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Let $P$ be the transition matrix of a finite irreducible Markov chain with at least two states, reversible with stationary distribution $\pi$. Its Dirichlet form of a Markov chain isThe Discrete Poincare inequality states that ${\mathrm{Var}}_{\pi }f\le C\mathcal{E}(f,f)$ for every real $f$. Its optimal constant is $C=(1-{\lambda }_2{)}^{-1}$, where ${\lambda }_2$ is the second largest eigenvalue of $P$. Indeed, detailed balance makes $P$ self-adjoint on $L^2(\pi )$. Expand $f-\pi (f)={\sum }_{j\ge 2}{a}_j{v}_j$ in an orthonormal eigenbasis. Then ${\mathrm{Var}}_{\pi }f=\sum a_j^2$ and $\mathcal{E}(f,f)=\sum (1-{\lambda }_j)a_j^2$, proving the inequality and its sharpness.

The useful constructive version is the canonical paths Poincare bound. Choose transition graph paths ${\gamma }_{xy}$ between each ordered pair, write $Q(u,v)=\pi (u)P(u,v)$, and definewhere $N_{uv}$ counts occurrences of the directed transition. The Cauchy-Schwarz inequality along each path in a graph gives $(f(x)-f(y){)}^2\le |{\gamma }_{xy}|{\sum }_{{\gamma }_{xy}}(f(u)-f(v){)}^2$. Multiply by $\pi (x)\pi (y)/2$ and sum. The identity ${\mathrm{Var}}_{\pi }f=\frac{1}{2}{\sum }_{x,y}\pi (x)\pi (y)(f(x)-f(y){)}^2$ proves ${\mathrm{Var}}_{\pi }f\le \rho \mathcal{E}(f,f)$.

For a lazy Markov chain this also proves the mixing time estimateTo see this, the centred initial density has $L^2(\pi )$ norm at most $\pi (x{)}^{-1/2}$; laziness puts all nonconstant eigenvalues between zero and $1-{\rho }^{-1}$. Their powers contract that norm, and the Cauchy-Schwarz inequality bounds total variation distance by half the density's $L^2(\pi )$ norm.

Here is a precise sufficient density condition for the permanent application: every row and every column has more than $n/2$ ones. Interpret the permanent of a matrix as the number $m_n$ of perfect matchings in its bipartite graph, with $n$ vertices per side. Write $m_k$ for the number of $k$-edge matchings in a graph.

First prove the short augmenting paths in a dense bipartite graph bound. Given a nonperfect matching in a graph, choose unmatched $u,v$ on opposite sides. If either has an unmatched neighbour, augment with one edge. Otherwise the matched partners of $N(u)$ and the neighbours $N(v)$ are subsets of the same $n$-element side whose total size exceeds $n$. Their intersection gives an alternating path in a graph $u,b,a,v$ of length three, with $ab$ a matching edge. Repeating proves $m_n>0$. Choose augmentations by a fixed deterministic rule. A reverse augmentation is specified by at most four vertices, so a next-size matching in a graph has at most $C_0=2n^4$ preimages. Therefore

Use the matching generating function $Z(\lambda )={\sum }_{k=0}^n{m}_k{\lambda }^k$ and the weighted matching Markov chain on all matchings in a graph, with stationary weights ${\pi }_{\lambda }(M)={\lambda }^{|M|}/Z(\lambda )$. It stays put with probability $1/2$; otherwise it chooses a graph edge uniformly and proposes its deletion, its addition if both endpoints are unmatched, or its exchange for the unique incident matching edge if precisely one endpoint is matched. Other proposals stay put. Accept with probability $\min (1,{\lambda }^{|M^{\prime }|-|M|})$. Reverse proposals use the opposite exchanged edge, so detailed balance holds. Deletions connect every state to the empty matching in a graph.

For completeness, the canonical paths for the weighted matching chain give a polynomial bound without assuming a mixing theorem. The symmetric difference $I△F$ consists of alternating graph paths and graph cycles. Order its components deterministically, orient open components from their smaller endpoint, and convert them from $I$ to $F$ by moving an unmatched endpoint with exchanges. If the first edge belongs to $I$, delete it first. For a graph cycle, delete one $I$-edge, move the unmatched endpoint around the opened cycle, then insert the last $F$-edge. Each intermediate state is a matching in a graph, and each entire transition path in a graph has length at most $4n$.

Fix a directed transition with initial state $M$. The complementary edge set $K=I△F△M$ agrees with one of the two matchings in a graph outside the active component and has at most two degree-two vertices on that component. Remove at most two offending edges to obtain a matching in a graph $K^{\prime }$. Store those edges and one bit specifying the initial alternating colour of the active component. This encoding is injective for the fixed transition: it restores $K$ and $I△F=M△K$, identifies the active component from the changed edges, and recovers the two colours using the fixed component order. Completed components have $F$'s colour in $M$, pending components have $I$'s colour, and the stored bit resolves the active component. Common edges are $M\cap K$.

There are at most $2(m+1{)}^2$ auxiliary records per $K^{\prime }$, where $m$ is the number of graph edges. If $d=|K|-|K^{\prime }|\le 2$, thenAlso $Q(M,M^{\prime })\ge {\pi }_{\lambda }(M)/(2m\max (\lambda ,{\lambda }^{-1}))$. Sum over the encodings and use the path length bound to obtainThis is polynomial whenever $\lambda$ and ${\lambda }^{-1}$ are polynomially bounded. Starting at the empty matching in a graph, $\log (1/{\pi }_{\lambda }(\varnothing ))=\log Z(\lambda )\le m\log (1+\lambda )$, so the Discrete Poincare inequality gives polynomial-time sampling to any prescribed total variation distance. The addition/deletion/exchange construction is also described in Section 12.4 of Jerrum and Sinclair's survey.

It remains to turn sampling into a relative count; merely sampling perfect matchings would not do that. For $0<\epsilon \le 1/2$, choose ${\lambda }_0=\epsilon /(8m)$ and ${\lambda }_{\ast }=8C_0/\epsilon$. The elementary bounds above giveChoose a schedule ${\lambda }_0<{\lambda }_1<\cdots <{\lambda }_J={\lambda }_{\ast }$ whose successive ratios are at most $1+1/n$. It has $J=O(n\log (n/\epsilon ))$ stages. The annealing ratios for a matching generating function satisfyEach observable lies in $[1,e]$, so the Hoeffding inequality gives relative accuracy $\eta =\epsilon /(10J)$ with $O({\eta }^{-2}\log (J/\delta ))$ independent samples per stage. Run independent copies of the chain from the empty matching in a graph; make the total variation distance per sample at most $\delta /(3JK)$, where $K$ is that sample count. The coupling characterization of total variation distance and the union bound make the entire sample collection coincide with exact stationary samples except with probability at most $\delta /3$. Choose the sample-count constant so all stationary sample means are accurate except with probability at most $\delta /3$.

Let $\hat{R}$ be the product of the sample means, and output $\hat{p}=\hat{R}/{\lambda }_{\ast }^n$. On the successful event its multiplicative error relative to $Z({\lambda }_{\ast })/Z({\lambda }_0)$ lies between $e^{-\epsilon /5}$ and $e^{\epsilon /5}$. Combining the two endpoint bounds provesActivities, sample counts, simulation lengths and rational bit lengths are polynomial in $n,{\epsilon }^{-1},\log ({\delta }^{-1})$. Thus this is a fully polynomial randomized approximation scheme. Small dimensions can be evaluated directly.

Write $[\mathbb{N}{]}^{\omega }$ for the space of infinite subsets of the natural numbers, with each subset identified with its increasing enumeration. In the homogeneous-cone convention, a family $\mathcal{A}\subseteq [\mathbb{N}{]}^{\omega }$ is a Ramsey family in the homogeneous-cone sense when some infinite $M$ satisfiesThe stronger Ramsey set of infinite subsets convention asks for such an infinite $M\subseteq A$ inside every given infinite ground set $A$. We will prove the open-set assertion in this stronger form, and our Ramsey-but-not-completely-Ramsey example will satisfy it as well. Thus the distinction in conventions does not affect any of the conclusions below.

For a non-Ramsey example, declare $X\sim Y$ if their symmetric difference is finite, and choose a representative $R$ from each equivalence class using the axiom of choice. Setfor the representative of the class of $X$. Removing one point reverses this colour, because its symmetric difference with $R$ changes by exactly one point. Consequently every cone $[M{]}^{\omega }$ contains the oppositely coloured sets $M$ and $M\setminus \{\min M\}$. Either colour class of this finite-symmetric-difference parity colouring is not a Ramsey family in the homogeneous-cone sense. No definability is claimed for the representative selection.

The ordinary topology on infinite subsets, denoted $\tau$, has basic cylinders $[s]=\{X:s⊏X\}$, where $s$ is a finite stem of the increasing enumeration. Let $O$ be a $\tau$-open family and fix an arbitrary infinite ground set $A$. We give a complete fusion proof of a homogeneous cone inside $A$.

Use the convention $\max \varnothing =0$. For finite $s$ and an infinite reservoir $B$ above $\max s$, say that $B$ accepts $s$ if $[s,B]\subseteq O$, whereSay that $B$ rejects $s$ if no infinite subset of $B$ accepts $s$. There is always a refinement deciding $s$: choose an accepting subreservoir if one exists, and otherwise retain $B$, which rejects. Both decisions persist under further thinning.

A first fusion produces an infinite $H\subseteq A$ deciding every finite $s\subseteq H$ on its tail $H/s=\{h\in H:h>\max s\}$. First decide the empty stem. Choose the first point, and thin the remaining reservoir successively to decide every subset of the selected prefix. Repeat after choosing each new point. There are finitely many subsets to decide at each stage. The diagonal sequence of chosen points has the required property, since its tail after the largest point of any stem lies in the reservoir where that stem was decided. The empty-stem decision persists too. This is deciding all finite stems by fusion.

If $H$ accepts the empty stem, $[H{]}^{\omega }\subseteq O$ and we are done. Suppose it rejects. For a rejected finite $s\subseteq H$, only finitely many $h\in H/s$ can have $H/h$ accepting $s\cup \{h\}$. Otherwise let $D$ be their infinite set. Every infinite $X\subseteq D$ starts with some such $h$, and its remaining tail lies in $H/h$, so $s\cup X\in O$. Then $D$ accepts $s$, contradicting rejection. This proves finitely many accepting extensions of a rejected stem.

A second fusion constructs $K\subseteq H$ whose every finite subset is rejected. If a finite prefix of chosen points has been selected, all its subsets are rejected by induction. Avoid the union of their finitely many accepting successor sets when choosing the next point. The first fusion already decides every successor, so every new subset is rejected. An infinite reservoir remains after each finite exclusion.

If some $X\in [K{]}^{\omega }$ belonged to $O$, openness would give an initial segment $s⊏X$ with $[s]\subseteq O$. Its reservoir $H/s$ would then accept $s$, contradicting the second fusion. Therefore $[K{]}^{\omega }\cap O=\varnothing$. In either case we have a homogeneous cone inside the arbitrary ground set $A$. Every $\tau$-open family is a Ramsey set of infinite subsets, under both conventions.

A completely Ramsey set requires more: for every finite $s$ and infinite reservoir $A$ above $\max s$, there is an infinite $B\subseteq A$ such thatThe finite stem must be kept fixed. These $[s,A]$ are the basic neighbourhoods of the Ellentuck topology, also called the star topology.

Let $E$ be the even positive integers, and construct the finite-symmetric-difference parity colouring $\epsilon$ on $[E{]}^{\omega }$. DefineFor any infinite ground set $A$, the set $A\setminus \{1\}$ is still infinite and its whole cone misses $\mathcal{R}$. Thus $\mathcal{R}$ is a Ramsey set of infinite subsets even in the stronger convention. But inside $[\{1\},E]$, every infinite refinement $B\subseteq E$ gives the two sets $\{1\}\cup B$ and $\{1\}\cup (B\setminus \{\min B\})$ of opposite colours. Hence no such refinement is homogeneous. The family $\mathcal{R}$ is Ramsey but not completely Ramsey. The stem-supported Ramsey family need not be completely Ramsey construction explains this example: forgetting the stem $\{1\}$ is precisely what loses the obstruction.

Finally take $D=[E{]}^{\omega }$. It is nonempty and star-open, since it equals $[\varnothing ,E]$. It is $\tau$-closed: if an infinite set contains an odd integer, an initial segment witnessing that integer has a cylinder disjoint from $D$. Its $\tau$-interior is empty: after any finite stem of even integers, append an arbitrarily large odd integer and then infinitely many further points. ThusThis is the cone on an infinite coinfinite ground set. Choose its subset $Y=\{X\in [E{]}^{\omega }:\epsilon (X)=0\}$. Since ${\overline{Y}}^{\tau }\subseteq D$, $Y$ is also $\tau$-nowhere dense. Nevertheless every $[\varnothing ,B]$ with infinite $B\subseteq E$ contains both $B$ and $B\setminus \{\min B\}$, of opposite colours. Hence $Y$ is not completely Ramsey. We have proved explicitly that ordinarily nowhere-dense sets need not be completely Ramsey; the topology in the nowhere-dense hypothesis cannot be silently changed to the star topology.

For a rational matrix $A$ with columns $a_1,\dots ,a_n$, partition regularity means that every finite colouring of the positive integers admits a positive monochromatic vector $x$ with $Ax=0$. Repeated coordinates are allowed. The columns condition is an ordered partition $\{1,\dots ,n\}=B_1\bigsqcup \cdots \bigsqcup B_s$ into nonempty blocks such thatRado's theorem says that $A$ is partition regular if and only if it satisfies the columns condition. We prove both directions.

For necessity, clear denominators so the columns are integer vectors. For every pair of disjoint index sets $I,B$ with $B\ne \varnothing$ for which ${\sum }_{i\in B}{a}_i$ is not in the rational linear span of the I-columns, choose an integer linear functional ${\ell }_{I,B}$ that vanishes on those I-columns but not on that sum. Such a functional exists by finite-dimensional linear algebra; clear its rational denominators. There are only finitely many pairs, so choose a prime number $p$ dividing none of the nonzero integersIf there are no offending pairs, any prime number suffices.

Colour a positive integer $x=p^{v_p(x)}u$, $p\nmid u$, by $u\mathrm{mod}p$. This is the last nonzero digit coloring, using $p-1$ colours. Partition regularity gives a positive monochromatic solution. Group its coordinate indices by their distinct P-adic valuations $e_1<\cdots <e_s$, writing $B_j=\{i:v_p(x_i)=e_j\}$. All units $x_i/p^{e_j}$ have the same nonzero residue $\alpha$ modulo $p$.

For each $j$, let $I=B_1\cup \cdots \cup B_{j-1}$. If its block sum failed the required span condition, apply the chosen ${\ell }_{I,B_j}$ to ${\sum }_i{a}_i{x}_i=0$. The earlier terms vanish exactly, not just modulo a power of $p$. Divide by $p^{e_j}$; terms with later valuations are still divisible by $p$. Reduction modulo $p$ then yieldscontrary to the choice of $p$. Thus every required span relation holds, and with $I=\varnothing$ the first block sum is zero. This is the finite separating-functional proof of the columns condition; it avoids an unjustified passage from congruences to rational equality.

For sufficiency, assume the columns condition. Choose rational coefficients ${\lambda }_{ij}$, for $j\ge 2$ and $i\in B_1\cup \cdots \cup B_{j-1}$, such thatChoose a positive integer $c$ clearing every denominator, and a positive integer $p\ge \max |c{\lambda }_{ij}|$, taking $p=1$ if the list is empty. By the allowed monochromatic m-p-c set theorem, there are positive generators $z_1,\dots ,z_s$ such that all numbersare positive and have one colour. This is a full m-p-c set; positivity is part of its definition, not a rule discarding negative expressions. For $i\in B_h$, putThese coordinates all belong to that monochromatic m-p-c set. Collecting coefficients of $z_j$ givesThus the Rado solution inside an m-p-c set proves partition regularity and completes the theorem.

To obtain the Finite sums theorem from Rado's theorem, introduce a positive variable $y_F$ for every nonempty $F\subseteq \{1,\dots ,k\}$ and imposePartition the columns of this system into $B_j=\{F:\min F=j\}$. The $B_1$-column sum is zero: in a row indexed by $F$ containing $1$, the contributions of $y_F$ and $y_{\{1\}}$ cancel, and other rows receive neither. For $j>1$, the $B_j$-column sum iswhere $a_F$ is the column of $y_F$. Those columns lie in earlier blocks. Hence the columns condition holds. A monochromatic positive solution gives $x_i=y_{\{i\}}$ and $y_F={\sum }_{i\in F}{x}_i$, provingFor $k=1$ the system has no rows and the conclusion is immediate. The construction is also visible directly in the sufficiency proof: a monochromatic $(k,1,1)$-set contains every nonempty sum of its generators, classified by the smallest chosen index.

For a finite alphabet $K$, a combinatorial line in $K^N$ has a nonempty active set $J\subseteq \{1,\dots ,N\}$, fixed letters outside $J$, and the same variable letter at every coordinate in $J$. The Hales-Jewett theorem states that for every alphabet size $k\ge 1$ and number of colours $r\ge 1$, some $N$ makes every r-colouring of $\{1,\dots ,k{\}}^N$ contain a monochromatic combinatorial line.

We first prove the alphabet insensitivity lemma with a concrete bound. Fix two distinct letters $a,b$, a desired dimension $d$, and a colouring $\chi :K^N\to \{1,\dots ,r\}$. A combinatorial subspace of dimension $d$ has disjoint nonempty active blocks $J_1,\dots ,J_d$ and fixed other coordinates. It is $(a,b)$-insensitive when exchanging $a,b$ in any one variable leaves the colour unchanged, whatever the other variable letters are.

Choose consecutive blocks of lengths $L_1,\dots ,L_d$ recursively byProcess these blocks from right to left. When block $i$ is reached, the later blocks already have one variable each; the earlier $P_i$ coordinates are still entirely unrestricted. For each of the $L_i+1$ words $a^t{b}^{L_i-t}$ in block $i$, form its complete colour profile over all $k^{P_i}$ earlier words and all $k^{d-i}$ assignments to the later variables. There are at most $r^{k^{P_i+d-i}}=L_i$ profiles. The pigeonhole principle gives $s<t$ with equal profiles. Fix the first $s$ coordinates of this block to $a$, the coordinates after $t$ to $b$, and make the interval from $s+1$ to $t$ variable. Substitution of $b$ gives the chain word indexed by $s$, and substitution of $a$ gives the one indexed by $t$; they have identical profiles. Thus this variable is insensitive to $a,b$ for every remaining context. Previously constructed later variables remain insensitive, because their colour equalities held for every possible earlier word. After all $d$ blocks, we have the required insensitive combinatorial subspace. This proves the explicit block bound for alphabet insensitivity.

Now induct on $k$, simultaneously for every $r$. The case $k=1$ is immediate with $N=1$. For $k\ge 2$, assume the Hales-Jewett theorem for the alphabet of size $k-1$, and take $d$ large enough for that theorem with $r$ colours. Apply the proved alphabet insensitivity lemma to the letters $k-1,k$, obtaining a d-dimensional combinatorial subspace $\varphi :K^d\to K^N$. On $\{1,\dots ,k-1{\}}^d$, the induced colouring $\chi \circ \varphi$ has a monochromatic combinatorial line. Its value at letter $k$ has the same colour as its value at $k-1$: change the active variables one at a time and use insensitivity. The union of their nonempty active blocks is the active set of a combinatorial line in the original word space. All its k points have one colour, completing the induction and the proof.

To deduce the Van der Waerden theorem, take $K=\{0,\dots ,m-1\}$ and pull a colouring of the positive integers back throughA combinatorial line with active set $J$ maps toHence every finite colouring has a monochromatic arithmetic progression of every prescribed finite length. In finite form, if $H(k,r)$ is a word-length bound for the Hales-Jewett theorem, then $W(r,m)\le 1+(m-1)H(m,r)$ is a valid interval bound.

The Strengthened Van der Waerden theorem also puts the common difference in the colour class:Here is a finite proof by induction on the number $r$ of colours. For $m=1$ choose $a=d=1$. For $r=1$ choose $a=d=1$ in $[m]$. For $m\ge 2,r\ge 2$, let $M$ be a bound for the strengthened theorem with $r-1$ colours, put $L=(m-1)M+1$, and let $W=W(r,L)$ be an ordinary Van der Waerden theorem bound. Work in $[MW]$. Its first $W$ points contain a monochromatic progressionof colour $c$. Every $jd$, $1\le j\le M$, lies in $[MW]$. If some $jd$ has colour $c$, then $a,a+jd,\dots ,a+(m-1)jd$, together with its difference $jd$, gives the answer. Otherwise the colouring of $\{d,2d,\dots ,Md\}$ uses only $r-1$ colours. Pull it back to $[M]$ and apply the induction hypothesis to get a monochromatic set $\{e,b,b+e,\dots ,b+(m-1)e\}$. Multiplication by $d$ gives the required configuration in the original colouring. This proves the strengthened theorem, including finite bounds.

The Gallai theorem for an integer lattice states that every finite colouring of ${\mathbb{Z}}^h$ contains a monochromatic homothetic copy $u+tF$ of every prescribed finite nonempty pattern $F\subseteq {\mathbb{Z}}^h$, where $t$ is a positive integer. The positive-lattice version holds for finite $F\subseteq {\mathbb{N}}^h$ as well. Write $F=\{v_1,\dots ,v_k\}$ and colour $w\in \{1,\dots ,k{\}}^N$ by the colour ofA monochromatic combinatorial line maps to $u+|J|F$, with $u=b+{\sum }_{j\notin J}{v}_{w_j}$. Thus the Hales-Jewett theorem supplies the copy, with nonzero dilation $|J|$. Choosing $b$ large and positive places all these word sums in the positive lattice if required. This sum map from words to homothetic copies also proves the real-vector-space version: replace ${\mathbb{Z}}^h$ by ${\mathbb{R}}^h$ and allow an arbitrary finite real pattern, while retaining positive integer dilations.

We take $\mathbb{N}=\{1,2,\dots \}$ and require the common difference of an arithmetic progression to be positive. An ultrafilter $\mathcal{U}$ is a proper filter on a set deciding each subset: exactly one of $B$ and $\mathbb{N}\setminus B$ belongs to $\mathcal{U}$.

For the fixed length $m$, let ${\mathcal{B}}_m$ be the family of sets containing no m-term arithmetic progression. Any acceptable ultrafilter must contain every complement $\mathbb{N}\setminus B$ with $B\in {\mathcal{B}}_m$. These mandatory sets have the finite intersection property. Otherwise finitely many $B_1,\dots ,B_r\in {\mathcal{B}}_m$ would cover $\mathbb{N}$. Colour an integer by the first $i$ for which it belongs to $B_i$. The Van der Waerden theorem gives a monochromatic m-term arithmetic progression, contained in the corresponding $B_i$, a contradiction.

Generate a proper filter from those complements and extend it by the ultrafilter lemma. If some $B\in \mathcal{U}$ were in ${\mathcal{B}}_m$, its complement would also belong to $\mathcal{U}$, forcing the empty set into the filter. ThusNotice that sets avoiding one fixed length need not be closed under finite unions; the argument uses the finite intersection property, not an unsupported ideal claim.

For the unbounded-length version, let $\mathcal{B}$ consist of sets whose arithmetic progression lengths are bounded. Each $B_i\in \mathcal{B}$ fails to contain a progression of some length $m_i$, and hence of every greater length. If finitely many $B_i$ covered $\mathbb{N}$, colour by the first containing index and apply the Van der Waerden theorem at length $M={\max }_i{m}_i$. Again a contradiction results. The complements of all members of $\mathcal{B}$ therefore have the finite intersection property. Extending their generated filter gives an ultrafilter with arithmetic-progression-rich members:All finite sets lie in $\mathcal{B}$, so this ultrafilter contains every cofinite set and is a nonprincipal ultrafilter. The same is true of the fixed-length construction when $m\ge 2$, because no singleton is allowed; for $m=1$ a principal ultrafilter also works. These constructions illustrate ultrafilter selection from a partition-rich family.

The infinite-length answer is no. Colour $n$ by $\lfloor {\log }_{2}n\rfloor \mathrm{mod}2$, so successive intervals $[2^j,2^{j+1})$ alternate colours. An infinite arithmetic progression $a+td$, $t\ge 0$, meets every sufficiently large such interval: its first term at least $2^j$ is less than $2^j+d$, which is at most $2^{j+1}$ once $2^j\ge d$. It therefore has terms of both colours. Neither colour class contains an infinite arithmetic progression, but every ultrafilter contains one of the two colour classes. This dyadic block obstruction to infinite arithmetic progressions rules out the proposed ultrafilter.

Use discrete Shannon entropy with base-two logarithms. The standard facts needed are the chain rule for information entropy, conditioning reduces entropy, nonnegativity of conditional entropy, and the fact that a uniform distribution on $M$ outcomes has Shannon entropy ${\log }_{2}M$.

Order the coordinates by their indices and put $h_j=H(X_j\mid X_1,\dots ,X_{j-1})\ge 0$. Applying the chain rule within a coordinate subset $A$ and then adding the missing earlier conditioning coordinates givesOn summing over the family, coordinate $j$ occurs at least $k$ times. Since each $h_j$ is nonnegative,This proves Shearer's inequality without any independence assumption on the coordinates. For countable alphabets, apply the finite-valued assertion to refining finite coordinate partitions; the standard monotone approximation of Shannon entropy gives the same result with extended values. Differential entropy is not the Shannon entropy intended here.

For the graph application, let its bipartition be $U,V$, with degrees $a$ on $U$ and $b$ on $V$, where $a,b\ge 1$. Choose an independent set uniformly from all $i(G)$ such sets, and let $X$ be its vertex-indicator vector. Then $H(X)={\log }_{2}i(G)$. For $v\in V$ put $Y_v=X_{N(v)}$ and $q_v=\mathrm{Pr}(Y_v=0)$.

Each vertex of $U$ belongs to exactly $a$ of these neighbourhoods. Shearer's inequality therefore gives $H(X_U)\le a^{-1}{\sum }_{v\in V}H(Y_v)$. Conditional on $X_U$, a vertex of $V$ is forced absent if any neighbor was selected, and otherwise is a free fair binary choice. There are no edges within $V$, and the conditional independent-set distribution is uniform on all these free choices. Henceand

There are $2^b$ possible neighborhood patterns. Give the zero pattern weight $w_0=2^a$ and every other pattern weight one. For its distribution $(p_y)$, concavity of the logarithm, or Jensen's inequality, givesThis establishes the local weighted-entropy estimate rather than just counting neighborhood patterns.

Edge counting gives $a|U|=b|V|$; together with $|U|+|V|=n$, it yields $|V|/a=n/(a+b)$. Exponentiating the Shannon entropy estimate proves the independent-set bound for biregular graphs:The complete bipartite graph $K_{b,a}$ has exactly $2^a+2^b-1$ independent sets, since one chooses a subset of one side and subtracts the double-counted empty set. Disjoint unions of such components attain equality. If exactly one degree parameter is zero, edge counting forces the graph to be edgeless and the displayed bound equals $2^n$, so it remains valid. If $a=b=0$, the denominator is undefined; the edgeless graph has exactly $2^n$ independent sets.

For $x\in \Omega$ and nonempty $A$, letThe Talagrand convex distance isEquivalently,For completeness, let $v_{\ast }$ be a closest point of the finite convex hull to zero. If $v_{\ast }\ne 0$, minimality gives $v_{\ast }\cdot (v-v_{\ast })\ge 0$ for every $v\in V$. Taking $\alpha =v_{\ast }/\Vert v_{\ast }\Vert$ proves that the supremum is at least $\Vert v_{\ast }\Vert$. Conversely each permitted $\alpha$ has ${\inf }_{v\in V}\alpha \cdot v\le \alpha \cdot v_{\ast }\le \Vert v_{\ast }\Vert$ by Cauchy-Schwarz. The weights are nonnegative because the hull has nonnegative coordinates. If $v_{\ast }=0$, both sides are zero. This proves the dual definition used below.

We prove Talagrand's convex distance inequality by induction on the number of factors. Write $\Omega ={\Omega }^{\prime }\times X_n$, with measure $\mu \otimes \nu$, and $F_A=d_T(\cdot ,A{)}^2$. For a last-coordinate value $z$, let $A_z=\{y^{\prime }:(y^{\prime },z)\in A\}$. Choose a section $B=A_b$ of largest probability $m=\mu (B)>0$, and put $a_z=\mu (A_z)/m\in [0,1]$. The total event probability is $p=\mathrm{Pr}(A)=m{\sum }_z\nu (z)a_z$.

If $A_z$ has positive probability, combine mismatch vectors from $A_z$ and from $B$. A weight $t$ on the second section contributes at most $t$ in the last coordinate. Convexity of the squared Euclidean norm gives, for every $0\le t\le 1$,The last-coordinate contribution is actually zero if $z=b$, so the displayed upper bound remains valid. Hölder's inequality and the induction hypothesis yieldFor a zero-probability section use $t=1$ directly, so no division by zero or exponent of an infinite quantity is needed.

The scalar estimate for Talagrand product induction isHere is its proof. For $a>0$ set $L=-\log a$. If $L\le 1/2$, take $t=2L$; the logarithm of the left expression is $L-L^2$. Put $h(L)=\log (2-e^{-L})-L+L^2$. Then $h(0)=h^{\prime }(0)=0$ andsince $(2u-1{)}^2-u=(4u-1)(u-1)\ge 0$ for $u=e^L\ge 1$. Thus $h\ge 0$. If $L\ge 1/2$, choose $t=1$. The already proved boundary case gives $e^{1/4}\le 2-e^{-1/2}\le 2-a$. The same direct choice covers $a=0$.

Average the resulting section bounds over $z$:where the last inequality is equivalent to $(m-p{)}^2\ge 0$. The induction starts on the zero-coordinate singleton, where every positive-probability event is the whole space and the moment is one. HenceWe assume $\mathrm{Pr}(A)>0$; a zero-probability event has no finite right-hand bound to prove. Markov's inequality also gives

For the subsequence application, let $L(x)$ be the longest increasing subsequence length for independent sequence coordinates. A witness of length $\ell$ uses only $\ell$ positions. If $L(x)=\ell \ge b$ and $y$ satisfies $L(y)\le a<b$, at least $\ell -a$ witness positions must change: all unchanged witness positions remain in increasing order. Put weights $1/\sqrt{\ell }$ on the witness and zero elsewhere. The dual distance definition giveswhere the latter expression increases with $\ell$ for $a\ge 0$. Therefore the increasing-subsequence certificate concentration bound isIf the lower event has probability zero the conclusion is immediate.

Let $m$ be an integer median, so $\mathrm{Pr}(L\le m),\mathrm{Pr}(L\ge m)\ge 1/2$. Taking the two thresholds in the bound gives, for $t>0$,For $t>m$ the lower event is empty. In a nonempty sequence $m\ge 1$. Integration of these tails gives an explicit mean-median estimate: use $m+t\le 2m$ for $t\le m$ and $m+t\le 2t$ for $t\ge m$, obtainingWriting $\mu =\mathbb{E}L$, the median property also gives $m\le 2\mu$. Consequently, for a universal constant $C$, all $u\ge 0$ satisfyThus the subsequence length is concentrated near its mean on the scale $\sqrt{\mu }$, rather than the generic coordinate-exposure scale $\sqrt{n}$. This argument applies directly to finite-valued independent sequences. Independent continuous samples follow by refining finite coordinate discretizations; the strict-order pattern stabilizes almost surely when ties have probability zero. A uniform random permutation is the rank sequence of independent continuous samples, so it has the same subsequence-length concentration. Arbitrarily dependent random sequences are not covered by the product-space assumption.

The assertion with the same $\epsilon$ in the degree interval and the final colour bound is false as printed. Fix $r=2$, $\epsilon =1/10$, let $G=K_{22m+1}$, and take $D=20m$. Every degree is $22m=(1+\epsilon )D$, while every pair hypergraph codegree is one. For any proposed $\delta >0$, choose $m$ large enough that $1\le \delta D$. Nevertheless a matching in a graph contains at most $11m$ edges, whereas there are $(22m+1)11m$ edges in total. Thus every edge coloring requires at least $22m+1$ colours, exceeding $(1+\epsilon )D=22m$. The obstruction persists with arbitrarily small relative hypergraph codegree.

The meaningful asymptotic target separates the output tolerance from the input regularity error. For fixed rank $r$ and target $\eta >0$, choose a sufficiently small $\delta =\delta (r,\eta )$ and require degrees in $[(1-\delta )D,(1+\delta )D]$ as well as pair hypergraph codegrees at most $\delta D$. The semi-random method then gives ${\chi }^{\prime }(G)\le (1+\eta )D$. Here is the requested proof essay for that corrected conclusion, including its random rounds, tracking requirements and completion step.

A useful way to run all colour classes together is the auxiliary matching construction for hypergraph edge colouring. Use $K=\lceil D\rceil$ initial colours. Create a task vertex for each original edge $e$, and a resource vertex $(v,c)$ for each original vertex and colour. An assignment $(e,c)$ is an $(r+1)$-edge consisting of its task and the resources $(v,c)$ for all $v\in e$. A matching in a hypergraph in this auxiliary system is a proper partial colouring: a task cannot get two colours, and two edges sharing $v$ cannot use the same resource $(v,c)$.

The auxiliary task degrees are $K$, resource degrees are $d_G(v)$, and its pair hypergraph codegrees are at most $\max (1,\delta D)$. It is therefore nearly regular with small hypergraph codegrees. For $r\ge 2$ and a nonempty original hypergraph, the hypergraph codegree condition also forces $D\ge 1/\delta$; taking $\delta$ small absorbs rounding and ensures a large working degree. Rank one is immediate by assigning distinct colours to edges at each vertex.

The extra balance condition is important. Let $S_v$ be the set of tasks corresponding to edges through original vertex $v$. A matching in a hypergraph missing a small fraction of all tasks need not leave few tasks in each $S_v$. The Rödl nibble must leave at most $\beta D$ unmatched tasks in every such star, where one may choose $\beta =\eta /(4r)$. This local requirement, rather than just a global almost-perfect matching in a hypergraph, is what permits a cheap final edge colouring.

Run a Rödl nibble on the auxiliary system, whose rank is $s=r+1$. In a round of working degree $D_j$, sample each surviving assignment independently with probability $\alpha /D_j$, for a small fixed $\alpha$. Accept the sampled assignments that meet no other sampled assignment. Delete every auxiliary vertex touched by any sampled assignment, including conflicted ones, before proceeding. The accepted assignments are a matching in a hypergraph, and later assignments cannot reuse a consumed task or resource. Record each discarded but uncoloured task as waste; it must be counted in the final residual graph.

For an auxiliary vertex of degree $(1\pm \xi )D_j$, the probability of remaining untouched isFor an auxiliary edge through a surviving vertex, the other $s-1$ vertices survive with probability approximately $e^{-(s-1)\alpha }$. Their incident trial families overlap in at most $O_s(\delta D)$ trials by the hypergraph codegree bound, so the overlap error in this product calculation is small. This leads to the degree trajectorywhile the surviving fraction of task vertices is approximately $e^{-\alpha }$ per round. For a sampled assignment, a union bound over at most $s(1+O(\xi ))D_j$ conflicting assignments gives collision probability at most $s\alpha (1+O(\xi ))$. More explicitly, for a fixed task the expected number of ordered sampled conflict pairs having the first assignment through that task is at mostThis bounds its waste probability. Thus productive deletion is first order in $\alpha$, and collision loss is second order.

The technical tracking step makes these expected value calculations locally simultaneous and keeps their errors from accumulating. Its requirements are: residual auxiliary degrees stay near their trajectory after controlled pruning; each original star has the predicted number of surviving tasks; and every star's total collision and pruning waste is small. The standard probabilistic tools enter at distinct points.

Within a fixed round, task-survival indicators are independent: trial families for different tasks are disjoint, even when their assignments share resources. Conditional Chernoff bounds therefore control the number of surviving tasks in each star. Collision loss has short certificates consisting of pairs of selected intersecting assignments. The number selected through one resource has a binomial distribution of bounded mean. Chernoff bounds truncate these numbers at a suitable logarithmic cutoff; after truncation one trial has only a logarithmically bounded effect on the certified collision count. A certifiable function concentration bound, or a corresponding exposure argument, then controls loss in each star. These rare local overload events have polynomially bounded local dependency neighborhoods in the working degree, so the Lovász local lemma can select a simultaneous outcome independent of the total number of vertices.

Residual-degree counts require additional care because they are not sums of independent indicators. Two survival tests share at most a hypergraph codegree's worth of trials; expanding their joint probabilities bounds their covariance by the corresponding relative overlap. For a residual-degree count $Z$ with $D_j$ summands, overlapping summands and shared survival trials give a variance bound of the formChebyshev's inequality then bounds $\mathrm{Pr}(|Z-\mathbb{E}Z|>\xi D_j)$ by $C_s(D_j^{-1}+\zeta )/{\xi }^2$. This gives a small exceptional fraction when the relative hypergraph codegree is sufficiently small; it does not by itself make every degree good. The tracking argument records and prunes these exceptional configurations, bounding their contribution to each star by codegree-weighted exposure and the same local-loss bookkeeping. Markov's inequality controls exceptional mass, and repeated conditional concentration controls the retained counts. One must not apply an independent Chernoff bound directly to a residual degree or infer a uniform star bound just from a global expected loss. Keeping the pruning loss explicitly is the key distinction between a valid Rödl nibble argument and those shortcuts.

Choose the parameters in this order: first $r,\eta$ and $\beta$; next $\alpha$ so small that $s\alpha \log (4/\beta )\ll \beta$; nextand finally the initial regularity/hypergraph codegree tolerance $\delta$ small enough for all $T$ tracking and pruning errors. This is a bounded number of rounds depending on the tolerances, not on the graph order. After $T$ rounds, the surviving task fraction is at most approximately $\beta /4$, while accumulated collision loss is $O_s(T{\alpha }^2)=O_s(\alpha \log (4/\beta ))$. Reserving the remaining margin for tracking and pruning leaves at most $\beta D$ uncoloured edges at every original vertex. The working degree is still a fixed positive multiple of $D$, namely about $D{e}^{-r\alpha T}$, so the large-degree estimates remain applicable throughout.

Finally use a fresh palette on the residual original hypergraph. Its maximum degree is at most $\beta D$; an edge meets at most $r(\beta D-1)$ others. Greedy colouring therefore needs at most $r\beta D+1$ fresh colours. The total is at mostfor sufficiently large $D$, already ensured by choosing $\delta$ small. This completes the passage from the locally balanced Rödl nibble to the hypergraph chromatic index bound. The degree tolerance must be the small input parameter $\delta$; replacing it by the entire output allowance $\eta$ removes the slack and leads precisely to the complete-graph counterexample above.

Expose the independent coordinates in order and let $M_i=\mathbb{E}[f\mid X_1,\dots ,X_i]$, $M_0=\mathbb{E}f$. This is the Doob exposure martingale. For fixed earlier coordinates, write $h_i(x)=\mathbb{E}[f\mid X_i=x,\text{earlier coordinates}]$. The remaining coordinates have the same distribution for all $x$, and the coordinate-change hypothesis gives $|h_i(x)-h_i(y)|\le c_i$. Thus the conditional range of $M_i-M_{i-1}$ has length at most $c_i$, and its conditional mean is zero.

Here is the Hoeffding lemma needed for the sharp constant. If a mean-zero random variable $Y$ lies in an interval of length $c$, set $\psi (s)=\log \mathbb{E}{e}^{sY}$. Its second derivative is the variance under the exponentially tilted law. Under any such law, variance is at most $c^2/4$, by comparison with squared distance from the interval midpoint. Since $\psi (0)={\psi }^{\prime }(0)=0$, integration gives $\psi (s)\le s^2{c}^2/8$ for all real $s$. Applying this conditionally and iterating the tower property yieldsThe exponential Markov bound now givesFor $C>0$ choose $s=4t/C$. Apply the same argument to $-f$ and add the two tails, proving the McDiarmid inequalityFor $C=0$, $f$ is constant almost surely, and every positive-deviation probability is zero.