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Induction with Pascal's identity gives the binomial upper bound for a Ramsey number $R(s,t)\le \left(\genfrac{}{}{0ex}{}{s+t-2}{s-1}\right)$. Thus $R(s)\le \left(\genfrac{}{}{0ex}{}{2s-2}{s-1}\right)\le 4^{s-1}$, proving $R(s)=O(4^s)$.

For the lower bound, take a binomial random graph $G(n,1/2)$. The expected total number of $s$-cliques and independent $s$-sets isFor $n=\lfloor 2^{s/2}/2\rfloor$, this is at most $2^{1-s/2}/s!<1$ for all sufficiently large $s$. Some graph therefore has neither forbidden set, so $R(s)>n$. This proves $R(s)=\Omega (2^{s/2})$ by the probabilistic method.

The binomial upper bound for a Ramsey number gives $R(3,t)\le \left(\genfrac{}{}{0ex}{}{t+1}{2}\right)=O(t^2)$. For the triangle-free Ramsey lower bound by alteration, a binomial random graph $G(n,p)$ has expected triangle count $\left(\genfrac{}{}{0ex}{}{n}{3}\right)p^3$ and expected independent $t$-set count $\left(\genfrac{}{}{0ex}{}{n}{t}\right)(1-p{)}^{\left(\genfrac{}{}{0ex}{}{t}{2}\right)}$. Choose a realization whose sum of these counts is at most its expectation. Delete one vertex from each surviving forbidden set, iterating through the originally listed sets. The remaining graph has neither a triangle nor an independent $t$-set. This random alteration method provesFor large $t$, set $n=\lfloor (t/(4\log t){)}^{3/2}\rfloor$ and $p=n^{-2/3}$. The triangle term is at most $n/6$. Using $\left(\genfrac{}{}{0ex}{}{n}{t}\right)\le (en/t{)}^t$ and $1-p\le e^{-p}$, the logarithm of the independent-set term divided by $t$ is at mostwhich tends to $-\infty$. That term is eventually below $1$. Thus $R(3,t)>5n/6-1$ and

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:

Use natural logarithms. A $K_t$ graph minor is represented by $t$ disjoint nonempty branch sets of a graph minor, each inducing a connected graph, with an edge between every pair. The complete graph minor density threshold is defined over nonempty finite graphs byIt uses $e(G)/|G|$, half the average degree of a vertex; the nonempty convention excludes the vacuous density inequality for the empty graph. The PDF has the intended definition with a universal implication over $G$; the TeX transcription corrupts that definition.

Probabilistic lower bound. Put $N=\lfloor t\sqrt{\log t}/4\rfloor$ and take the binomial random graph $G(N,1/2)$. Its edge count has mean $N(N-1)/4$. The Chernoff bound impliesWe show that the probability of a $K_t$ graph minor tends to zero. Ignore connectedness, which only enlarges the set of possible models, and assign each of the $N$ vertices one of $t$ branch labels or an unused label. There are at most $(t+1{)}^N$ assignments.

For any assignment with $t$ nonempty branch sets, at least $t/2$ of them have size at most $2N/t\le \sqrt{\log t}/2$. Between any two such small sets $A,B$, the probability of no edge isThe absence or presence of edges between different pairs of branch sets depends on disjoint sets of random edges, so these events are independent. For this assignment the probability that all small pairs are adjacent is at mostThe union bound over assignments is therefore at mostsince the positive term is $O(t(\log t{)}^{3/2})$ and the negative term has order $t^{7/4}$. There is consequently a graph with no $K_t$ graph minor and with $e(G)/N\ge N/8$. Any constant having the universal forcing property must exceed this ratio. Since $N/8\ge t\sqrt{\log t}/64$ for large $t$,

The auxiliary density lemma contains a typo. Both the original PDF and the TeX print $e(G)\le 11k|G|$ in the hint. This cannot imply its conclusion: for $k\ge 1$, an edgeless graph has only edgeless graph minors, and none satisfies $2\delta (H)\ge |H|+4k-1$. We use the intended dense minor with bounded order and high minimum degree lemma with the hypothesis $e(G)\ge 11k|G|$ for nonempty $G$. The upper-bound argument below depends on that corrected supplied lemma; the literal printed hint is false.

Probabilistic upper bound and the constant seven. Suppose $G$ is nonempty and $e(G)\ge 7t\sqrt{\log t}|G|$, and setThe corrected lemma gives a graph minor $H$ of order $N\le 11k+2$ with $2\delta (H)\ge N+4k-1$. Since $\delta (H)\le N-1$, we also have $N\ge 4k+1$. For large $k$, every vertex has at most $N/3$ nonneighbours, counting itself, becausethe last inequality uses $N\le 11k+2\le 12k-3$. Moreover any two distinct vertices have at leastcommon neighbours.

Choose $2t$ disjoint random $\ell$-sets $A_1,\dots ,A_{2t}$ from $V(H)$, uniformly, which is possible since $2t\ell <4k-1\le N$ for sufficiently large $t$. For a set $A$, let $U(A)$ be the set of vertices having no neighbour in $A$. For each vertex, the probability that all members of a uniformly sampled $\ell$-set are its nonneighbours is at most $3^{-\ell }$, even when sampling without replacement. By linearity of expected value,Call $A$ good if $|U(A)|\le N{3}^{-19\ell /20}$. The Markov inequality shows that a random set is bad with probability at most $3^{-\ell /20}=o(1)$. Thus the expected number of bad sets among our $2t$ samples is $o(t)$.

Conditional on a fixed good $A_i$, the marginal distribution of $A_j$ is uniform among $\ell$-sets outside $A_i$. For there to be no edge between them, all its members must lie in $U(A_i)$. HenceHere $(N/(N-\ell ){)}^{\ell }=\exp (O({\ell }^2/N))=1+o(1)$. Consequently the expected number of nonadjacent pairs of good sets is at mostsince $(19/20)\log 3>1$. Applying the Markov inequality to both counts shows that there exists a choice with fewer than $t$ bad sets and at most $t$ nonadjacent pairs of good sets. Select $t$ of the good sets; they still have at most $t$ missing adjacencies.

Make each selected set connected by fixing one of its vertices as a root and, for each of its other members, adding one unused common neighbour of that member and the root. This uses at most $t(\ell -1)$ additional vertices. Then for each pair with no original edge, choose one vertex in each set and add an unused common neighbour to one of the sets. The added vertex is joined to that set and to the other set, so it preserves connectedness and repairs their adjacency. At most $t$ such repairs are required.

Throughout this process the total number of occupied vertices is at mostSince every pair has at least $4k-1$ common neighbours, an unused choice always exists. The final sets are disjoint connected branch sets of a graph minor with every pair adjacent. They represent $K_t$ in $H$, and hence in $G$, by transitivity of graph minors. This random branch-set construction of a complete graph minor proves

The sharp linear threshold for a four-vertex complete minor. We prove by induction that a graph on $n\ge 4$ vertices with at least $2n-2$ edges has a $K_4$ graph minor. The case $n=4$ is $K_4$ itself. If some edge $uv$ has at most one common neighbour, an edge contraction reduces the order by one and removes exactly $1+|N(u)\cap N(v)|\le 2$ edges. The contracted graph has at least $2(n-1)-2$ edges, so induction applies.

Otherwise every edge is in at least two triangles in a graph. Choose a vertex $v$ incident with an edge. Each vertex $u\in N(v)$ has at least two neighbours of a vertex in $N(v)$, since these are the common neighbours of $u,v$. Thus $G[N(v)]$ has minimum degree of a graph at least two. A longest path in this finite graph has an endpoint adjacent to an earlier nonconsecutive path vertex, and hence contains a cycle in a graph. Together with $v$, this cycle in a graph gives a wheel graph as a subgraph. Divide its rim into three consecutive nonempty connected arcs; these arcs and $\{v\}$ form four pairwise adjacent branch sets of a graph minor. Therefore

With $2n-3$ edges the answer is no. Take the join of graphs of $K_2$ and an independent set of $n-2$ vertices. It has $1+2(n-2)=2n-3$ edges. Among four disjoint putative branch sets of a graph minor, at most two can contain the two vertices of $K_2$. Any connected branch set avoiding them is a singleton in the independent set. At least two such singleton sets would therefore have no edge between them. This excludes a $K_4$ graph minor for every $n\ge 4$ and proves sharpness.

The binomial random graph $G(n,p)$ has vertex set $[n]=\{1,\dots ,n\}$, and each unordered pair of distinct vertices is included as an edge independently with probability $p$.

Let $X_t$ count copies of the complete graph $K_t$. Each $t$-element vertex set forms a copy with probability $p^{\left(\genfrac{}{}{0ex}{}{t}{2}\right)}$, so linearity of expectation givesThe event $E_t$ is $\{X_t\ge 1\}$. By the first moment method,This is the clique count in a binomial random graph.

Let $X$ count the triangles of the binomial random graph $G(n,p)$. ThenTwo distinct triangle indicators are dependent only when the triangles share an edge. Hence their ordered Janson dependency sum is

If $0<p\le n^{-1/2}$, then $\Delta =O(\mu )$. The first Janson inequality givesFor the reverse bound, the events that individual triangles are absent are decreasing, so Harris' inequality giveswhere $p<1/2$ keeps the logarithmic estimate uniform. Thus the probability is $e^{-\Theta (p^3{n}^3)}$.

If $n^{-1/2}\le p<1/2$, the extended Janson bound givesFor a lower bound, fix a balanced bipartition of the vertices and require every edge inside either part to be absent. The resulting graph is bipartite, hence triangle-free, and this event has probabilityCombining the bounds proves the second regime.

Letwhere the positive constant $c$ will be chosen small, and sample the binomial random graph $G(N,p)$. If $X$ counts its copies of $C_4$, thenIf $Y$ counts its independent $k$-sets, thenHere $pk\sim c^{-3/4}\log k$. Choosing $c$ sufficiently small makes the negative exponential term dominate $k\log (eN/k)$, so $\mathbb{E}Y=o(1)$. By Markov inequality, with positive probability $X<N/2$ and $Y=0$.

Starting from such a graph, delete one vertex from each remaining $C_4$. This random alteration method removes fewer than $N/2$ vertices, destroys every $C_4$, and cannot create an independent set of order $k$. The resulting graph has at least $N/2$ vertices, so its edges and nonedges give a red-blue colouring with neither a red $C_4$ nor a blue $K_k$. Consequently

In a binomial random graph with fixed edge probability below one, the displayed first-moment bound tends to zero for $t=C\log n$ and sufficiently large fixed $C$. At the same time the edge density concentrates near $p$. Thus a positive fixed density cannot force balanced complete multipartite subgraphs whose part size grows faster than logarithmically.

For $X_j$ the number of tree components of order $j$ in a binomial random graph, choose their vertices, choose one of $j^{j-2}$ labelled trees by the Cayley formula, require its $j-1$ edges, and exclude both the remaining internal edges and all crossing edges. For $j=1$ interpret $j^{j-2}=1$, recovering the isolated vertex count. If $p=\lambda /n$ for fixed positive $\lambda$ and $j=O(\log n)$, the Stirling formula gives $\mathbb{E}{X}_j\sim n(\lambda e^{1-\lambda }{)}^j/(\lambda \sqrt{2\pi }{j}^{5/2})$. Distinct overlapping vertex sets cannot both be graph components; for disjoint sets their joint occurrence gains the factor $(1-p{)}^{-j^2}$ relative to the product, because their between-set edges must be absent only once.

Start with the empty graph and reveal the edges of the complete graph in a uniformly random order. After $m$ steps its edge set is a uniform $m$-element subset. Independent uniform edge labels couple this process with all binomial random graphs: $G(n,p)=G_{n,M(p)}$, where $M(p)$ has a binomial distribution with parameters $\left(\genfrac{}{}{0ex}{}{n}{2}\right),p$. For a monotone graph property, concentration of $M(p)$ transfers a threshold estimate between edge probability and edge count.