Proof. Consider an r-edge colouring of $K_n$ . Our goal is to show that there exist s colours such that there are at least $1+\frac {s(n-1)}{d}$ vertices incident to an edge coloured in one of these s colours or at least $n-n \cdot \binom {r-d-1}{s}/\binom {r}{s}$ .

Suppose first that there is a vertex v for which there are at most d different colours used on the edges incident to v. In this case, the s colours with most edges incident to v account for at least $\frac {s(n-1)}{d}$ edges incident to v. Counting, in addition, v this implies that there are at least $1+\frac {s(n-1)}{d}$ vertices incident to an edge coloured using one of these s colours, as desired.

Hence, we may assume that for every vertex v, there are at least $d+1$ different colours used on the edges incident to v. We will choose our s colours uniformly at random. The probability that we did not pick any of the at least $d+1$ colours incident to v is at most ${\binom {r-d-1}s}/{\binom {r}s}$ . So the expected number of vertices not incident to an edge using one of our random s colours is at most $n \cdot {\binom {r-d-1}s}/{\binom {r}s}$ . In particular, there exists a choice of s colours for which there are at most this many vertices not incident to an edge using one of them. We conclude that there are at least $n-n \cdot \binom {r-d-1}{s}/\binom {r}{s}$ vertices incident to an edge using one of these colours.

Proof. Our goal is to colour the edges of $K_n$ into r colours in such a way that for any s colours, there are quite a few vertices of $K_n$ not incident to an edge coloured using one of the s colours. To do this, we will exploit the existence of the auxiliary hypergraph H.

Let $m:=|E(H)|$ , and let us partition the vertices of $K_n$ into m parts $A_e$ , one for each edge $e \in E(H)$ with each $A_e$ of size either $ \left \lfloor n/m \right \rfloor $ or $ \left \lceil n/m \right \rceil $ . We note that the r vertices of H represent the r colours we are going to use, while an edge $e \in E(H)$ should be viewed as the set of colours we are ‘permitted’ to use on the edges of our $K_n$ incident to a vertex belonging to $A_e$ . With this in mind, we now colour all the edges of $K_n$ as follows. Given any two distinct vertices belonging to $A_e$ and $A_{e'}$ , respectively (with possibly $e=e'$ ), we colour the edge between them using an arbitrary colour in $e \cap e'$ . We know such a colour always exists since $e \cap e'\neq \emptyset ,$ by the assumption that H is intersecting – namely, property 1.

Now to show this colouring satisfies the desired property suppose S is an arbitrary set of s colours. By property 2, we know the set of $r-s$ colours in $V(H) \setminus S$ contains at least t edges, the set of which we denote by T. Observe now that any vertex v belonging to $A_e$ with $e \in T$ is not incident to an edge coloured in a colour in S. Indeed, by the definition of our colouring, we are only ever colouring an edge incident to a vertex $v \in A_e$ by a colour which belongs to e, and since $e \in T$ , we know $e \subseteq V(H) \setminus S,$ so, in particular, no colour in S is allowed to be used. This implies the desired claim since there are at least $t \cdot \left \lfloor n/m \right \rfloor $ vertices of $K_n$ contained in one of the sets $A_e$ with $e \in T$ .

Proof. Let $r=p^2+p+1$ . If $s\ge p+1$ , then the inequality is implied by the trivial upper bound of $g(n,r,s)\le n$ , so we may assume $s \le p$ . By the assumption that p is a power of a prime, we know there exists a projective plane of order p. In particular, there exists an r-vertex, r-edge, $(p+1)$ -uniform hypergraph H in which every vertex belongs to exactly $p+1$ edges and every pair of vertices belongs to a unique edge.

Observe first that if we fix a set S of s vertices of H, it intersects at most $sp+1$ edges of H. To see this, fix some $v \in S$ . Note that v is contained in precisely $p+1$ edges, but each $w \in S \setminus \{v\}$ belongs to at most p edges not already accounted for (since one of the edges containing w contains v as well). Therefore, there are at most $p+1+(s-1)p=sp+1$ edges intersecting S in total.

This in particular implies that any set of $r-s$ vertices contains at least $r-sp-1$ edges of $H,$ so Lemma 2.2 implies

$$ \begin{align*}g(n,r,s)\le n-(r-sp-1)\cdot \frac{n}{r}= (sp+1)\cdot \frac{n}{r},\end{align*} $$

as desired.

Proof. Note that the desired inequality holds by the trivial upper bound of $g(n,r,s) \le n$ if $s \ge \sqrt {r}$ , so we may assume $s<\sqrt {r}.$ Let $r'$ be the largest integer smaller or equal to r of the form $p^2+p+1$ where p is a prime. By well-known results on the difference between consecutive primes (see, for example, [Reference Baker, Harman and Pintz4]), we know that $p=(1-o_r(1))\sqrt {r}.$ Let finally $n'=n- (n \bmod r')$ so that $r' \mid n'$ and $n'=(1-o_r(1))n$ . We now have

$$ \begin{align*}g(n,r,s) \le g(n,r',s) \le g(n',r',s)+n-n'\le (sp+1)\cdot \frac{n'}{r'}+r'\le (1+o_r(1))\cdot \frac{s}{\sqrt{r}}\cdot n,\end{align*} $$

where in the penultimate inequality we used Proposition 2.3 and in the final the assumption that $n \gg r^{3/2}/s$ .

Proof. Let us apply Lemma 2.1 with $d= \left \lfloor \sqrt {r} \right \rfloor \ge s$ . This shows that either $g(n,r,s) \ge 1+\frac {s(n-1)}{ \left \lfloor \sqrt {r} \right \rfloor }\ge \frac {sn}{ \left \lfloor \sqrt {r} \right \rfloor }$ in which case we are done or

$$ \begin{align*} \frac{g(n,r,s)}n&\ge 1-\frac{\binom{r-d-1}{s}}{\binom{r}{s}}\ge 1-\left(1-\frac{d+1}{r}\right)^s \ge \frac{s(d+1)}{r}-\binom{s}{2} \cdot\frac{(d+1)^2}{r^2}\\ &=\frac{s(d+1)}{r} \cdot \left(1 - \frac{(s-1)(d+1)}{2r}\right)\ge \frac{s}{\sqrt{r}}\cdot \left(1 - \frac{sd}{2r}\right), \end{align*} $$

which in turn is at least $(1-o_r(1)) \cdot \frac {s}{\sqrt {r}}$ for $s \ll \sqrt {r},$ and at least $\frac 12 \cdot \frac {s}{\sqrt {r}}$ for $s \le \sqrt {r}$ , as desired.

Proof. Let us apply Lemma 2.1 with $d=s$ . Since the first term is equal to n and hence always at least as large as the second, we always have

$$ \begin{align*}g(n,r,s) \ge n-n \cdot \binom{r-s-1}{s}\Big/\binom{r}{s} \ge n -n \cdot \left(1-\frac{s+1}{r}\right)^s \ge n \cdot \left(1-e^{-\frac{s(s+1)}r}\right),\end{align*} $$

as desired.

Proof. Let $u:=8s$ . Let H be a random hypergraph obtained by picking $m= \left \lfloor e^{u^2/(2r)} \right \rfloor $ random edges $e_1,\ldots , e_m \subseteq [r]$ , each sampled uniformly at random among all subsets of $[r]$ of size u, independently from all other edges.

First, observe that for any $1\le i,j \le m$ , the probability that $e_i$ and $e_j$ do not intersect is at most

$$ \begin{align*}\binom{r-u}{u}\Big/\binom{r}{u} = \frac{(r-u)\cdot \ldots \cdot (r-2u+1)}{r \cdot \ldots \cdot (r-u+1)}< \left(1-\frac{u}{r}\right)^u < e^{-u^2/r}.\end{align*} $$

So by a union bound, the probability that there exists a non-intersecting pair of edges is less than

$$ \begin{align*}\binom{m}{2}\cdot e^{-u^2/r}\le \frac{m^2}{2}\cdot e^{-u^2/r} \le \frac12.\end{align*} $$

However, given a fixed set of $r-s$ vertices, the probability that $e_i$ is contained in it equals

$$ \begin{align*}\binom{r-s}{u}\Big/\binom{r}{u} \ge \left(1-\frac{s}{r-u+1}\right)^u \ge \left(1-\frac{4s}{3r}\right)^u \ge e^{-{3su}/{r}+1},\end{align*} $$

where in the penultimate inequality we used $u = 8s \le \frac r4$ and in the final inequality we used that $1-x \ge e^{-2x}$ for any real $0\le x \le \frac 12$ and that $\frac {su}{3r}\ge 1$ .

Hence, the probability that none of our random m edges is contained in our set of size $r-s$ is at most

$$ \begin{align*}\left(1-e^{-{3su}/{r}+1}\right)^m \le \exp\left(-m\cdot e^{-{3su}/{r}+1} \right)\le \exp\left(-e^{{su}/{r}} \right).\end{align*} $$

Since there are $\binom {r}{r-s}=\binom {r}{s}\le \frac {r^s}{s!}\le \frac {r^s}2$ subsets of size $r-s,$ another union bound implies that the probability that there exists a set of $r-s$ vertices not containing any edge is at most

$$ \begin{align*}\frac{r^s}2\cdot \exp\left(-e^{{su}/{r}} \right)=\frac{1}2\cdot \exp\left(s \log r -e^{{su}/{r}} \right)\le \frac12,\end{align*} $$

where the final inequality follows since $su/r=8s^2/r \ge 8 \log r \ge \log s + \log \log r$ .

A final union bound implies that there exists an outcome in which the hypergraph we obtain is intersecting and any subset of $r-s$ vertices contains an edge; in other words, there is no cover of size at most s, as desired.

Proof. Proposition 2.7 provides us with an r-vertex hypergraph with at most $e^{O(s^2/r)}$ edges which is intersecting, and any set of $r-s$ vertices contains at least one edge. This allows us to apply Lemma 2.2 with $t=1$ to obtain

$$ \begin{align*}g(n,r,s) \le n - \left\lfloor \frac{n}{e^{O(s^2/r)}} \right\rfloor \le \left(1-e^{-O(s^2/r)}\right)\cdot n,\end{align*} $$

as desired.

Proof. Let $\varepsilon>0,$ and let us also for convenience set

$$ \begin{align*}x:=s+1 \hspace{1cm} \text{ and } \hspace{1cm} d:=r/2-x=r/2-s-1=o((r/\log r)^{1/3}).\end{align*} $$

Our general strategy is similar to the one we used in Proposition 2.7. However, in order to obtain a result at our desired level of precision, we cannot afford to simply sample every edge with an appropriate probability. What we do is for each $X\in \binom {[r]}{x}$ , sample a Bernoulli random variable $z_X$ with probability $p=\frac 1{\binom {r-x}{x}} \le \frac 12,$ where the inequality holds since $r-x>x$ . We now choose X to be an edge if $z_X=1$ , but $z_{X'}=0$ for all $X'\in \binom {[r]\backslash X}{x}$ . Let $A_X$ be the indicator random variable for whether we selected X as an edge, so

$$ \begin{align*}A_X = z_X\cdot\prod_{X'\in\binom{[r]\backslash X}{x}}(1-z_{X'}).\end{align*} $$

Note first that H is intersecting. Indeed, if $A_X=1$ for some $X\in \binom {[r]}{x},$ then $z_X=1$ and $z_{X'}=0$ for all $X' \in \binom {[r]\backslash X}{x},$ which in turn guarantees $A_{X'}=0,$ so H contains no edge disjoint from X.

Let B be the number of edges of H, so $B=\sum _{X\in \binom {[r]}{x}}A_X.$ Note that since for any $X\in \binom {[r]}{x}$ , $\mathbb {P}(z_X=1)=p$ , and since the variables $z_X$ are independent, we have

$$ \begin{align*}\mathbb{P}(A_X=1) =p(1-p)^{\binom{r-x}{x}}\ge p \cdot e^{-2p\binom{r-x}{x}}=\frac p{e^2}, \end{align*} $$

where in the inequality, we used that $1-p \ge e^{-2p}$ which holds since $p \le \frac 12$ . This combined with the linearity of expectation implies

$$ \begin{align*}\mathbb{E}[B] = \binom{r}{x} \cdot \mathbb{P}(A_X=1)\ge \binom{r}{x} \cdot \frac p{e^2}. \end{align*} $$

Our probability space is given by $\binom {r}{x}$ independent random variables, and changing one of them changes the value of B by at most $\binom {r-x}{x}$ . In other words, B is $\binom {r-x}{x}$ -Lipschitz, implying by Azuma’s inequality (see, for example, [Reference Alon and Spencer3, Chapter 7]) that

(1) $$ \begin{align} \mathbb{P}\left(B>{(1+\varepsilon)\mathbb{E}[B]}\right)&\leq \exp\left(-\frac{\varepsilon^2\mathbb{E}[B]^2}{2\sum_{X\in\binom{[r]}{x}}\binom{r-x}{x}^2}\right) \le \exp\left(-\binom{r}{x}\cdot\frac{\varepsilon^2p^4}{2e^4}\right)\le \exp\left(-\binom{r}{x}\cdot\Omega(r^{-8d})\right), \end{align} $$

where we used that $\frac 1p = \binom {r-x}{x} = \binom {r-x}{r-2x}=\binom {r-x}{2d} \le r^{2d}$ .

We next turn to the number of edges contained by vertex subsets of size $r-s$ . For each $T\in \binom {[r]}{r-s}$ , let $B_T$ count the number of edges of the subgraph of H induced by T, so $B_T = \sum _{X\in \binom {T}{x}}A_X$ . Hence, for any T,

(2) $$ \begin{align} \mathbb{E}[B_T] = \binom{|T|}{x} \cdot \mathbb{P}(A_X=1)\ge \binom{r-s}{x} \cdot \frac p{e^2}= \frac {\binom{r-s}{x}}{e^2 \cdot \binom{r-s-1}{x}}=\frac 1{e^2} \cdot \frac{r-s}{r-s-x}> \frac 1{2e^2} \cdot \frac{r}{2d+1}. \end{align} $$

Now, let us fix some $T\in \binom {[r]}{r-s}$ . Note that the random variable $B_T$ is increasing in the product probability space given by random variables $z_X$ for $X\in \binom {T}{x}$ and $1-z_X$ for $X\in \binom {[r]}{x}\backslash \binom {T}{x}$ since $r-s<2x$ , and therefore, for no $X\in T$ does $A_X$ depend on $z_{X'}$ for some $X'\in \binom {T}{x}\setminus \{X\}$ . This will allow us to apply Janson’s inequality (see [Reference Alon and Spencer3, Chapter 8]). Before applying it, we first need to understand the dependency structure.

Given some $X,X'\in \binom {T}{x}$ , $A_X$ and $A_{X'}$ are functions of disjoint sets of independent variables if and only if $r-|X\cup X'|< x$ since this is equivalent to there being no $X"\in \binom {[r]}{x}$ disjoint from both X and $X'$ (and $r-s<2x$ guarantees X and $X'$ are not disjoint). It follows that $A_X$ is not independent of $A_{X'}$ only if

$$ \begin{align*}r-x\geq|X\cup X'|=|X|+|X'|-|X\cap X'| = 2x-|X\cap X'|.\end{align*} $$

This is equivalent to $|X\cap X'|\geq 3x-r=x-2d$ . Therefore, for each $X\in \binom {T}{x}$ , $A_X$ is independent of all but at most

$$ \begin{align*} \sum_{i=x-2d}^{x}\binom{x}{i}\binom{r-s-x}{x-i}&=\sum_{i=x-2d}^{x}\binom{x}{x-i}\binom{2d+1}{x-i}\\ &= \sum_{i=0}^{2d}\binom{x}{i}\binom{2d+1}{i}\\ &\le (2d+1)\cdot \sum_{i=0}^{2d}\binom{x}{x-i}\binom{2d}{i}\\ &=(2d+1)\cdot \binom{x+2d}{x}= (2d+1)\cdot \binom{r-x}{x} =\frac{2d+1}{p} \end{align*} $$

events $A_{X'}$ . The above allows us to calculate the following term, with the goal of applying the lower tail version of Janson’s inequality [Reference Janson32]:

$$ \begin{align*} \Delta(B_T) = \sum_{\substack{X,X'\in \binom{T}{x}\\ \text{s.t. } A_X\ \sim\ A_{X'}}}\mathbb{E}[A_XA_{X'}] \leq \mathbb{E}[B_T] + \binom{r-s}{x}\cdot \frac{2d+1}{p} \cdot \mathbb{P}[A_X]\cdot p=(2d+2)\cdot \mathbb{E}[B_T], \end{align*} $$

where in the inequality, we used that $\mathbb {P}(A_X \cap A_{X'}) \le \mathbb {P}[A_X] \cdot p,$ which holds since for $A_{X'}$ to happen, we need to have $z_{X'}=1,$ which happens with probability p, independently of $A_X$ since $r-s-x<x$ . We get

$$ \begin{align*} \frac{\mathbb{E}[B_T]^2}{\Delta(B_T)}\geq \frac{\mathbb{E}[B_T]}{2d+2} \ge \frac{r}{2e^2 (2d+1) (2d+2)}, \end{align*} $$

where we used (2) in the final inequality. Finally, the lower tail version of Janson’s inequality [Reference Janson32] tells us

(3) $$ \begin{align} \mathbb{P}\left [B_T\leq {(1-\varepsilon)\mathbb{E}[B_T]}\right ]\leq \exp\left(-\frac{\varepsilon^2\mathbb{E}[B_T]^2}{2\Delta(B_T)}\right) \le \exp\left(-\Omega\left(\frac r{d^2}\right)\right). \end{align} $$

The last step is to apply the quantitative version of the Lovász Local Lemma (see [Reference Alon and Spencer3, Chapter 5]) with ‘bad events’ being $B_T< (1-\varepsilon ){\mathbb {E}[B_T]}$ for $T \in \binom {[r]}{r-s}$ to show that all these events can be avoided with probability which is greater than the probability that $B>(1+\varepsilon )\mathbb {E}[B]$ estimated in (1). Towards this, we first need to estimate, given $T\in \binom {[r]}{r-s}$ , for how many $T'\in \binom {[r]}{r-s}$ do $B_T$ and $B_{T'}$ both depend on the same $z_X$ for some $X\in \binom {[r]}{x}$ , since $B_T$ is independent of all other $B_{T'}$ . Towards this, note that $B_T=\sum _{X \in \binom {T}{x}} A_X,$ and that each $A_X$ only depends on $z_{X}$ and $z_{X'}$ for $X'$ disjoint from X. This means that $B_T$ depends only on $z_X$ when $X \subseteq T$ or $|T \setminus X|\ge x$ (as otherwise X is not disjoint from any subset of T of size x). To summarise, $B_T$ (and similarly $B_{T'}$ ) depends on $z_X$ if and only if one of the following holds:

• • $X\subseteq T$ or

• • $|T\backslash X|\geq x,$ or equivalently $|T\cap X|\leq r-s-x=2d+1$ .

It follows that there exists a $z_X$ such that both $B_T$ and $B_{T'}$ depend on it only if one of the following holds:

1. 1. $|T\cap T'| \geq x$ (there exists $X\in \binom {[r]}{x}$ such that $X\subseteq T \cap T'$ );

2. 2. $|T\cap T'| \leq 4d+2$ (there exists X with $|X\cap T| \leq 2d+1$ and $X\subseteq T'$ or vice versa);

3. 3. $|T\cap T'| \geq r/2-3d$ (there exists X with $|X\cap T|,|X\cap T'| \leq 2d+1$ ),

where 2 follows since $|X \cap T| \le 2d+1$ and $X \subseteq T'$ imply $x-(2d+1) \le |X \cap (T'\setminus T)| \le r-s-|T \cap T'|$ . Similarly, 3 follows since $|X\cap T|,|X\cap T'| \leq 2d+1$ imply that $x \le r-|T \cup T'|+|X \cap T| + |X \cap T'|$ , which is equivalent to $r/2-3d \le |T \cap T'|$ .

If we fix T of size $r-s,$ then there are

$$ \begin{align*}\binom{r-s}{i} \binom{s}{r-s-i}=\binom{r-s}{i} \binom{s}{2s-r+i}=\binom{r-s}{i} \binom{s}{i-2d-2}\end{align*} $$

choices for $T'$ of size $r-s$ such that $|T \cap T'|=i$ .

Since $x=r/2-d$ , 1 is implied by 3, so $B_T$ is independent of all but at most

$$ \begin{align*} \sum_{i=2d+2}^{4d+2} \binom{r-s}{i} \binom{s}{i-2d-2}+\sum_{i=r/2-3d}^{r-s} \binom{r-s}{r-s-i} \binom{s}{r-s-i} &\le \\ (2d+1)\binom{r-s}{4d+2} \binom{s}{2d}+\sum_{i=0}^{4d+2} \binom{r-s}{i} \binom{s}{i} &\le \\ 2(4d+2) \binom{r-s}{4d+2} \binom{s}{4d+2}&\le r^{8d+4} \end{align*} $$

$B_T'$ . Note next that since $d^3 = o(r/\log r)$ , we get from (3)

$$ \begin{align*}\mathbb{P}[B_T\leq (1-\varepsilon)\mathbb{E}[B_T]]\leq \exp\left(-\Omega\left(\frac r{d^2}\right)\right) = r^{-\omega(d)}\le r^{-8d-4}(1-r^{-8d-4})^{r^{8d+4}}, \end{align*} $$

allowing us to apply the Lovász Local Lemma to conclude

$$ \begin{align*} \mathbb{P}\Bigg[\bigcap_{T\in\binom{[r]}{r-s}} B_T> (1-\varepsilon)\mathbb{E}[B_T]\Bigg] &\ge (1-r^{-8d-4})^{\binom{r}{r-s}}\ge \exp\left(-2\binom{r}{r-s}/r^{8d+4}\right)> \exp\left(-\binom{r}{x}\cdot\Omega(r^{-8d})\right)\\&\ge \mathbb{P}\left(B>{(1+\varepsilon)\mathbb{E}[B]}\right). \end{align*} $$

This implies that there exists an outcome in which for each $T \in \binom {[r]}{r-s}$ , we have $B_T> (1-\varepsilon ) \mathbb {E}[B_T],$ and $B \le (1+\varepsilon )\mathbb {E}[B]$ . In other words, there exists an $(s+1)$ -uniform intersecting hypergraph with r vertices such that any $r-s$ vertices contain at least $(1-\varepsilon ) \mathbb {E}[B_T]$ edges, while the whole hypergraph has at most $(1+\varepsilon )\mathbb {E}[B]$ edges. The conclusion follows from

$$ \begin{align*}\frac{(1-\varepsilon)\mathbb{E}[B_T]}{(1+\varepsilon)\mathbb{E}[B]}\geq (1-\varepsilon)^2 {\frac{\binom{r-s}{x}}{\binom{r}{x}}}=(1-\varepsilon)^2\frac{\binom{r-s-1}{s}}{\binom{r}{s}}\end{align*} $$

and letting $\varepsilon \to 0$ .

Proof. Proposition 2.9 provides us with an r-vertex intersecting hypergraph H such that any set of $r-s$ vertices contains at least $\displaystyle (1-o_r(1))\frac {\binom {r-s-1}{s}}{\binom {r}{s}}\cdot |E(H)|$ edges. Lemma 2.2 applied with $\displaystyle t=(1-o_r(1))\frac {\binom {r-s-1}{s}}{\binom {r}{s}}$ now implies

$$ \begin{align*}g(n,r,s)\le n- (1-o_r(1)) \frac{\binom{r-s-1}{s}}{\binom{r}{s}} \cdot |E(H)| \cdot \left\lfloor \frac{n}{|E(H)|} \right\rfloor \le n- (1-o_r(1)) n \cdot \frac{\binom{r-s-1}{s}}{\binom{r}{s}},\end{align*} $$

where we used the assumption that n is sufficiently large compared to r (and hence also $|E(H)|$ ).

Proof. Let $s= \left \lfloor r/2 \right \rfloor $ . For the upper bound, we use Lemma 2.2 with the auxiliary hypergraph being the complete $(s+1)$ -uniform hypergraph on $2s+1$ vertices.

For the lower bound, we use Lemma 2.1 with $d=s,$ which shows

$$ \begin{align*}g(n,2s+1,s)\ge {\left(1-\frac{1}{\binom{2s+1}{s}}\right)n},\end{align*} $$

and since $g(n,2s+1,s)$ is always an integer, we may put a ceiling function here.

Proof. Curiously, $g(n,5,1)$ is the only result on the above list for which the lower bound does not follow from Lemma 2.1 directly, so we point the reader to, for example, [Reference Shiloach, Vishkin and Zaks45]. For the upper bound, one may take the (multi) hypergraph with vertex set $[5]$ and edge (multi) set $\{145,145,145,234,234,235,235,12,13\}$ . This is an intersecting hypergraph in which any four vertices contain at least $4$ edges, implying via Lemma 2.2 that $g(n,5,1) \le n- 4 \cdot \left \lfloor \frac n9 \right \rfloor = \frac 59 n$ if $9 \mid n$ .

The lower bounds for $g(n,6,1), g(n,6,2), g(n,7,1)$ and $g(n,7,2)$ all follow from Lemma 2.1 by choosing $d=2$ . The upper bound on $g(n,6,1)$ follows by taking the Fano plane and removing a vertex to obtain a six-vertex hypergraph with four edges in which every vertex has degree two so any five vertices contain at least two edges. The upper bound on $g(n,6,2)$ follows by applying Lemma 2.2 to the hypergraph with vertex set $[6]$ and edge set $\{123,124,346,345,256,135,245,236,146,156\},$ which has the property that any four vertices contain at least two edges. The upper bounds on $g(n,7,1)$ and $g(n,7,2)$ follow from Proposition 2.3. The results for $g(n,5,2)$ and $g(n,7,3)$ follow from Proposition 2.11.

Proof. Let $V=\mathbb {Z}_2^d$ , so $|V|=r+1$ . Let us consider the complete graph with vertex set V and let us colour the edge between distinct $\textbf {x},\textbf {y} \in \mathbb {Z}_2^d$ in colour identified with $\textbf {x}-\textbf {y}$ . This gives a colouring of an $(r+1)$ -vertex complete graph using r colours.

The key property of this colouring is that in any s colours, the graph consisting only of edges using these s colours is a vertex disjoint union of connected components, each consisting of at most $2^s$ vertices. Indeed, any set of s colours corresponds to a subset $S \subseteq \mathbb {Z}_2^d$ of size s, and the components correspond to cosets of $\text {Span}(S)$ which partition the vertices and have size at most $2^s$ as claimed.

Now let us consider a complete graph $K_n$ and partition its vertices into r parts of sizes as equal as possible, so each of size $ \left \lfloor \frac {n}{r+1} \right \rfloor $ or $ \left \lceil \frac {n}{r+1} \right \rceil .$ We then identify the parts with V and colour the edges between parts according to the colouring described above. We colour the edges inside the parts in an arbitrary colour used between it and some other part. In this colouring for any s colours, at most $2^s$ parts become connected showing that the maximum size of a connected component in the graph consisting only of edges using these s colours is at most $2^s \cdot \left \lceil \frac {n}{r+1} \right \rceil $ , as claimed.

Proof. We will construct our subgraphs greedily. Let $V_1$ be the vertex set of a largest k-connected subgraph of G. In general, given $V_1,\ldots , V_{i-1}$ , we let $V_i$ be the vertex set of the largest k-connected subgraph of $G \setminus (V_1 \cup \ldots \cup V_{i-1})$ , so long as such a subgraph exists of size more than $\frac {n}{2r}+1$ . Let $V_1 \cup \ldots \cup V_t$ be the disjoint sets that the above process produced, and let $V'=V(G) \setminus (V_1 \cup \ldots \cup V_{t})$ . Let us also write $s_i=|V_i|$ .

Since by adding to a k-connected graph a vertex adjacent to at least k of its vertices keeps the graph k-connected, we know that the total number of edges in $G \setminus (V_1 \cup \ldots \cup V_{i-1})$ with at least one endpoint in $V_i$ is at most $\binom {s_i}{2}+n(k-1)$ . Turning to $V',$ if the average degree in $G[V']$ is at least $\frac {n}{2r}+2k-2$ , then we claim it contains a k-connected subgraph with at least $\frac {n}{2r}+1$ vertices, contradicting the definition of $V'$ . If $k=1$ , this holds by taking a vertex of maximum degree and all of its at least $\frac {n}{2r}$ neighbours. For $k \ge 2,$ by Mader’s Theorem (Theorem 3.2), it contains a subgraph with connectivity at least k, since $\frac {n}{8r} \ge k,$ and average degree larger than $\frac {n}{2r}$ . Hence, we may conclude that the average degree of $G[V']$ is less than $\frac {n}{2r}+2k-2\le \frac {n-1}{r}$ (which guarantees $V' \neq V(G)$ and $|V'|\le n-1$ ) and in particular $G[V']$ has at most $\frac {n(n-1)}{4r}+(k-1)n$ edges.

Using this estimate on the number of edges, we get that there are at most

$$ \begin{align*}\frac 1r \binom{n}{2} \le \frac{n(n-1)}{4r}+(k-1)n+\sum_{i=1}^t \left(\binom{s_i}{2}+n(k-1) \right) \le \frac{1}{2r}\binom{n}{2}+(t+1)(k-1)n+\sum_{i=1}^t \binom{s_i}{2}\end{align*} $$

edges in G. Since $s_i> \frac {n}{2r},$ we have $t < 2r$ . So $(t+1)(k-1)n \le 2r(k-1)n \le \frac {1}{4r}\binom {n}{2},$ using that $n>16r^2(k-1)$ . This implies

$$ \begin{align*}\sum_{i=1}^t \binom{s_{i}}{2} \ge \frac{1}{4r} \cdot \binom{n}{2},\end{align*} $$

which in turn implies $s_1^2+\ldots +s_t^2 \ge \frac {n^2}{4r}$ .

Since for all $ i \in [t]$ we have $n \ge s_i \ge \max \left \{\frac {n}{2r},2\right \}$ , we know that there exists $2\le j \le \left \lceil \log r \right \rceil +2$ such that the set $I \subseteq [t]$ consisting of all i such that $s_i \in \left [2^{j-1}\cdot \frac {n}{4r},2^{j}\cdot \frac {n}{4r}\right ]$ satisfies

$$ \begin{align*}\frac{n^2}{4r (\log r +2)} \le \sum_{i \in I} s_i^2 \le |I|\cdot 2^{2j}\cdot \frac{n^2}{16r^2}.\end{align*} $$

This implies $|I| \ge \frac {r}{ 4^{j}\log r}$ . Finally, note that since $s_i \ge 2$ , we may also assume that $2^{j-1} \cdot \frac {n}{4r} \ge 2$ .

Proof. Pick an arbitrary perfect matching and observe it contains more than $r(k-1)$ edges. By the pigeonhole principle, this means some k of these edges have the same colour, providing us with the desired matching.

Proof. We will prove that the desired inequality holds for $s \le \log \log r-\log (4+3\log \log r).$

Let $n_0=n$ and let us apply Lemma 3.3 to the graph consisting of the edges in the majority colour $c_1$ to find $j_1\ge 2$ such that there exists at least $n_1 = \left \lceil \frac {r}{4^{j_1} \log r} \right \rceil $ k-connected subgraphs, each with at least $s_1\ge 2^{j_1-3}\cdot \frac nr \ge 2$ vertices. We now contract each of these graphs into a single vertex to obtain a (multi) coloured $K_{n_1}$ , where each vertex corresponds to a set of at least $s_1$ vertices in our original graph, spanning a k-connected subgraph in colour $c_1$ . Next, we keep an edge of only one colour c between any two vertices $u,v\in V(K_{n_1}),$ where c is selected in the following way. Let $U,V\subseteq V(K_n)$ be the sets of vertices in our original graph corresponding to u and v, respectively. We choose c to be such that the original graph contains a matching of size k in c between U and V. Note that there exists such a colour by Observation 3.4 since $s_1 \ge \frac {n}{2r}> r(k-1)$ . We then repeat, applying Lemma 3.3, but from now on always with $k=1$ , to the subgraph consisting of the majority colour in the current $K_{n_i}$ , for as long as we have at least two vertices left. So, after step i for some $j_i\ge 0$ , we have at least $n_i = \left \lceil \frac {r}{4^{j_i} \log r} \right \rceil $ vertices each corresponding to at least $s_1 \cdot s_2 \cdots s_i$ vertices of our original $K_n,$ which induce a k-connected graph using the edges of colours $c_1,\ldots ,c_i$ and where $s_i\ge 2^{j_i-3}\cdot \frac {n_{i-1}}{r}\ge 2$ . Note that placing a matching of size k between two k-connected graphs produces a new k-connected graph. Therefore, our choice for the colouring of $K_{n_1}$ guarantees that a connected component in any colour c in $K_{n_i}$ , when $i \ge 2$ , corresponds to a k-connected subgraph of our original graph using colours $c_1,\ldots ,c_{i-1}$ and c.

Let us suppose that the process runs for t steps so that $n_t=1$ and $n_i \ge 2$ for all $i < t$ . Since $n_t=1$ , we have $\frac {r}{\log r} \le 4^{j_t},$ or after taking a logarithm that $j_t \ge \frac 12 \log r -\frac 12 \log \log r$ . Using that $n_{i-1} \ge 2$ for all $i \le t,$ we get

$$ \begin{align*}2 \le 2^{j_i-3}\cdot \frac{n_{i-1}}{r}\le 2^{j_i-3}\cdot \frac{ \left\lceil \frac{r}{4^{j_{i-1}} \log r} \right\rceil }{r}\le 2^{j_i-3}\cdot \frac{2}{4^{j_{i-1}} \log r}.\end{align*} $$

After taking a logarithm of both sides, this implies $j_i \ge 2j_{i-1}+3+\log \log r$ .

If $t \ge s,$ then we know that after s steps, we connected at least $s_1 \cdot s_2 \cdots s_s \ge 2^{s-2} \cdot \frac nr$ vertices, since $s_1 \ge \frac 12 \cdot \frac nr$ and $s_i \ge 2$ for all i, and are done. So we may assume that the total number of steps t satisfies $t<s,$ and hence, it suffices to show that $s_1 \cdot s_2 \cdots s_t \ge 2^{s-2} \cdot \frac nr.$

Finally, we prove by induction on i that for $i \le t$ , we have $s_1 \cdot s_2 \cdots s_i \ge 2^{j_i/2^{i-1}-(1-1/2^{i-1})(3+\log \log r)} \cdot \frac n{8r}.$ Indeed, the base case $i=1$ holds since $s_1 \ge 2^{j_1-3} \cdot \frac nr$ , and for $i>1$ , we get

$$ \begin{align*} s_1 \cdot s_2 \cdots s_i &\ge 2^{j_{i-1}/2^{i-2}-(1-1/2^{i-2})(3+\log \log r)} \cdot \frac n{8r} \cdot s_i \\ &\ge 2^{j_{i-1}/2^{i-2}-(1-1/2^{i-2})(3+\log \log r)} \cdot \frac n{8r} \cdot \frac{2^{j_i-3}}{4^{j_{i-1}} \log r}\\ &=2^{j_i-(2-1/2^{i-2})(j_{i-1}+3+\log \log r)} \cdot \frac n{8r}\\ &\ge 2^{j_i/2^{i-1}-(1-1/2^{i-1})(3+\log \log r)} \cdot \frac n{8r}, \end{align*} $$

where in the second inequality, we used $s_i \ge 2^{j_i-3}\cdot \frac {n_{i-1}}{r} = 2^{j_i-3}\cdot \frac { \left \lceil \frac {r}{4^{j_{i-1}} \log r} \right \rceil }{r}\ge \frac {2^{j_i-3}}{4^{j_{i-1}} \log r},$ and in the final inequality, we used $(1-1/2^{i-1})j_i \ge (1-1/2^{i-1})(2j_{i-1}+3+\log \log r).$ This completes the induction step.

Applying this inequality for $i=t$ gives $s_1 \cdot s_2 \cdots s_t \ge 2^{j_t/2^{t-1}-3-\log \log r} \cdot \frac n{8r}>2^{s-2} \cdot \frac nr$ , since

$$ \begin{align*}\frac{j_t}{2^{t-1}}-3-\log \log r \ge \frac{\log r - \log \log r}{2^{t}}-3-\log \log r \ge \frac{\log r}{2^{t}}-3-2\log \log r> \log \log r +1 \ge s+1,\end{align*} $$

where in the penultimate inequality, we used $t<s \le \log \log r-\log (4+3\log \log r)$ .

Proof. We will prove the stronger bound by induction on s. The basis, for $s=1$ , the desired lower bound of $\frac {n}{r}$ follows by looking at a vertex and the colour which appears the most often on the edges incident to it. Suppose now that the result holds for $s-1$ colours, and let S be the largest component one can obtain using $s-1$ colours.

For any vertex v outside of $S,$ if v sends edges of at most $s-1$ different colours towards S, then we can connect $S \cup \{v\}$ using the edges of the star between v and S and their at most $s-1$ colors, contradicting the maximality of S. Hence, every vertex outside of S sends edges of at least s different colours to S. Now by picking a uniformly random new colour, we will connect any such vertex to S with probability at least $\frac {s}{r-s+1}$ . This means that the expected number of vertices we connect to S is $\frac {s}{r-s+1}\cdot (n-|S|)$ . Therefore, we can choose a colour in such a way that the new set of s colours connects at least

$$ \begin{align*}|S|+\frac{s}{r-s+1}\cdot (n-|S|)=n-(n-|S|)\cdot \left(1- \frac s{r-s+1}\right)\end{align*} $$

vertices, which after invoking the induction assumption on $|S|$ gives the desired bound.

Proof. We may assume that $8(s+1) \log r\le \frac rs,$ or the inequality is trivial. This implies $\frac {r}{\log r} \ge 8(s+1)s\ge 48$ , which in turn implies $\log r \ge 6.$ Let us define $m:= \left \lfloor \frac {r}{6s} \right \rfloor \ge \frac {r}{7s}$ . To start with, we want to show that there exists an r-edge colouring of $K_m$ with no s colours connecting more than $ (s+1)\log r$ vertices. To do so, let us consider a random colouring in which every edge of our $K_m$ chooses one of the r colours uniformly at random, independently between different edges.

If there are $k = \left \lceil (s+1)\log r \right \rceil $ vertices connected using up to s colours, then there exists a k-vertex tree using up to s colours. There are $\binom {m}{k} \cdot k^{k-2}$ possible choices for the tree, using Cayley’s formula, and $\binom {r}{s}$ choices for the set of s colours used on the tree. Given these choices, the probability a given tree is coloured using the given set of s colours is $\left (\frac sr \right )^{k-1}$ . Now a union bound implies that the probability that there exists a k-vertex tree using up to s colours is at most

$$ \begin{align*}\binom{m}{k} \cdot k^{k-2} \cdot \binom {r}{s} \cdot \left(\frac sr \right)^{k-1} \le r^{s+1} \cdot \left(\frac{ems}{r}\right)^k \le 2^k\cdot \left(\frac{e}{6}\right)^k < 1,\end{align*} $$

where in the first inequality, we used the standard estimates $\binom {x}{y} \le \min \left \{\left (\frac {ex}{y}\right )^y, x^y\right \}$ . This shows that indeed there exists a desired colouring of $K_m$ .

To complete the proof, we now take a blow-up of the above colouring of $K_m$ where we replace every vertex with a set of $ \left \lfloor \frac nm \right \rfloor $ or $ \left \lceil \frac nm \right \rceil $ vertices to obtain an n-vertex graph. Let $V(K_m)=\{v_1,\ldots ,v_m\}$ , and we denote the corresponding sets by $V_1,\ldots ,V_m.$ For $1\le i<j\le m$ , we colour the edges between $V_i$ and $V_j$ with the same colour as the edge between $v_i$ and $v_j$ . For $1\le i\le m$ , we colour the edges inside $V_i$ with an arbitrary colour appearing on an edge incident to $v_i$ . This shows that

$$ \begin{align*}f(n,r,s) < (s+1)\log r \cdot \left\lceil \frac nm \right\rceil \le (s+1)\log r \cdot \left\lceil \frac n{r/(7s)} \right\rceil \le \frac{8s(s+1)\log r}{r} \cdot n,\end{align*} $$

completing the proof.