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The box topology on ${\prod }_{i\in I}{X}_i$ has a basis of a topology consisting of all Cartesian products ${\prod }_i{U}_i$ with every $U_i$ an open set in $X_i$; infinitely many coordinates may be restricted. It contains the product topology and agrees with it for finite $I$.^[ref]

For copies of the usual real line, the box topology is strictly finer when $I=\mathbb{N}$. The box $W={\prod }_{j\ge 1}(-1/2,1/2)$ is not an open set in the product topology: a basic cylinder set containing zero leaves some coordinate $j$ unrestricted, so it contains $e_j\notin W$. This is a construction for every finite coordinate set, not evidence from testing a finite prefix.

Cantor space is the binary sequence space with the product topology of discrete two-point factors. Here the Bernoulli-space synonym refers to this topology, without selecting a particular probability measure. The metric $d(x,y)={\sum }_{j\ge 1}{2}^{-j}|x_j-y_j|$ induces the topology. A diagonal subsequence argument gives compactness. Fixing a finite prefix gives a clopen cylinder set. The map $x\mapsto {\sum }_{j}2x_j{3}^{-j}$ is a homeomorphism to the usual Cantor set.

For a topological space $Y$ and topological spaces $X_i$, a function $f:Y\to {\prod }_i{X}_i$ is continuous into the product topology exactly when every coordinate function $f_i={\pi }_i\circ f$ is continuous.^[ref]

Proof outline: one direction uses function composition with the continuous projection maps. For the other, the preimage of a basic cylinder set isa finite set intersection of open sets in $Y$. This gives continuity on every basic test. It explains, for example, why $t\mapsto (t,t^2,t^3,\dots )$ from the real line to ${\mathbb{R}}^{\mathbb{N}}$ is continuous in the product topology: each coordinate is a polynomial function. The general coordinate criterion fails for the box topology, where infinitely many preimages can be required at once. For example, $g(t)=(t,t,\dots )$ has continuous coordinate functions, but the preimage of the box ${\prod }_{j\ge 1}(-1/j,1/j)$ is $\{0\}$, which is not an open set in the real line.

A continuous function on Cantor space is uniformly continuous. Replacing it by one constant on each length-$n$ prefix cylinder set changes it by at most its oscillation on sets of diameter $2^{-n}$. This tends uniformly to zero. Each approximant is a finite linear combination of continuous indicator functions, proving density in the supremum norm.

Every open set in Cantor space is a countable disjoint union of prefix cylinder sets. Select all finite words whose cylinder lies in the open set but whose immediate predecessor's cylinder does not. These words are prefix-free, so their cylinders are disjoint. Every point of the open set has a shortest qualifying prefix. The whole space is represented by the empty prefix.

Extend a finite coloring of the positive integers to a point $x$ of a two-sided full shift. A minimal subsystem of its forward orbit closure, together with the proximal-minimal existence theorem, supplies a minimal point $y$ proximal to $x$. Put $q=y(0)$. If all sums in $\{0\}\cup \mathrm{FS}(a_1,\dots ,a_r)$, the augmented finite-sums set, have color $q$ in $y$, their coordinate constraints define a cylinder set containing $y$. The joint return lemma for a proximal minimal pair chooses a new positive term so that all new sums have color $q$ in both $x$ and $y$. Mathematical induction gives an infinite monochromatic finite-sums set. The new term can exceed the sum of all previous terms, giving unique representations. The proximal-minimal existence theorem is a substantive input to this proof.

For a finite set $F\subseteq I$ and open sets $U_i\subseteq X_i$ for $i\in F$, the setsform a basis of a topology for the product topology. The empty set intersection is $X$. Each basic open set is a cylinder set; all coordinates outside $F$ are free. Arbitrary open sets are unions of these basic tests and need not themselves depend on one fixed finite collection of coordinates.^[ref]

Why finitely many? Any topology making the projection maps continuous must contain their preimages of open sets, and hence finite set intersections of those preimages. Taking arbitrary unions supplies exactly the product topology. Intersecting two basic cylinder sets still restricts only the finite union of their coordinate sets.

On two copies of the real line with their usual topologies, let $U=(1,2)$ and $V=(2,4)$. The coordinate projection maps giveThe first two open sets are unbounded strips; their set intersection is an open rectangle. The point $(1.4,3)$ passes both coordinate tests, whereas $(2,3)$ fails the first because the endpoint $2$ is excluded.

Every point of an open set in the Euclidean plane has such an open rectangle around it inside that open set. Conversely, every open rectangle contains a small Euclidean ball around each of its points. Thus the product topology here is the usual metric topology of the Euclidean plane. With finitely many factors all coordinates may be restricted; on an infinite Cartesian product a basic cylinder set must leave all but finitely many factors free.

We prove uniqueness in the sense of at most one Gibbs measure satisfying the printed conditional distributions. No shift-invariance assumption is needed in advance. Write a history as $\eta =({ϵ}_{s-1},{ϵ}_{s-2},\dots )$. The variation hypothesis makes $U$ continuous and bounded on the finite-alphabet sequence space; let $u_{-}\le U\le u_{+}$ and $V=u_{+}-u_{-}$. The conditional distribution of the first spin, obtained by marginalizing the prescribed two-spin law, must beIn particular $q(a\mid \eta )\ge \delta :=e^{-2V}/k>0$ for every symbol and history. Any two candidate measures have this same time-homogeneous one-step kernel, by conditional marginalization and the tower property.

If $\eta ,{\eta }^{\prime }$ agree in their most recent $r$ symbols, each two-spin energy in this formula changes by at mostBoth numerator and denominator therefore change by factors between $e^{-D{\lambda }^{-r}}$ and $e^{D{\lambda }^{-r}}$. Consequently the total variation distance of the two kernels is at most $e^{2D{\lambda }^{-r}}-1\le B{\lambda }^{-r}$, where $B=e^{2D}-1$. Uniform positivity also bounds it by $1-k\delta$. PutThese bounds are decreasing, strictly below one, and summable. We now give the summable-memory coupling argument rather than assuming a small one-step contraction coefficient.

Starting from any two histories at a remote time, use a maximal coupling of the next-spin laws. If the actual histories have agreed for $r$ consecutive symbols, their probability of disagreement is at most $b_r$. Use shared independent uniform random numbers to construct a dominating match-length process $R$: it increases from $r$ to $r+1$ with probability $1-b_r$ and resets to zero with probability $b_r$. Since $b_r$ decreases, the actual consecutive-match length can be kept at least $R$ at every step. When the dominating process does not reset, the actual spins can be coupled to agree.

A new run from zero has probabilityof never resetting. The product is positive because $b_r<1$ and $\sum b_r<\infty$: its tail logarithm converges absolutely, and its finite initial factors are positive. After each reset a fresh run has the same success probability, so the probability of $m$ failed runs is $(1-p_{\ast }{)}^m$. There are almost surely only finitely many resets. Every failed run has finite duration, and thus the time of the final reset is finite almost surely. Its distribution depends only on $(b_r)$, not on the two starting histories. It follows that the probability of any disagreement in a fixed future block tends uniformly to zero as the block's distance from the initial time tends to infinity.

Now take two candidate Gibbs measures, sample their respective histories before time $-N$, and couple their forward kernels as above. Their finite-block laws really are obtained by these kernels: successive conditional expectations use the common conditional distribution $q$. For any fixed coordinate block $[s,t]$, the coupling inequality for total variation bounds the distance between the two candidate block laws by the dominating probability of a reset at or after step $N+s$. That bound tends to zero as $N\to \infty$. Hence all probabilities of finite cylinder sets agree; cylinder sets generate the product Borel sigma-algebra, so the measures agree.

Thus there is at most one measure satisfying these conditional specifications. There is a separate compatibility issue: for an arbitrary unnormalized symbolic potential the finite-horizon conditional distributions printed in the question need not be consistent as the right endpoint changes. The argument proves the requested uniqueness whenever such a law exists; it does not silently infer existence from the variation estimate. A normalized one-sided kernel or the usual compatible Gibbs specification supplies the standard existence formulation. For a concrete incompatibility, take two symbols and $U(a,\eta )=\log W_{{\eta }_1,a}$ with $W=\left(\begin{array}{cc}1& 1\\ 1& 2\end{array}\right)$. This has zero variation once two symbols agree. From a history ending in symbol one, the first-symbol probability obtained from a two-spin block is $2/5$, but from a three-spin block it is $5/13$. Thus the printed finite-horizon specification can indeed have no measure, even though its uniqueness implication remains valid.

A qualification is necessary: an arbitrary symbolic Gibbs measure does not satisfy the claim. Consider the doubling map with its two binary branches and a Bernoulli symbolic Gibbs measure with probabilities $2/3$ and $1/3$. This is the Gibbs law for a locally constant symbolic potential, while every level-$n$ interval has length $2^{-n}$. Along the nonboundary alternating itinerary, for even $n$,No finite $g$ can approximate these $g_n$ with a decaying uniform error. This proves the reusable fact that an arbitrary Gibbs measure need not be absolutely continuous.

For the intended geometric version, take the Gibbs measure for the geometric potential of an expanding map $-\log |T^{\prime }|$, with the standard sufficiently smooth finite full-branch hypotheses. The allowed transfer-operator result supplies a strictly positive Lipschitz continuous invariant probability density function $h$, normalized by $\int h=1$, and its symbolic weights of cylinder sets are $\nu (C_w)={\int }_{{\Delta }_w}h(x)dx$. State the needed regularity explicitly: $h\ge h_{\ast }>0$ and $|h(x)-h(y)|\le L_h|x-y|$. Expansion gives $\mathrm{diam}{\Delta }_w\le {\lambda }^{-n}$.

Let $\pi (ϵ)$ denote the point coded by the infinite sequence, and define $g(ϵ)=\log h(\pi (ϵ))$. The expression defining $g_n$ is the logarithm of the average of $h$ over the cylinder. For $x=\pi (ϵ)\in {\Delta }_w$,Since the logarithm is Lipschitz continuous with constant $1/h_{\ast }$ on this positive range, the cylinder averages of an invariant density giveThis proves the intended estimate. Expansion alone and an unspecified Gibbs symbolic potential are insufficient; the geometric choice and the regularity producing a positive Lipschitz continuous probability density function are substantive hypotheses.

In bond percolation on the square lattice, every nearest-neighbour edge independently has state one, meaning open, with probability $p$, and state zero, meaning closed, with probability $1-p$. Thus the configuration law is the product measure ${\mathbb{P}}_p={⨂}_{e\in E}\mathrm{Bernoulli}(p)$. An open path in bond percolation uses only open edges. An increasing event $A$ satisfies $\omega \in A,{\omega }^{\prime }\ge \omega \Rightarrow {\omega }^{\prime }\in A$: opening further edges cannot destroy the event.

The Harris-FKG inequality for two increasing events isFor disjoint occurrence of increasing events, write $A\square B$ when disjoint sets of open edges separately witness $A$ and $B$, so each event is forced by its own witness regardless of the other edges. The Van den Berg-Kesten inequality statesFor events given by cylinder sets these are finite-coordinate assertions. Connection events have finite open-path witnesses; truncating to finite boxes and taking increasing limits gives the same disjoint-occurrence bound for them. Sharing a vertex between the two witness paths is permitted; sharing an edge is not.

For $1\le m<n$, take a simple open path from the origin to the radius-$n$ boundary and let $x$ be its first vertex on the radius-$m$ boundary. Its initial segment witnesses $0\leftrightarrow x$. If $z$ is its final vertex, then $\Vert z-x{\Vert }_{\infty }\ge n-m$. Consequently the remaining segment reaches $x+\partial {\Lambda }_{n-m}$, and its initial portion until that visit witnesses $x\leftrightarrow x+\partial {\Lambda }_{n-m}$. The two segments use disjoint edges. HenceThe union bound, Van den Berg-Kesten inequality and translation invariance of the product measure now give the weighted BK boundary-splitting estimateFor $m=n$ the same bound follows directly from the union bound, with $g_0=1$. With the conventional $\partial {\Lambda }_0=\{0\}$, $m=0$ gives equality.

Write $C(0)$ for the percolation cluster of the origin. By summing its membership indicators, the percolation susceptibility isIf it is finite, then $a_m\to 0$. For $0<p<1$, choose $m$ with $0<a_m=a<1$. Iterating the boundary-splitting estimate with this fixed $m$ givesFor $k\ge 2m$, $\lfloor k/m\rfloor \ge k/(2m)$, so $g_k\le \exp [-(-\log a)k/(2m)]$. To handle all smaller positive radii without an unspecified prefactor, observe that $g_k\le g_1=1-(1-p{)}^4<1$. ThereforeFor $1\le k<2m$, the second chosen rate gives $e^{-\gamma k}\ge g_1$, and for larger $k$ the first rate applies. At $k=0$ both sides equal one. The endpoint $p=0$ has $g_k=0$ for every $k\ge 1$, so any positive rate works; $p=1$ is excluded by finite susceptibility. This proves exponential one-arm decay from finite susceptibility with the required unit prefactor.

The weighted binary-sequence metric is finite because $0\le d(ϵ,\eta )\le {\sum }_{n\ge 1}{2}^{-n}=1$. It is symmetric and nonnegative. If $d(ϵ,\eta )=0$, every nonnegative summand is zero, so every coordinate agrees; conversely agreement makes the sum zero. For three sequences, apply $|{ϵ}_n-{\zeta }_n|\le |{ϵ}_n-{\eta }_n|+|{\eta }_n-{\zeta }_n|$ in each coordinate and sum. This proves the triangle inequality, so $d$ is a metric.

If two sequences differ in one of their first $n$ coordinates, their distance is at least $2^{-n}$. Thus the open ball $B_d(\eta ,2^{-n})$ is contained in $C_n(\eta )$. For every $\eta \in C_n(ϵ)$, the latter cylinder equals $C_n(\eta )$, so each cylinder set is open in the metric topology. Every set in the specified topology, being a union of cylinder sets, is therefore metric-open.

Conversely, agreement in the first $n$ coordinates givesGiven a metric-open set containing $ϵ$, choose a ball $B_d(ϵ,r)$ inside it, then choose $n$ with $2^{-n}<r$. The bound gives $C_n(ϵ)\subseteq B_d(ϵ,r)$, establishing the required prefix-neighbourhood condition. The two topologies coincide. The strict inequality $2^{-n}<r$ avoids incorrectly treating the closed tail bound as a strict ball bound.

For a sequence $ϵ$, let $C_n(ϵ)$ be its prefix cylinder set, consisting of sequences agreeing with $ϵ$ in the first $n$ coordinates. The specified collection consists exactly of unions of such cylinder sets.

The empty set meets the defining condition vacuously, and $X$ meets it because every cylinder set is contained in $X$. For a union ${\bigcup }_{\lambda }{Y}_{\lambda }$, a point lies in one $Y_{\lambda }$ and has a prefix cylinder set contained in that set, hence contained in the union. For a finite intersection, a point has a cylinder neighbourhood in every constituent. Taking the largest of the finitely many prefix lengths gives one cylinder set contained in all of them. Therefore the collection contains the empty and whole sets, is closed under arbitrary unions and finite intersections, and is a topology on $X$, the prefix topology of Cantor space.

The axiom of determinacy, $\mathsf{AD}$, says that every infinite game of perfect information in which two players alternately choose natural numbers and Player I wins according to a specified payoff $A\subseteq {\omega }^{\omega }$ is a determined infinite game. A strategy specifies a move at every finite position where its player moves; it is winning if every compatible infinite play is won by that player.

Finite games are determined by backward induction. For infinite games, open determinacy has a direct proof. Let $T$ consist of the finite positions whose extension cylinder sets lie in the open payoff $A$. Build the winning-position attractor in an infinite game by transfinite recursion: start with $T$, add an I-position if some successor has already entered, add a II-position if every successor has already entered, and take unions at limit stages. The process stabilizes because there are only countably many finite positions.

At a position in the attractor but outside $T$, I can choose a successor with strictly smaller entry rank, and every move of II has smaller entry rank. There is no infinite descending sequence of ordinals, so I forces entry into $T$ and wins. Outside the attractor, every I-move stays outside it, while II can choose a successor outside it; that strategy avoids $T$ forever. Since $A$ is open, a play belongs to $A$ exactly when some prefix belongs to $T$. Thus II wins from outside the attractor. This proves determination for open payoffs; swapping the players proves it for closed payoffs too. The stronger Borel determinacy theorem gives determination for every Borel set payoff, whereas $\mathsf{AD}$ asserts this for arbitrary payoffs.

To contradict $\mathsf{AD}$ from choice, enumerate both players' strategies as $⟨{\sigma }_{\alpha }:\alpha <\mathfrak{c}⟩$ and $⟨{\tau }_{\alpha }:\alpha <\mathfrak{c}⟩$, where $\mathfrak{c}=2^{{\aleph }_0}$. There are $\mathfrak{c}$ strategies because the set of finite positions is countable; there are also $\mathfrak{c}$ plays following any fixed strategy because the other player's moves are arbitrary. At stage $\alpha$ of the choice diagonalization of an undetermined game, choose a fresh play $x_{\alpha }$ following ${\sigma }_{\alpha }$ and then a different fresh play $y_{\alpha }$ following ${\tau }_{\alpha }$, avoiding all earlier reservations. This is possible because fewer than $\mathfrak{c}$ plays have been reserved at every $\alpha <\mathfrak{c}$; regularity of $\mathfrak{c}$ is not required.

Set $A=\{y_{\alpha }:\alpha <\mathfrak{c}\}$. Every ${\sigma }_{\alpha }$ loses on $x_{\alpha }\notin A$, and every ${\tau }_{\alpha }$ loses on $y_{\alpha }\in A$. Thus neither player has a winning strategy, and AD and AC are inconsistent together.

In an infinite game of perfect information, the players alternately choose elements of a nonempty set $A$, producing $x\in A^{\omega }$. Player I wins when $x$ belongs to the payoff set $B$. A strategy in an infinite game chooses a move at each position where its player moves, and a winning strategy in an infinite game wins against every possible opposing play. The Gale-Stewart theorem states that such a game is determined when $B$ is open in the product topology of discrete $A$. The same holds for closed payoff sets and, more strongly, for every $G_{\delta }$ payoff set.

For open determinacy, let $P=A^{<\omega }$ be the positions and $D=\{s:[s]\subseteq B\}$ the prefixes certifying a win for I, where $[s]$ is the extension cylinder set. Form the winning-position attractor in an infinite game by transfinite recursion: start at $W_0=D$, at a successor stage add I-positions with some successor already in $W_{\alpha }$ and II-positions with all successors in $W_{\alpha }$, and take unions at limits. This increasing sequence stabilizes, since its positions form a set. Call the stable set $W$.

Each position in $W\setminus D$ has a least entry ordinal, necessarily a successor. I chooses a successor with smaller entry ordinal; every II successor has smaller entry ordinal. An infinite strictly decreasing sequence of ordinals is impossible, so play reaches $D$ in finitely many moves. This gives I a winning strategy in an infinite game from every position in $W$. Outside $W$, all I moves stay outside, while II can choose a move outside. The axiom of choice selects those moves simultaneously. Such a play never reaches $D$, so it is outside the open payoff set. Thus II wins from the complement of $W$, proving determinacy. Interchanging the players and complementing the payoff gives closed determinacy.

We prove the $G_{\delta }$ strengthening via Buchi games, including their determinacy argument. On a set of positions $S$ with nonempty successor sets, putFor a specified set $F$ of marked positions, I's objective is to visit $F$ infinitely often. For each $X\subseteq S$, let $H(X)$ be the least fixed point obtained by starting $Y_0=\varnothing$ and iteratingwith unions at limits. Its inner entry ranks let I force a visit to $F\cap \mathrm{Pre}(X)$. The operation $H$ is monotone. Starting at $X_0=S$, iterate $X_{\alpha +1}=H(X_{\alpha })$ with intersections at limits. This descending sequence stabilizes at $W=H(W)$.

From $W$, the inner ranks force a marked position with a next move still in $W$. I chooses such a move, and the construction repeats. At II-positions all necessary moves stay in the indicated inner attractor, or in $W$ at a marked position. Every round takes finitely many moves by decreasing inner ranks. I therefore forces infinitely many marked visits.

For $s\notin W$, let $\alpha$ be its removal rank: $s\in X_{\alpha }\setminus X_{\alpha +1}$. Since $s\notin H(X_{\alpha })$, at I's turn all successors lie outside $H(X_{\alpha })$, and at II's turn there is a successor outside it. II chooses such successors. If also $s\in F$, then $s\notin \mathrm{Pre}(X_{\alpha })$; II can choose outside $X_{\alpha }$, or every I move lies outside $X_{\alpha }$. Along this strategy removal ranks never increase and strictly decrease after each marked visit. Infinitely many visits would give an infinite ordinal descent, so II wins. This proves Buchi game determinacy on a set-sized arena without any countability restriction.

Now write the original payoff as $B={\bigcap }_{n<\omega }{U}_n$, with each $U_n$ open. Add a deterministic monitor to the finite position, initially $n=0$. At each new prefix, if its cylinder is contained in $U_n$, increment the monitor and mark that step. For a play $x$, the monitor increments infinitely often exactly when $x\in B$. In one direction every increment certifies the next required open set. In the other, after the monitor reaches $n$, openness of $U_n$ supplies a later certifying prefix, so it eventually reaches $n+1$. The expanded game is a Buchi game; its monitor is determined by the observed finite play, so its winning strategy transfers directly to the original game. ConsequentlyComplementing and interchanging players also gives determinacy for $F_{\sigma }$ payoff sets. These arguments work on pruned game subtrees as well, and assigned terminal winners can be incorporated by replacing each terminal position by an absorbing position with the corresponding outcome.

Use the convention that level $k$ counts the $k$ nonzero shifts, hence an arithmetic progression of length $k+1$. The SZ property at level $k$ means that every measurable set $A$ of positive measure satisfiesThis is the averaged multiple-recurrence property, stronger than merely finding one positive intersection. Shifting the indexing convention from $k+1$ terms to $k$ terms does not change the all-levels assertion. We work with probability measure-preserving systems.

Here is the Furstenberg correspondence principle construction proving the Szemerédi theorem. Let $D\subseteq \mathbb{N}$ have positive upper density $\delta ={\mathrm{limsup}}_N|D\cap \{1,\dots ,N\}|/N$. Put $z\in \{0,1{\}}^{\mathbb{Z}}$ equal to the indicator of $D$ on the positive coordinates and zero elsewhere, and let $S$ be the left shift. The binary full shift $\Omega =\{0,1{\}}^{\mathbb{Z}}$ is a compact metric space in the product topology, and $S$ is a homeomorphism. Choose $N_i\to \infty$ attaining the upper density and define the empirical measuresWe use the standard weak compactness fact that every sequence of Borel probability measures on a compact metric space has a weakly convergent subsequence, where weak convergence means convergence of integrals of continuous functions. Pass to a subsequence with limit $\mu$. For every continuous $F$, telescoping givesPassing to the limit proves $S_{\ast }\mu =\mu$. The cylinder set $A=\{w:w(0)=1\}$ is clopen, so its indicator is continuous andIf every measure-preserving system has the SZ property at level $k$, there is some $n\ge 1$ for which $\mu (C_n)>0$, where $C_n={\bigcap }_{j=0}^k{S}^{-jn}A$. This is also a clopen cylinder set, so ${\mu }_i(C_n)\to \mu (C_n)>0$. For some sufficiently large $i$, at least one term of the empirical average is positive: $S^{m}z\in C_n$ for some $1\le m\le N_i$. Unwinding the coordinates givesTherefore $D$ contains an arithmetic progression of length $k+1$ with positive common difference. Having the SZ property at every level gives the infinite-density form of the Szemerédi theorem.

The finite uniform form follows from the same argument. If, for fixed $\delta >0$ and $k$, arbitrarily large $N_i$ admitted sets $D_i\subseteq \{1,\dots ,N_i\}$ with $|D_i|\ge \delta N_i$ and no such arithmetic progression, replace $z$ above by the indicator $z_i$ of $D_i$, extended by zero outside that interval. Their empirical measures still have an invariant weak limit with $\mu (A)\ge \delta$. For every fixed positive $n$, however, ${\mu }_i(C_n)=0$, since a point contributing to that cylinder would be an actual progression in $D_i$. Thus $\mu (C_n)=0$ for all positive $n$, contradicting the SZ property. Consequently a sufficiently large $N$ depending only on $\delta ,k$ forces the required progression.

A compact measure-preserving system is one for which every observable $f\in L^2(\mu )$ has relatively compact forward orbit $\{U^{n}f:n\ge 0\}$ under the Koopman operator $Uf=f\circ T$. Equivalently, each of these orbits is totally bounded in the global $L^2$ norm; compactness refers to observables, not just to the underlying topological space. For an invertible system it is usual to use all integer times. One may also define compactness by density of almost periodic observables: the two definitions coincide because $\Vert U^n(f-g){\Vert }_2=\Vert f-g{\Vert }_2$, so a finite net for an approximating observable is a finite net, with slightly larger radius, for $f$.

To prove the SZ property, fix $A$ with $a=\mu (A)>0$, put $f=1_A$, and chooseCompactness gives a finite $\epsilon$-net for its orbit with centers $U^{r_1}f,\dots ,U^{r_s}f$, where $r_i\ge 0$. Let $M={\max }_i{r}_i$. For every integer $n\ge M+1$, some center satisfies $\Vert U^{n}f-U^{r_i}f{\Vert }_2<\epsilon$. Since $U$ is an isometry, cancellation givesThus the return set $R=\{r\ge 1:\Vert U^{r}f-f{\Vert }_2<\epsilon \}$ meets every interval $\{n-M,\dots ,n\}$ for $n\ge M+1$. In particular, $R$ is a syndetic set with lower density at least $1/(M+1)$. This proves the needed syndetic near returns of an almost periodic observable without any assumption of ergodicity.

For $r\in R$, telescoping and the isometry property giveAs $f$ is an indicator, $\Vert U^{jr}f-f{\Vert }_2^2=\mu (A△T^{-jr}A)$. The union bound therefore givesAll other multiple-intersection measures are nonnegative. Averaging over the bounded-gap set $R$ yieldsThus every compact measure-preserving system has the SZ property at every level, including nonergodic systems and, under the forward-orbit definition, noninvertible transformations.

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.

Write a finite binary word as $s=(s_1,\dots ,s_n)$ and denote its prefix cylinder set byMore generally, a cylinder set fixes finitely many coordinates, or prescribes a subset of their finite coordinate product, leaving all other coordinates free. Each coordinate factor is discrete, so finite-coordinate inverse images are both open and closed. Thus cylinders are clopen. The basic open sets of the product topology restrict only finitely many coordinates, so cylinders form a base. Any finite-coordinate cylinder is a finite union of prefix cylinders of one sufficiently long common length; the prefix cylinders therefore also form a base.

The Bernoulli space here is Cantor space, with compatible metricIt is a compact metric space: from any sequence, choose successively subsequences constant in the first, second and subsequent coordinates, and take the diagonal subsequence. Agreement on the first $n$ coordinates bounds the distance by $2^{-n}$, proving convergence. Compactness also follows from Tychonoff theorem.

Every cylinder indicator function is continuous because the cylinder is clopen. Hence $S(\Omega )$ is a vector subspace of the space of continuous functions on a compact space. To prove density, fix $f\in C_{\mathbb{R}}(\Omega )$ and $\epsilon >0$. By uniform continuity, choose $n$ so that $d(x,y)\le 2^{-n}$ implies $|f(x)-f(y)|<\epsilon$. Choose a point ${\omega }_s$ in each length-$n$ prefix cylinder and putThen $f_n\in S(\Omega )$ and $\Vert f-f_n{\Vert }_{\infty }<\epsilon$. This proves cylinder-function density in Cantor space:

Positivity of $\varphi$ implies monotonicity. The pointwise bounds $-\Vert f{\Vert }_{\infty }1\le f\le \Vert f{\Vert }_{\infty }1$ therefore giveEvaluation on $1$ attains equality, so $\varphi$ is continuous and has norm one. This is the norm of a positive functional on C(K).

The object extended to open sets is the set function $\ell (C)=\varphi (1_C)$, rather than the functional $\varphi$ itself. Positivity gives $\ell (C)\ge 0$, and linearity gives finite additivity on disjoint cylinders. To construct its extension directly, use the disjoint cylinder decomposition of open subsets of Cantor space. For an open $U$, take the shortest prefixes $s$ for which $[s]\subset U$. They are prefix-free, their cylinders are disjoint, and their union is $U$. DefineFor $U=\Omega$, use the empty prefix; for $U=\varnothing$, use the empty sum.

This value is independent of the chosen disjoint prefix-cylinder decomposition. Indeed, compare two such decompositions $(C_i)$ and $(D_j)$. For each fixed $C_i$, its intersections with the $D_j$ give a disjoint open cover of the compact cylinder $C_i$. Compactness reduces this to finitely many nonempty intersections. Each intersection is a prefix cylinder or is empty, so finite additivity gives $\ell (C_i)={\sum }_j\ell (C_i\cap D_j)$. Sum over $i$ and rearrange the nonnegative double sum. Doing the same with each $D_j$ proves equality of the two totals.

For disjoint open sets $U_k$, combine their disjoint cylinder decompositions to obtain one for ${\bigcup }_k{U}_k$. Rearrangement of nonnegative sums proves countable additivity on this family of open sets. The extension is unique because any countably additive extension must have the prescribed sum on every disjoint cylinder decomposition. Open sets are not themselves a sigma-algebra; countable additivity here concerns disjoint open families and their open union.

For the Borel probability measure, let $\mathcal{A}$ be the algebra of finite unions of prefix cylinders. It is also the algebra of clopen sets, because every clopen set is compact and has a finite cylinder cover. Define $\ell (A)=\varphi (1_A)$ on $\mathcal{A}$. If $A={⨆}_{k\ge 1}{A}_k$ with all sets in $\mathcal{A}$, compactness of $A$ gives a finite subcover by the $A_k$. Disjointness makes all remaining members empty. Thus finite additivity already establishes the premeasure condition.

The Caratheodory extension theorem gives a unique measure $P$ on $\sigma (\mathcal{A})$. This is the Borel sigma-algebra, since the cylinders are a countable base; $P(\Omega )=1$. Its values on open sets agree with the extension above. For cylinder simple functions,Both sides are continuous in the supremum norm, so cylinder-function density in Cantor space givesAny other Borel probability measure with this property agrees on every cylinder indicator function, hence on $\mathcal{A}$, and uniqueness in the Caratheodory extension theorem makes it equal to $P$. This proves the Cantor-space representation of positive functionals without assuming the general representation theorem.

Now let $h:\Omega \to X$ be the supplied continuous surjection. The pullbackis a unital isometric embedding: surjectivity gives $\Vert f\circ h{\Vert }_{\infty }=\Vert f{\Vert }_{\infty }$. On the vector subspace $J(C_{\mathbb{R}}(X))$, define $L(Jf)=\varphi (f)$. This is well-defined, has norm one and satisfies $L(1)=1$.

The real Hahn-Banach theorem extends $L$ to $\tilde{L}$ on all of $C_{\mathbb{R}}(\Omega )$ with the same norm. Norm preservation alone does not automatically mean positivity, so verify it. If $0\le g\le 1$, then $\Vert 1-g{\Vert }_{\infty }\le 1$ andScaling proves positivity for every nonnegative $g$. This is the unital contraction positivity criterion, giving a positive extension from a unital subspace of C(K).

Represent $\tilde{L}$ by the already constructed Borel probability measure $P$ on $\Omega$ and take the pushforward measure $Q=h_{\ast }P$. Continuity of $h$ makes its Borel inverse images measurable, and $Q(X)=1$. The pushforward integral identity givesThis is compact-metric representation by Cantor-space pullback. The hypothesis $\varphi (1)=1$ already excludes an empty $X$. The measure is on the Borel sets; no uniqueness claim on an unspecified larger collection of subsets is needed.

The Szemerédi theorem says that for every integer $k\ge 2$ and every $\delta >0$, there is $N_0(k,\delta )$ such that, for $N\ge N_0$, every $A\subseteq \{1,\dots ,N\}$ with $|A|\ge \delta N$ contains a nonconstant arithmetic progressionEquivalently, every subset of the positive integers with positive upper asymptotic density contains arithmetic progressions of every finite length. Length one is immediate.

The Furstenberg multiple recurrence theorem states that for every probability measure-preserving system, every measurable $B$ with $\mu (B)>0$, and every integer $k\ge 2$, there is $n\ge 1$ withHere $T^{-jn}B$ means a preimage, so invertibility is unnecessary. The Furstenberg multiple recurrence theorem also does not assume an ergodic transformation.

We prove the implication to the finite Szemerédi theorem by constructing the Furstenberg correspondence principle explicitly. Suppose, to the contrary, that for fixed $k$ and $\delta >0$ there are $N_{\ell }\to \infty$ and $A_{\ell }\subseteq \{1,\dots ,N_{\ell }\}$ with $|A_{\ell }|\ge \delta N_{\ell }$, but with no length-$k$ arithmetic progression of positive common difference. In the binary full shift $\Omega =\{0,1{\}}^{\mathbb{Z}}$, encode $A_{\ell }$ as $x_j^{(\ell )}=1_{A_{\ell }}(j)$, taking all coordinates outside $A_{\ell }$ to be zero. Let $S$ be the left shift, $(S\omega {)}_j={\omega }_{j+1}$, and setThe cylinder set $C$ is a clopen set, and ${\nu }_{\ell }(C)=|A_{\ell }|/N_{\ell }\ge \delta$. The binary full shift is a compact metric space, so compactness of probability measures on a compact metric space gives a subsequence of these empirical measures with weak convergence of probability measures to a Borel probability measure $\nu$.

For every continuous $h$ on the full shift,Passing to the weak limit shows that $\nu$ is an invariant measure for the continuous left shift. Thus $(\Omega ,\mathcal{B},\nu ,S)$ is a probability measure-preserving system. Since the indicator function of $C$ is continuous, $\nu (C)\ge \delta$.

Apply the Furstenberg multiple recurrence theorem to $C$. For some $n\ge 1$, the clopen sethas $\nu (D)>0$. Its continuous indicator function gives ${\nu }_{\ell }(D)\to \nu (D)>0$. For large $\ell$, at least one $S^a{x}^{(\ell )}$ in the defining empirical measure therefore belongs to $D$. The choice of left shift impliesso $a,a+n,\dots ,a+(k-1)n$ lies in $A_{\ell }$. This contradicts the assumed absence of arithmetic progressions and proves the finite Szemerédi theorem. Applying the finite Szemerédi theorem on intervals where an infinite set has density bounded below proves the stated upper asymptotic density formulation.

For multiple recurrence for circle rotations, write $\mathbb{T}=\mathbb{R}/\mathbb{Z}$ and $R_{\alpha }x=x+\alpha$, with normalized Lebesgue measure $m$, which is the Haar measure of the circle group. Fix a measurable $B\subseteq \mathbb{T}$ with $m(B)>0$ and $k\ge 2$. If $\alpha$ is rational, there is $q\ge 1$ with $R_{\alpha }^q$ the identity, and $n=q$ gives an intersection of measure $m(B)$.

For irrational $\alpha$, the pigeonhole principle applied to $0,\alpha ,\dots ,Q\alpha$ in $Q$ equal arcs gives $1\le n\le Q$ with $\Vert n\alpha {\Vert }_{\mathbb{T}}\le 1/Q$. In particular there are arbitrarily small nonzero returns to zero for the irrational rotation of the circle.

We also need translation continuity in L1 on the circle. Given a measurable $B$, approximate its indicator function in the L1 norm by a continuous $g$ on the circle group. Such approximation follows from regularity of Lebesgue measure, or approximation by finite unions of intervals. Translation invariance and uniform continuity givewhere the approximation error is first made arbitrarily small. Equivalently, $m(B△(B-t))\to 0$.

Choose a return time $n\ge 1$ so small modulo one that for each $1\le j<k$,This is possible because multiplication by each fixed $j$ is continuous on the circle group and translation continuity in L1 on the circle applies to the finitely many translates. The union bound now givesThis proves the Furstenberg multiple recurrence theorem for every circle rotation with its normalized Lebesgue measure, including both rational and irrational angles.

For a finite position $s\in A^{<\omega }$, let $[s]$ be the cylinder set of infinite plays extending $s$. Because the factors have the discrete topology, these cylinder sets form a basis for the product topology. PutMembership in $D$ means that I has already secured the outcome, regardless of future moves. Since $G$ is an open set, a play belongs to $G$ exactly when one of its prefixes belongs to $D$.

Build the winning-position attractor in an infinite game by transfinite recursion. Start with $W_0=D$, use unions at limit ordinals, and setThis sequence is increasing: the defining operation is order-preserving, and $W_0\subseteq W_1$. It stabilizes at some ordinal. Indeed, if it strictly increased at every successor stage below the successor cardinal number of $|A^{<\omega }|$, choosing one newly added position per stage would inject that larger cardinal into $A^{<\omega }$. Denote the stable set by $W$. This constructs the needed fixed point directly, without an appeal to an unstated fixed-point theorem.

Every position in $W$ has a least entry ordinal, its rank. Rank zero means membership in $D$. A positive rank is a successor $\alpha +1$: at an I-position some extension belongs to $W_{\alpha }$, while at a II-position every extension does. If the empty word belongs to $W$, I always chooses an extension of smaller rank until $D$ is reached. II's choices also decrease rank before that happens. There is no infinite strictly decreasing sequence of ordinals, so some finite prefix lies in $D$. This is a winning strategy in an infinite game for I.

If the empty word is outside $W$, the fixed point equation says that every extension of an I-position outside $W$ is still outside $W$, and that a II-position outside $W$ has at least one extension outside $W$. II chooses such an extension. Every resulting prefix avoids $D$, hence the whole play avoids $G$ by openness. This is a winning strategy in an infinite game for II.

The choices can be made simultaneously by fixing a well-order of $A$, as allowed by the usual axiom of choice. The argument does not assume that $A$ is countable. Every such open game is determined: