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Let $X_t$ count $t$-cliques, and put ${\mu }_t=\left(\genfrac{}{}{0ex}{}{n}{t}\right)p^{\left(\genfrac{}{}{0ex}{}{t}{2}\right)}$. Since ${\mu }_{r+1}\to 0$, the Markov inequality gives $\mathrm{Pr}(X_{r+1}>0)\to 0$. Any larger clique contains an $(r+1)$-clique, so it remains to prove $X_r>0$ with high probability.

We give the required growing-clique overlap bound explicitly. Set $\mu ={\mu }_r$ and, for $2\le s\le r$,The quantity $a_s$ is the probability that a uniformly chosen $r$-set overlaps a fixed $r$-set in exactly $s$ vertices. Two clique indicators share exactly $\left(\genfrac{}{}{0ex}{}{s}{2}\right)$ edge requirements, so their joint-to-product ratio is $p^{-\left(\genfrac{}{}{0ex}{}{s}{2}\right)}$. Overlap zero or one gives independence. Hence the overlap formula for the variance of a clique count impliesWe must bound this sum with $r$ allowed to grow.

Eventually ${\mu }_r\ge 1$ and ${\mu }_{r+1}\le 1$. Using $(n/t{)}^t\le (\genfrac{}{}{0ex}{}{n}{t})\le (en/t{)}^t$, these inequalities giveFor $2\le s\le r/2$, a union bound over the $s$ selected shared vertices gives $a_s\le \left(\genfrac{}{}{0ex}{}{r}{s}\right)(r/(n-r){)}^s$. Put $a=(s-1)/(r-1)<1/2$. Since $n\ge r^3$ and $n-r\ge n/2$,For each fixed $s$, the first bound tends to zero if $r$ stays bounded, because its power of $n$ is negative. If $r\to \infty$, the middle bound tends to zero because $a\to 0$. Thus each fixed-$s$ term tends to zero along arbitrary admissible sequences. The bounding series ${\sum }_{s\ge 2}(2e^2/s{)}^s$ converges. Splitting at a fixed large index, then taking $n\to \infty$ and the index to infinity, proves that the entire lower-half sum is $o(1)$.

For the upper half write $s=r-j$. The exact diagonal comparison isFor $1\le j\le r/2$, the second bound on $p$ givesThe last inequality uses $n\ge r^3$: the exponent of $n$ in the bracket is negative, so substituting $n=r^3$ bounds it above. Uniformly for this range of $j$, $r^{2(j+1)/r}\le 9j$. Indeed $r^{2/r}\le 3$, and concavity of $\log (3j)-(2j/r)\log r$, checked at $j=1$ and $j=r/2$, gives $r^{2j/r}\le 3j$. It follows thatThus $\mathrm{Var}{X}_r=o({\mu }^2)$. The second moment method gives $\mathrm{Pr}(X_r=0)\to 0$. Combining this with the absence of larger cliques provesThe restriction on $r$ is used in both overlap ranges; divergence of a clique count in a binomial random graph alone would not suffice for arbitrary growing $r$.

We need independent sets in every sufficiently large remaining induced subgraph, rather than just one in the initial random graph. The useful concentration variable is the maximum edge-disjoint clique packing in the complement graph; changing one edge changes that variable by at most one. This avoids the much larger sensitivity of the ordinary clique count in a binomial random graph.

First work in a binomial random graph $H\sim G(m,1/2)$, take $r=r_0(m)$, and let $X$ count its $r$-cliques. Put $\mu =\mathbb{E}X\ge m^{9/5}$ and let $Z$ be its maximum edge-disjoint clique packing size. Form the clique conflict graph whose vertices are these cliques, adjacent when they share an edge. If $Q$ counts unordered conflicting pairs, the Caro-Wei bound gives, for each realization,To verify the bound, randomly order the vertices of a finite graph and retain each vertex preceding all its graph neighbours. These retained vertices are an independent set, with expected size ${\sum }_v(1+\mathrm{deg}v{)}^{-1}\ge X^2/(X+2Q)$ by the Cauchy-Schwarz inequality. Interpret the ratio as zero when $X=0$. A second application of the Cauchy-Schwarz inequality, now to expectations, givesCounting ordered pairs of $r$-sets with $j$ common vertices, including identical sets when $j=r$, yieldsThe shared edges account for the factor $2^{\left(\genfrac{}{}{0ex}{}{j}{2}\right)}$. We claimHere are the estimates, including the large overlaps. For $2\le j\le \lfloor r/2\rfloor$, the probability that a random $r$-set contains any specified $j$ elements is at most $(r/(m-r){)}^j$. Summing over their choices inside the first $r$-set givesThe logarithm of $T_j$ is a convex function, a quadratic in $j$, so the maximum on this interval is at an endpoint. We have $T_2=O(r^4/m^2)$. As $r=(2+o(1)){\log }_{2}m$, the other endpoint satisfiesand is smaller than $T_2$ for sufficiently large $m$. Summing at most $r$ terms gives $O(r^5/m^2)$.

For the other half of the overlaps, write $j=r-s$, where $0\le s\le \lceil r/2\rceil$. The exact identity and an upper bound areThe quantity in parentheses is $m^{-1/2+o(1)}$, which tends to zero. The geometric series is thus at most $2/\mu$ for sufficiently large $m$. This proves the claim. In fact $r^5/m^2=o(m^{-9/5})$ and $1/\mu \le m^{-9/5}$, so eventuallyNow expose the $N=\left(\genfrac{}{}{0ex}{}{m}{2}\right)$ independent edge indicators one at a time. The edge-exposure martingale $M_i=\mathbb{E}[Z\mid \text{first }i\text{ indicators}]$ starts at $\mathbb{E}Z$ and ends at $Z$. Deleting one edge destroys at most one member of any edge-disjoint clique packing. Hence changing one indicator changes $Z$ by at most one; coupling the remaining indicators shows $|M_i-M_{i-1}|\le 1$.

Precisely, the Azuma-Hoeffding inequality says that if $(M_i{)}_{i=0}^N$ is a martingale and $|M_i-M_{i-1}|\le c_i$ almost surely for deterministic $c_i$, then, for $t>0$,If all $c_i=0$, the martingale is constant. Applying the lower-tail inequality with $c_i=1$ and $t=\mathbb{E}Z$ givesfor sufficiently large $m$. This is the only martingale concentration inequality needed.

Return to the original binomial random graph $G\sim G(n,1/2)$ and set $m=\lceil n/({\log }_{2}n{)}^2\rceil$. For each fixed $m$-element vertex set $U$, the complement graph of $G[U]$ has distribution $G(m,1/2)$. The union bound therefore givessince $m^{8/5}/n\to \infty$. Thus with high probability every vertex set of size at least $m$ contains an independent set of sizeApply greedy colouring by removing independent sets: remove such an independent set, give it a fresh colour, and repeat while at least $m$ vertices remain. Give each of the fewer than $m$ final vertices its own colour. This uses at mostcolours, since $m=o(n/{\log }_{2}n)$. Together with the lower bound, this gives the chromatic number of the half-density binomial random graph:

For the clique count in a binomial random graph, writeFix $0<\epsilon <1$. At $r_{-}=\lfloor (2-\epsilon )L\rfloor$, the estimates $\left(\genfrac{}{}{0ex}{}{n}{r}\right)\ge (n/r{)}^r$ and $r=O(L)$ giveThis eventually exceeds $(9/5)L$. At $r_{+}=\lceil (2+\epsilon )L\rceil$, the upper bound $\left(\genfrac{}{}{0ex}{}{n}{r}\right)\le n^r$ instead givesso ${\mu }_{r_{+}}<n^{9/5}$. Moreoverso no later clique size can regain the threshold. The maximum defining $r_0$ exists for sufficiently large $n$, since ${\mu }_2=\left(\genfrac{}{}{0ex}{}{n}{2}\right)/2\ge n^{9/5}$ then. We have $r_{-}\le r_0<r_{+}$. Letting $\epsilon$ decrease to zero proves the asymptotic size:

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.