For $k\ge m$, $\lfloor k/m\rfloor \ge k/(2m)$, so $g_k\le \exp [-(-\log a)k/(2m)]$. To include the finitely many smaller indices without a prefactor, note that $p<1$ under finite susceptibility andTherefore one may takeFor $k<m$ this follows from $\mu (p)k\le -\log g_1$, and for $k\ge m$ it follows from the iterated estimate. This proves exponential one-arm decay from finite susceptibility with exactly the requested unit prefactor.

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.