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It suffices to prove the assertion after replacing $\omega$ by $\min \{\omega ,\sqrt{\log n}\}$: that replacement gives a tighter interval. Thus assume $\omega \to \infty$ and $\omega =o(\log n)$, and putAt $p_{-}$, let $J$ count isolated vertices in $L$. The same indicator random variable calculation as for all isolated vertices givesBy the Chebyshev inequality, $J>0$ with high probability. Since $\ell \ge 2$ eventually, the marked vertices are then not all connected.

At $p_{+}$, the logarithmic-regime giant component result applies with $k=2$ and any fixed ${\gamma }_0\in (1/3,1/2)$. Its unique giant component misses $(1+o(1))n^{1/2}{e}^{-\omega /4}$ vertices. On the event that it misses at most twice that number, exchangeability makes the missed vertex set uniform conditional on its size. A union bound therefore givesThus all of $L$ is connected at $p_{+}$ with high probability.

For the uniform random graph process, assign uniform labels to the edges of the complete graph as independent random variables and reveal them in label order. The binomial random graph $G(n,p)$ is the prefix containing the $M(p)$ labels at most $p$, where $M(p)$ has a binomial distribution with parameters $N=\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ and $p$. Its variance at $p_{\pm }$ is $O(n\log n)$, whereas the gap between $\mathbb{E}M(p_{\pm })$ and the relevant endpoint $m_{\pm }$ is $(1+o(1))\omega n/8$. The Chebyshev inequality gives $M(p_{-})\ge m_{-}$ and $M(p_{+})\le m_{+}$ with high probability, with harmless integer rounding. The monotone graph property that all marked vertices are connected now impliesThe marked-set connectivity threshold is lower than the threshold for connecting every vertex, because only about $\sqrt{n}$ specified vertices need to avoid the small graph components.

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.

Write $K_s(t)$ for the balanced graph blow-up of a complete graph, with $s$ classes of size $t$; containment here is as a subgraph, not necessarily an induced subgraph. The logarithmic form of the Erdős-Stone theorem is: for fixed $r\ge 1$ and $0<ϵ<1/r$, there are $d=d(r,ϵ)>0$ and $n_0$ such thatThus the guaranteed balanced part size is at least $\lfloor d\log n\rfloor$, with $d$ independent of $n$. We prove the logarithmic Erdős-Stone theorem by a density-to-cliques step and a constructive dense clique family blow-up lemma.

First choose a fixed $m$ large enough that $t_r(m)/\left(\genfrac{}{}{0ex}{}{m}{2}\right)\le 1-1/r+ϵ/2$, where $t_r(m)$ is the edge count of the Turan graph. A uniformly chosen $m$-vertex subset has expected edge count at least $(1-1/r+ϵ)\left(\genfrac{}{}{0ex}{}{m}{2}\right)$. If its induced graph contains no $K_{r+1}$, the Turan theorem bounds that count by $t_r(m)$. Since the count is always at most $\left(\genfrac{}{}{0ex}{}{m}{2}\right)$, the probability that the subset contains an $(r+1)$-clique is at least $ϵ/2$. Double counting the incidences between such subsets and their cliques gives, for $s=r+1$,for a fixed $\eta >0$ and all sufficiently large $n$. Here $(x{)}_s=x(x-1)\cdots (x-s+1)$. This is clique supersaturation by sampling.

We now prove the needed dense clique family blow-up lemma, with the following stronger induction invariant. Given any family $\mathcal{M}$ of at least $\eta n^s$ distinct $s$-cliques, there is a complete $s$-partite subgraph with each class of size $\lfloor a_s(\eta )\log n\rfloor$, containing that many pairwise vertex-disjoint transversal members of $\mathcal{M}$, each with one vertex in every class. Extra edges within its classes can be ignored. We may suppose $0<\eta <1/2$. For $s=1$, any $\eta n$ singletons suffice, and we may take $a_1(\eta )=1$ once $n$ is large.

For $s\ge 2$, put $\theta =\eta /2$. Repeatedly delete every member of the current family containing an $(s-1)$-clique whose number of extensions is at most $\theta n$. Each such face is processed at most once, and there are at most $n^{s-1}$ possible faces. At most $\theta n^s$ members are deleted. The remaining family $\mathcal{L}$ has size at least $(\eta /2)n^s$, and every face that remains has more than $\theta n$ extensions. Its family of $(s-1)$-faces has size at least $s|\mathcal{L}|/n\ge (\eta /2)n^{s-1}$, since each face is in at most $n$ members.

Apply the induction hypothesis to these faces. It gives a complete $(s-1)$-partite subgraph with $m=\lfloor a_{s-1}(\eta /2)\log n\rfloor$ vertices in each class and a matching $R_1,\dots ,R_m$ of transversal $(s-1)$-faces from the family. Form a bipartite graph whose left vertices are the $R_i$ and whose right vertices are the original $n$ vertices; join $R_i$ to $v$ when $R_i\cup \{v\}\in \mathcal{L}$. Every left degree exceeds $\theta n$.

The following common neighbourhood from bipartite density estimate is elementary. In a bipartite graph with class sizes $m,n$ and at least $\theta mn$ edges, averaging over left subsets $S$ of size $u$ and using convexity of the integer sequence $j\mapsto \left(\genfrac{}{}{0ex}{}{j}{u}\right)$ giveswhenever $1\le u\le \theta m/2$. The last inequality follows by comparing the $u$ factors in the two binomial coefficients; the integer convexity follows from the nondecreasing first differences $\left(\genfrac{}{}{0ex}{}{j}{u-1}\right)$.

ChooseFor large $n$, $m\ge \frac{1}{2}{a}_{s-1}(\eta /2)\log n$, so $u\le \theta m/4$. The estimate supplies $u$ selected faces with a common extension set of size at least $n(\theta /2{)}^u\ge \sqrt{n}$. This extension set is disjoint from the selected faces: a vertex in any one of them cannot extend that face. Their union has $u$ vertices in each of the old $s-1$ classes, with all required cross edges; choose $u$ distinct common extension vertices as a new class. Attaching one different extension vertex to each selected face also gives $u$ disjoint members of $\mathcal{L}$. This completes the induction, with positive constants independent of $n$. Apply it to the clique family above and take $d=a_{r+1}(\eta )$.

The logarithmic order cannot be increased in a uniform forcing result. Put $\rho =1-1/r+ϵ<1$, choose $p$ strictly between $\rho$ and one, and take a binomial random graph. Its edge density exceeds $\rho$ with probability tending to one, by the variance estimate for a sum of independent edge indicators. With $s=r+1$, the expected number of ordered embeddings of $K_s(t)$ is at mostFor $t=\lceil C\log n\rceil$ and $C>2/((s-1)\log (1/p))$, the logarithm of this bound is negative of order $(\log n{)}^2$. Markov's inequality therefore makes the probability of any such copy tend to zero. Both events hold simultaneously for some graphs of every sufficiently large order. There are graphs meeting the density hypothesis whose largest balanced blow-up has part size $O(\log n)$. This upper bound concerns what density can force; it is not an upper bound for every dense graph, since a complete graph has much larger blow-ups.

Finally, $H_k$ is the square of a cycle, for $k\ge 3$. Any three consecutive vertices form a triangle in a graph. In a proper three-colouring, once the first three colours are fixed, each subsequent vertex must repeat the colour three positions earlier. Closing the cycle is possible precisely when $3\mid k$. Conversely, repeating the three colours gives such a colouring whenever $3\mid k$. In that case $H_k$ embeds in $K_3(k/3)$, and the density $0.51\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ exceeds the bipartite Turan theorem threshold by a fixed amount, so the theorem forces $H_k$ eventually. If $3\nmid k$, the balanced Turan graphs $T_3(n)$ have density at least $2/3$ and contain no graph of chromatic number greater than three. They give a counterexample sequence. Exactly the cycle lengths $k\ge 3$ divisible by three have the required property.

Use the identification of two-element subsets of $[n]$ with the edges of the complete graph $K_n$. A colour class in the Kneser graph $KG(n,2)$ is an intersecting two-element set family, so its edges must pairwise meet.

Such a class is contained either in a star graph or in a triangle in a graph. Indeed, if not all edges have a common vertex, take two meeting edges $ab,ac$. An edge not containing $a$ must then be $bc$. Any further edge meeting all three of $ab,ac,bc$ is one of these three. Classes with at most two edges already lie in a star graph.

Assume that a graph colouring used $c\le n-3$ colours. Classify $s$ classes as stars and the remaining $t=c-s$ classes as triangles. Delete a chosen centre of each star class, leaving $m\ge n-s\ge t+3$ vertices. Every edge between those remaining vertices must belong to a triangle class, of which each covers at most three edges. HenceButa contradiction. Thus $\chi (KG(n,2))\ge n-2$. For the matching upper bound, assign a two-set the colour of its smaller element if that element is at most $n-3$, and otherwise give it colour $n-2$. The last class comprises the three pairs on the last three elements, an intersecting family. Consequently an entirely combinatorial argument gives

For each unordered triple of distinct vertices, let $I_A$ be the indicator variable that its three edges are present. This is a binomial random graph, so edge independence gives $\mathbb{E}[I_A]=p^3$. The triangle count in a binomial random graph is $T={\sum }_A{I}_A$. There are $\left(\genfrac{}{}{0ex}{}{n}{3}\right)$ such triples; all are triangles in the complete graph.

The combinatorial maximum and expectation areThe maximum is attainable with positive probability if $p>0$; for $p=0$, the random variable is identically zero. The linearity of expectation requires no independence between different triangle indicators.

Let $X_s$ count the copies of the complete graph $K_s$ in the binomial random graph $G_{n,p}$. By linearity of the expected value and independence of distinct edges,Throughout, $n$ is sufficiently large that $k=\lfloor \log \log n\rfloor \ge 2$. We have $k\to \infty$, $k=O(\log \log n)$ and $k\log k=o(\log n)$.

First, since $p=n^{-2/k}$,The first moment method proves $\mathbb{P}(X_{k+1}>0)=o(1)$.

For the existence of a $k$-clique, put $X=X_k$ and $\mu =\mathbb{E}X$. Since $p^{\left(\genfrac{}{}{0ex}{}{k}{2}\right)}=n^{-(k-1)}$,This alone is insufficient: we must control the variance via the overlap formula for the variance of a clique count. Let $I_S$ indicate that a given $k$-set $S$ spans a clique. If $|S\cap T|=j$, the two copies share $\left(\genfrac{}{}{0ex}{}{j}{2}\right)$ edges, soFor $j=0,1$ their edge sets are disjoint and the covariance vanishes, even when they share one vertex. Counting ordered pairs of $k$-sets gives the exact formulaThis includes the diagonal case $j=k$.

To bound the overlap factor, regard it as the probability that a uniformly chosen $k$-set $T$ intersects a fixed $S$ in exactly $j$ vertices. A union bound over the possible specified $j$-subsets of $S$ yieldsHere $(a{)}_j=a(a-1)\cdots (a-j+1)$ is a falling factorial, and the last bound holds for $n\ge 2k$. ThereforeFor $2\le j\le k$, the concave quadratic $j(k-j+1)$ is minimized at an endpoint. Its endpoint values are $2(k-1)$ and $k$, both at least $k$. Thus every power of $n$ in the last sum is at most $n^{-1}$, andbecause the logarithm of its numerator is $O(k\log k)=o(\log n)$. The second moment method, or Chebyshev's inequality directly, gives $\mathbb{P}(X=0)=o(1)$. Finally, every larger clique contains a $(k+1)$-clique, so the absence of $K_{k+1}$ excludes all larger sizes. The clique number is exactly $k$ with high probability:

Pull the given finite coloring $\chi :\mathbb{N}\to [k]$ back to a finite coloring of unordered pairs by puttingThe ordering $a<b$ makes this a well-defined function on the edges of a complete graph. By Ramsey's theorem, some infinite set $X=\{x_1<x_2<\cdots \}$ has all its pairs of one color $r$ under $\tilde{\chi }$. ConsequentlyPossible coincidences between these products cause no difficulty, since the original finite coloring assigns each positive integer a single color.

Write $\mathbb{N}=\{1,2,\dots \}$ and let $\chi$ be the given finite coloring of the edges of the complete graph on the positive integers. Construct an increasing sequence $(v_i)$ and nested infinite sets $A_0\supset A_1\supset \cdots$ as follows. Set $A_0=\mathbb{N}$. Having chosen $A_{i-1}$, take $v_i=\min A_{i-1}$, partition $A_{i-1}\setminus \{v_i\}$ by the color of $\{v_i,w\}$, and use the infinite pigeonhole principle to choose an infinite set $A_i$ on which this color is constant, say $r_i$. Every element of $A_i$ exceeds $v_i$.

The infinite pigeonhole principle applied again to $(r_i)$ gives an infinite set of indices $I$ and a color $r$ such that $r_i=r$ for all $i\in I$. For $i<j$ in $I$, nesting gives $v_j\in A_i$, so $\chi (\{v_i,v_j\})=r$. Thus the infinite set $X=\{v_i:i\in I\}$ is a monochromatic complete graph:This proves the required infinite case of Ramsey's theorem; finiteness of the color set is used at both applications of the infinite pigeonhole principle.

Let $R$ interchange the two complete graphs, sending $v_i$ to $v_i^{\prime }$ and fixing $w$. This involution preserves the transition matrix of the simple random walk. A mirror coupling under an involution starts $X$ at $v_i$, keeps $Y_t=R(X_t)$ until the hitting time $\tau$ of $w$, and then makes the two walks move identically. Each coordinate has the correct transition probabilities, and they coalesce precisely at $\tau$. The coupling inequality for total variation gives

There is a genuine error in the printed expected hitting time bound, as well as an unspecified initial law. For $n\ge 2$, write $a={\mathbb{E}}_{v_1}\tau$ and $b={\mathbb{E}}_{v_i}\tau$ for $i\ne 1$. From every clique vertex there is a positive chance of reaching $w$ within two steps, uniformly over the finite state space for each fixed $n$, so $\tau$ has finite expected value. First-step analysis, using the degree of a vertex $n$ at $v_1$ and $n-1$ at the other clique vertices, givesSolving these linear equations yieldsIn particular even the attachment vertex exceeds the printed bound by one. The valid uniform replacement is ${\max }_x{\mathbb{E}}_x\tau \le n^2$, the hitting time of the bridge vertex between two cliques formula.

We must also handle arbitrary starting states, including $w$, before applying the mirror argument. Identify related vertices and consider the lumped Markov chain $Z$ on $\{1,\dots ,n,w\}$. Its transition matrix $Q$ has $Q(1,j)=1/n$ for $j\ne 1$, $Q(1,w)=1/n$, $Q(i,j)=1/(n-1)$ for $i\ne 1$ and $j\ne i$, and $Q(w,1)=1$, with all other entries zero. Here the row for $i\ne 1$ only ranges over $j\in \{1,\dots ,n\}$.

For $n\ge 3$ and $j\in \{2,\dots ,n\}$, direct two-step calculation givesAll these entries are at least $1/(2n)$. Thus $Q^2(z,\cdot )\ge \alpha \nu (\cdot )$ for every $z$, where $\nu$ is the uniform distribution on a finite set $\{2,\dots ,n\}$ and $\alpha =(n-1)/(2n)\ge 1/3$. This is a Doeblin condition. At each two-step block couple the two quotient endpoints to the same sample from $\nu$ with probability $\alpha$, using the residual probability distributions otherwise. The first successful block has expected value at most $1/\alpha$, so the alignment time $A$ has $\mathbb{E}A\le 6$. Lift each endpoint coupling of probability distributions to the original walks by their conditional two-step path laws; this preserves every marginal path law. The success test uses fresh block randomness and so permits restarting the walks with their original transition matrix at the aligned endpoints.

At alignment the original walks are equal or related. Use identical transitions in the first case and the mirror coupling under an involution in the second. The resulting coalescing coupling has coalescence time $T$ satisfying $\mathbb{E}T\le 6+n^2$ for every pair of initial states. Therefore Markov inequality and the pairwise mixing diameter implyFor the hint's particular starting states, the quotient one-step laws from two different clique indices have a common component of mass at least $1-2/n$: for two ordinary indices it is $(n-2)/(n-1)$; for one attachment index it is $(n-2)/n$. A maximal coupling therefore aligns those indices with probability $1-O(1/n)$, as suggested. The two-step argument above additionally covers $w$.

The nonlazy walk is an aperiodic Markov chain for $n\ge 3$, because the connected graph contains a triangle. For $n=1,2$ it is a periodic Markov chain on a bipartite graph; each bipartition class has stationary mass $1/2$, and the walk stays on a single class at each time. Hence $d(t)\ge 1/2$ and the $1/4$ mixing time is infinite. The asymptotic conclusion is valid for $n\ge 3$.

Use the ratio definition of cutoff from part (a), and write $T_n=t_{\mathrm{mix}}^{(n)}(1/4)$. In the reversible Markov chain setting of the preceding parts let ${\gamma }_n=1-{\lambda }_{2,n}$ and ${\gamma }_{\ast ,n}=1-{\max }_{j\ge 2}|{\lambda }_{j,n}|$. In particular ${\gamma }_{\ast ,n}\le {\gamma }_n$.

If ${\gamma }_n{T}_n$ did not tend to infinity, there would be a subsequence and $B>0$ for which ${\gamma }_n{T}_n\le B$. Along it part (c) implies, for every fixed $0<\epsilon <1/2$,Since $T_n\to \infty$, choose fixed $\epsilon$ so small that $\log (1/(2\epsilon ))>B$. The lower limit of this ratio is then greater than one, contradicting cutoff for Markov chains. Thus the Peres product condition holds:In fact the argument with ${\gamma }_{\ast ,n}$ proves the stronger absolute-gap product condition. Part (d) does not repeat the reversible Markov chain hypothesis; the ordinary real ordering of eigenvalues above uses that hypothesis. If arbitrary chains are intended, define the spectral gap explicitly: the eigenvalue lower bound for total variation mixing works for complex eigenfunctions too, and proves the same assertion for ${\gamma }_{\ast }$ or for $1-{\max }_{j\ge 2}\mathrm{Re}{\lambda }_j$, which is at least ${\gamma }_{\ast }$. No real ordering of complex eigenvalues is implicit.

For the requested counterexample take nonlazy simple random walk on the complete graph $K_n$, $n\ge 3$. Let $\Pi$ have every row equal to the uniform distribution on a finite set, and let $a=-1/(n-1)$. ThenThus $d_n(0)\to 1$ and $d_n(1)=1/n\to 0$: every fixed-level mixing time is eventually one. This is a bounded-time jump satisfying the ratio definition of cutoff. The ordinary spectral gap is $n/(n-1)$ and the absolute one is $(n-2)/(n-1)$, so both products with $T_n=1$ tend to one, not infinity. The divergence assumption on $T_n$ is therefore essential under this convention.

A graph Ramsey number $R(s,t)$ is the least $N$ such that every red-blue edge colouring of the complete graph $K_N$ has a red $K_s$ or a blue $K_t$. At a vertex in $K_{R(s-1,t)+R(s,t-1)}$, either at least $R(s-1,t)$ neighbours have red incident edges or at least $R(s,t-1)$ have blue incident edges. In the first case find a red $K_{s-1}$ and adjoin the vertex, or a blue $K_t$; the second case is symmetric. ThusStarting with $R(2,t)=t$ and $R(s,2)=s$, induction gives the binomial upper bound for a Ramsey numberFor the lower construction, partition $2t-2$ vertices into two classes of size $t-1$, colour inside-class edges red and between-class edges blue. Red cliques have at most $t-1$ vertices, and the blue graph is bipartite, so has no odd cycle.

Conversely, a graph without an odd cycle is bipartite: in each connected component colour vertices by the parity of the length of a path from a root; differing path parities would give an odd closed walk, hence an odd cycle. On $2t-1$ vertices one of the two blue bipartition classes has at least $t$ vertices. Every edge within it is red, giving a red $K_t$. Thus $2t-1$ is the exact threshold for this alternative.

A $k$-colouring assigns one of $k$ colours to every vertex so that adjacent vertices have different colours. The chromatic number $\chi (G)$ is the least such $k$. An independent set contains no adjacent pair, and the independence number $\alpha (G)$ is the maximum size of such a set.

For a concrete first example, start with the five-cycle $C_5$ and apply the Mycielski construction $r-2$ times. Each application raises the chromatic number by one and preserves triangle-freeness. The resulting graph has chromatic number $r+1>r$ and contains no triangle, hence no complete graph $K_r$ for $r\ge 3$.

We next prove the stronger high-girth assertion by the probabilistic method. For a large integer $n$, takeIf $X$ counts cycles of lengths $3,\dots ,g$, thenBy the first moment method, with probability tending to one $X<n/2$.

Let $m=\lfloor n/(2k)\rfloor$. The expected number of independent sets of size $m$ is at mostbecause $p{m}^2=\Theta (n^{1+1/(2g)})$ dominates $m\log (en/m)=O(n)$. Hence with positive probability $G$ has fewer than $n/2$ short cycles and $\alpha (G)<m$.

Delete one vertex from each cycle of length at most $g$. The remaining graph $H$ has no such cycle, has at least $n/2$ vertices, and still has $\alpha (H)<n/(2k)$. Since each colour class is independent,This proves that there are graphs of arbitrarily high girth and chromatic number.

For the final claim, begin instead withThe expected number $T$ of triangles isso with probability tending to one $T<n$. Put $m=\lceil n^{0.7}\rceil$. The expected number of independent $m$-sets is bounded bybecause the positive term is $O(n^{0.7}\log n)$ whereas the negative term has order $n^{0.71}$.

Thus some such $G$ has fewer than $n$ triangles and no independent set of size $m$. Delete one vertex from every triangle; more than $n$ vertices remain and the graph is triangle-free. Take any induced subgraph on exactly $n$ vertices. Vertex deletion cannot increase the independence number, so the resulting graph satisfiesThis is the triangle-free graph with sub-power independence number construction.

The diagonal Ramsey number $R(s)$ is the least positive integer $N$ such that every red-blue edge colouring of the complete graph $K_N$ contains a monochromatic copy of $K_s$.

For the existence and bound, write $R(r,s)$ for the least $N$ forcing either a red $K_r$ or a blue $K_s$. Given a vertex $v$ of $K_{R(r-1,s)+R(r,s-1)}$, the pigeonhole principle says that at least $R(r-1,s)$ incident edges are red or at least $R(r,s-1)$ are blue. In the first case those red neighbours contain a red $K_{r-1}$, which joins to $v$ to make a red $K_r$, or a blue $K_s$; the second case is symmetric. HenceStarting from $R(1,s)=R(r,1)=1$ and using Pascal's identity gives the binomial upper bound for a Ramsey numberThereforewhich proves both existence and the required estimate.

To determine $R(3)$, color a cycle graph $C_5$ red and its complement graph, also a $C_5$, yellow. This colors $K_5$ without a monochromatic triangle, so $R(3)>5$. At any vertex of a red-yellow $K_6$, at least three of the five incident edges have one color. If an edge among their three other endpoints has that color, it completes a monochromatic triangle with the chosen vertex; if none does, those three endpoints themselves form a triangle of the other color. Thus $R(3)\le 6$, and

The same local argument proves uniqueness on $K_5$. No vertex can have three incident edges of one color: their other endpoints would either contain an edge of that color or form a triangle of the other color. Every vertex therefore has vertex degree at most two in each color, and because the two degrees sum to four, each is exactly two. Each color class is consequently a regular graph of degree two on five vertices, hence a single $C_5$. Up to relabelling, the coloring is exactly the cycle-and-complement coloring above.

Finally, the monochromatic-triangle counting formula gives, for a coloring of $K_6$ with red degrees $d_v$,The bound is attained by coloring all edges of $K_{3,3}$ red and the two complementary copies of $K_3$ yellow: there are no red triangles and exactly two yellow triangles. The least requested number is therefore