Proof of Theorem 1.4. Recall that ${\mathcal{L}}(n,k,2)$ is 2-complete intersecting and $f(n,k,2)\leq f^*(n,k,2)$ . It follows that $|{\mathcal{L}}(n,k,2)|\leq f(n,k,2)\leq f^*(n,k,2)$ . Thus we are left to show $f^*(n,k,2)\leq |{\mathcal{L}}(n,k,2)|$ .

Let ${\mathcal{F}}\subset \binom{[n]}{k}$ be an intersecting family. Let ${\mathcal{F}}^*$ be the family of 2-complete sets in $\mathcal{F}$ and let ${\mathcal{H}}=\partial{{\mathcal{F}}^*}$ . Note that this guarantees that every member of $\mathcal{H}$ is contained in at least two members of $\mathcal{F}$ .

By Theorem 1.6 and Claim 1, for $n\geq (5\times 3+13)k=28k$ we have $|{\mathcal{F}}^*|\leq |{\mathcal{K}}({\mathcal{H}})|\leq |{\mathcal{K}}({\mathcal{E}}(n,k,3))|=|{\mathcal{L}}(n,k,2)|$ . The uniqueness follows from Theorem 1.6.

Proof. Let ${\mathcal{F}}_0\subset{\mathcal{F}}_1\subset \cdots \subset{\mathcal{F}}_s\subset \binom{Y}{\ell }$ be overlapping families. Let $ t = \lfloor |Y|/\ell \rfloor \geq\lfloor d_{\vec{p}}\rfloor +1\geq s+1$ and choose a random matching $F_1,F_2,\ldots,F_t$ from $\binom{Y}{\ell }$ . Consider the weighted bipartite graph $G$ on partite sets $\{F_1,F_2,\ldots,F_t\}$ and $\{{\mathcal{F}}_0,{\mathcal{F}}_1, \cdots,{\mathcal{F}}_s\}$ where we have an edge $(F_i,{\mathcal{F}}_j)$ iff $F_i\in{\mathcal{F}}_j$ . This edge gets weight $p_j$ .

Since ${\mathcal{F}}_{0}\subset{\mathcal{F}}_{1}\subset \cdots \subset{\mathcal{F}}_{s}$ are overlapping, $G$ has matching number at most $s$ . Applying the König-Hall Theorem we can find $s$ vertices covering all edges of the bipartite graph $G$ . Let $F_1,\ldots,F_q$ be the vertices of the covering set chosen from the random matching and ${\mathcal{F}}_{q+1}, \ldots,{\mathcal{F}}_s$ the remaining $s-q$ chosen from the families.

The total weight of the edges covered by $F_i$ is at most $p_0+\ldots +p_s$ . The total weight of the edges covered by ${\mathcal{F}}_j$ is at most $tp_j$ . Thus, the total weight of the edges in $G$ is at most

(14) \begin{align} q(p_0+\cdots +p_s)+&t(p_{q+1}+\cdots +p_s)\nonumber \\ &= t(p_{1}+\cdots +p_s)- t(p_1+\cdots +p_q)+q(p_0+\cdots +p_s). \end{align}

Note that $p_1\geq \ldots \geq p_s$ implies

\begin{equation*} \frac {i+1}{p_1+\cdots +p_{i+1}}\geq \frac {i}{p_1+\cdots +p_{i}}. \end{equation*}

It follows that

(15) \begin{align} \frac{q}{p_1+\cdots +p_{q}}(p_0+\cdots +p_s)\leq \frac{s}{p_1+\cdots +p_s}(p_0+\cdots +p_s)=d_{\vec{p}}\lt t. \end{align}

By (14) and (15), the total weight of the edges in $G$ is at most $t(p_{1}+\ldots +p_s)$ .

Since the probability

\begin{equation*} Pr(F_i\in {\mathcal {F}}_j) =\frac {|{\mathcal {F}}_j|}{\binom {|Y|}{\ell}}, \end{equation*}

the expected value of the total weight of the edges in $G$ is $\sum _{j=0}^s tp_j \frac{|{\mathcal{F}}_j|}{\binom {|Y|}{\ell}}$ . Thus (13) follows. In case of equality $q=0$ . Then for every $t$ -matching $F_1,F_2,\ldots,F_t$ in $Y$ , ${\mathcal{F}}_0$ has degree 0 and ${\mathcal{F}}_i$ has degree $t$ in $G$ for $i=1,\ldots,s$ . Hence the equality holds iff ${\mathcal{F}}_{1}= \cdots ={\mathcal{F}}_{s}=\binom{Y}{\ell }$ , ${\mathcal{F}}_0=\emptyset$ .

Proof. We prove the statement by defining an injective map $\sigma$ from ${\mathcal{K}}({\mathcal{F}})\setminus{\mathcal{K}}(S_{ij}({\mathcal{F}}))$ to ${\mathcal{K}}(S_{ij}({\mathcal{F}}))\setminus{\mathcal{K}}({\mathcal{F}})$ . Let $K\in{\mathcal{K}}({\mathcal{F}})\setminus{\mathcal{K}}(S_{ij}({\mathcal{F}}))$ . Clearly $j\in K$ and $i\notin K$ , and we define $\sigma (K)=K'=(K\setminus \{j\})\cup \{i\}$ . We show that $\sigma$ is well-defined by checking $K'\in{\mathcal{K}}(S_{ij}({\mathcal{F}}))\setminus{\mathcal{K}}({\mathcal{F}})$ . Firstly, suppose that $K'\notin{\mathcal{K}}(S_{ij}({\mathcal{F}}))$ and let $F'\in \binom{K'}{k}$ be an edge not in $S_{ij}({\mathcal{F}})$ . If $i\notin F'$ then $F'=K\setminus \{j\}$ and $S_{ij}(F')=F'$ , implying that $F'\in S_{ij}({\mathcal{F}})$ , a contradiction. If $i\in F'$ , then $F=(F'\setminus \{i\})\cup \{j\}\subset K$ is an edge of $\mathcal{F}$ since $K\in{\mathcal{K}}({\mathcal{F}})$ . Hence after shifting we have $F'\in S_{ij}({\mathcal{F}})$ , a contradiction. This shows $K'\in{\mathcal{K}}(S_{ij}({\mathcal{F}}))$ . Secondly, if $K'\in{\mathcal{K}}({\mathcal{F}})$ then $K\in{\mathcal{K}}({\mathcal{F}})$ implies $K\in{\mathcal{K}}(S_{ij}({\mathcal{F}}))$ , contradicting the assumption that $K\notin{\mathcal{K}}(S_{ij}({\mathcal{F}}))$ . Thus $K'\in{\mathcal{K}}(S_{ij}({\mathcal{F}}))\setminus{\mathcal{K}}({\mathcal{F}})$ and $\sigma$ is indeed a map from ${\mathcal{K}}({\mathcal{F}})\setminus{\mathcal{K}}(S_{ij}({\mathcal{F}}))$ to ${\mathcal{K}}(S_{ij}({\mathcal{F}}))\setminus{\mathcal{K}}({\mathcal{F}})$ . Clearly, $\sigma$ is injective and the proposition follows.

Proof of Theorem 1.6. Since the shifting operator does not increase the matching number and does not decrease the size of ${\mathcal{K}}({\mathcal{F}})$ , we may assume that $\mathcal{F}$ is shifted. Let ${\mathcal{K}}={\mathcal{K}}({\mathcal{F}})$ and ${\mathcal{K}}^*={\mathcal{K}}({\mathcal{E}}(n,k,s))$ . For any $S\subset [s+1]$ and a family ${\mathcal{H}}\subset \binom{[n]}{h}$ , define

\begin{equation*} {\mathcal {H}}(S) \;:\!=\; \left \{H\setminus [s+1]\colon H\in {\mathcal {H}},\ H\cap [s+1]=S\right \}. \end{equation*}

Clearly ${\mathcal{H}}(S)\subset \binom{[s+2,n]}{h-|S|}$ .

For $|S|\geq 3$ , we have ${\mathcal{K}}^*(S)=\binom{[s+2,n]}{k+1-|S|}$ . It follows that

(16) \begin{align} \sum _{S\subset [s+1],|S|\geq 3} |{\mathcal{K}}(S)|\leq \sum _{S\subset [s+1],|S|\geq 3} |{\mathcal{K}}^*(S)|. \end{align}

We are left to compare $|{\mathcal{K}}(S)|$ with $|{\mathcal{K}}^*(S)|$ for all $S\subset [s+1]$ with $|S|\leq 2$ .

Note that for any $K\in{\mathcal{K}}(\emptyset )$ we have $\binom{K}{k}\subset{\mathcal{F}}(\emptyset )$ . It follows that $\partial{\mathcal{K}}(\emptyset )\subset{\mathcal{F}}(\emptyset )$ . Since $\nu ({\mathcal{K}}(\emptyset )) \leq s$ , by (8) we have

(17) \begin{align} s|{\mathcal{F}}(\emptyset )| \geq s|\partial{\mathcal{K}}(\emptyset )|\geq |{\mathcal{K}}(\emptyset )|. \end{align}

By Claim 2,

\begin{equation*} \sum _{1\leq i\leq s+1}|{\mathcal {K}}(\{i\})|= (s+1)|{\mathcal {F}}(\emptyset )| \end{equation*}

and

\begin{equation*} \sum _{1\leq i\lt j\leq s+1}|{\mathcal {K}}(\{i,j\})|=\sum _{2\leq j\leq s+1}(j-1)|{\mathcal {F}}(\{j\})|. \end{equation*}

It follows that

(18) \begin{align} \sum _{S\in [n],|S|\leq 2} |{\mathcal{K}}(S)| &=|{\mathcal{K}}(\emptyset )|+\sum _{1\leq i\leq s+1}|{\mathcal{K}}(\{i\})|+\sum _{1\leq i\lt j\leq s+1}|{\mathcal{K}}(\{i,j\})|\nonumber \\ &\overset{(17)}{\leq } s|{\mathcal{F}}(\emptyset )|+(s+1)|{\mathcal{F}}(\emptyset )|+\sum _{2\leq j\leq s+1}(j-1)|{\mathcal{F}}(\{j\})|\nonumber \\ &\leq (2s+1)|{\mathcal{F}}(\emptyset )|+\sum _{2\leq j\leq s+1}(j-1)|{\mathcal{F}}(\{j\})|. \end{align}

Again by shiftedness $\partial{\mathcal{F}}(\emptyset )\subset{\mathcal{F}}(\{s+1\})$ , and using (8) we infer

(19) \begin{align} s|{\mathcal{F}}(\{s+1\})| \geq s|\partial{\mathcal{F}}(\emptyset )|\geq |{\mathcal{F}}(\emptyset )|. \end{align}

Substituting (19) into (18), we arrive at

\begin{align*} \sum _{S\in [n],|S|\leq 2} |{\mathcal{K}}(S)| &\leq \sum _{2\leq j\leq s}(j-1)|{\mathcal{F}}(\{j\})|+ s|{\mathcal{F}}(\{s+1\})|+(2s+1)s|{\mathcal{F}}(\{s+1\})|\\[5pt] &= |{\mathcal{F}}(\{2\})|+\sum _{3\leq j\leq s}(j-1)|{\mathcal{F}}(\{j\})|+2s(s+1)|{\mathcal{F}}(\{s+1\})|\\[5pt] &\leq \frac{1}{2}|{\mathcal{F}}(\{1\})|+\frac{1}{2}|{\mathcal{F}}(\{2\})|+\sum _{3\leq j\leq s}(j-1)|{\mathcal{F}}(\{j\})|+2s(s+1)|{\mathcal{F}}(\{s+1\})|. \end{align*}

By shiftedness, ${\mathcal{F}}(\{1\})\supset \cdots \supset{\mathcal{F}}(\{s+1\})$ are overlapping families. Set

\begin{equation*} {\mathcal {F}}_0={\mathcal {F}}(\{s+1\}),\ {\mathcal {F}}_1={\mathcal {F}}(\{s\}),\ \ldots, {\mathcal {F}}_{s-2}={\mathcal {F}}(\{3\}),\ {\mathcal {F}}_{s-1}={\mathcal {F}}(\{2\}),\ {\mathcal {F}}_s={\mathcal {F}}(\{1\}) \end{equation*}

and set

\begin{equation*} p_0=2s(s+1),\ p_1=s-1,\ldots,p_{s-2}=2,\ p_{s-1}=\frac {1}{2},\ p_{s}=\frac {1}{2}. \end{equation*}

Then $p_0\geq p_1\geq \ldots \geq p_s$ and by $s\geq 3$ ,

\begin{equation*} d_{\vec {p}} = \frac {s(p_0+\cdots +p_s)}{(p_1+\cdots +p_s)}=\frac {4s(s+1)+s(s-1)}{s-1}= 5s+ 8 +\frac {8}{s-1}\leq 5s+12. \end{equation*}

By Lemma 3.2, for $n-s-1\geq (5s+13)(k-1)\geq (d_{\vec{p}}+1)(k-1)$ we have

(20) \begin{align} \sum _{S\subset [s+1],|S|\leq 2} |{\mathcal{K}}(S)|&\leq (p_1+p_2+\ldots +p_s)\binom{n-s-1}{k-1}\nonumber \\[5pt]&= \binom{s}{2}\binom{n-s-1}{k-1}\nonumber \\ &= \sum _{S\subset [s+1],|S|\leq 2} |{\mathcal{K}}^*(S)|. \end{align}

Adding (16) and (20), we conclude that

\begin{equation*} |{\mathcal {K}}({\mathcal {F}})|=\sum _{S\subset [s+1]} |{\mathcal {K}}(S)|\leq \sum _{S\subset [s+1]} |{\mathcal {K}}^*(S)| =|{\mathcal {K}}({\mathcal {E}}(n,k,s))|. \end{equation*}

Let $\mathcal{F}$ be a family with $\nu ({\mathcal{F}})\leq s$ and $|{\mathcal{K}}({\mathcal{F}})|= |{\mathcal{K}}({\mathcal{E}}(n,k,s))|$ . If $\mathcal{F}$ is shifted, then by Lemma 3.2 we have ${\mathcal{F}}(\{s+1\})=\emptyset$ . It follows that ${\mathcal{F}}={\mathcal{E}}(n,k,s)$ . Now assume that $\mathcal{F}$ is not shifted. Then it changes to ${\mathcal{E}}(n,k,s)$ by applying shifting repeatedly. Let $\mathcal{G}$ be the last family that is not isomorphic to ${\mathcal{E}}(n,k,s)$ in this process. That is, $\mathcal{G}$ is not isomorphic to ${\mathcal{E}}(n,k,s)$ but $S_{ij}({\mathcal{G}})$ is isomorphic to ${\mathcal{E}}(n,k,s)$ for some $1\leq i\lt j\leq n$ . By symmetry, we may assume that ${\mathcal{G}}\neq{\mathcal{E}}(n,k,s)$ and $S_{s,s+1}({\mathcal{G}})={\mathcal{E}}(n,k,s)$ . Let

\begin{align*} &{\mathcal{G}}(s\overline{(s+1)}) =\left \{E\in \binom{[n]\setminus \{s,s+1\}}{k-1}\colon E\cup \{s\}\in{\mathcal{G}} \right \},\\[5pt] &{\mathcal{G}}(\overline{s}(s+1)) =\left \{E\in \binom{[n]\setminus \{s,s+1\}}{k-1}\colon E\cup \{s+1\}\in{\mathcal{G}} \right \}. \end{align*}

Since $S_{s,s+1}({\mathcal{G}})={\mathcal{E}}(n,k,s)$ , we see ${\mathcal{G}}(s\overline{(s+1)})\cup{\mathcal{G}}(\overline{s}(s+1)) =\binom{[n]\setminus \{s,s+1\}}{k-1}$ and ${\mathcal{G}}(s\overline{(s+1)})\cap{\mathcal{G}}(\overline{s}(s+1)) =\emptyset$ . It follows that for each $E\in \binom{[n]\setminus \{s,s+1\}}{k-1}$ , exactly one of $E\cup \{s\}\in{\mathcal{G}}$ and $E\cup \{s+1\}\in{\mathcal{G}}$ holds. Now consider a graph $G$ on the vertex set $\binom{[n]\setminus \{s,s+1\}}{k-1}$ where $(E_1,E_2)$ forms an edge if and only if $|E_1\cap E_2|=k-2$ . It is easy to see that $G$ is a connected graph. Since $\mathcal{G}$ is not isomorphic to ${\mathcal{E}}(n,k,s)$ , we infer that ${\mathcal{G}}(s\overline{(s+1)})\neq \emptyset$ and ${\mathcal{G}}(\overline{s}(s+1))\neq \emptyset$ . Then there exists an edge $(E_1,E_2)$ in $G$ such that $E_1\cup \{s\}\in{\mathcal{G}}$ and $E_2\cup \{s+1\}\in{\mathcal{G}}$ . Let $F\;:\!=\;E_1\cup E_2\in \binom{[n]\setminus \{s,s+1\}}{k}$ . Then $F\cup \{s+1\}\notin{\mathcal{K}}({\mathcal{G}})$ and $F\cup \{s\}\notin{\mathcal{K}}({\mathcal{G}})$ . But $S_{s,s+1}({\mathcal{G}})={\mathcal{E}}(n,k,s)$ implies $F\cup \{s\} \in{\mathcal{K}}(S_{s,s+1}({\mathcal{G}}))$ . Moreover, for any $K\in{\mathcal{K}}({\mathcal{G}})\setminus{\mathcal{K}}(S_{s,s+1}({\mathcal{G}}))$ , we have $(K\setminus \{s+1\})\cup \{s\}\in{\mathcal{K}}(S_{s,s+1}({\mathcal{G}}))\setminus{\mathcal{K}}({\mathcal{G}})$ by the injective map defined in Proposition 3.3. Hence $|{\mathcal{K}}(S_{s,s+1}({\mathcal{G}}))|\gt |{\mathcal{K}}({\mathcal{G}})|\geq |{\mathcal{K}}({\mathcal{F}})|=|{\mathcal{K}}({\mathcal{E}}(n,k,s))|$ , a contradiction. Thus up to isomorphism ${\mathcal{E}}(n,k,s)$ is the only family attaining equality.

Proof. Suppose that $\mathcal{F}$ is an intersecting $k$ -graph and each $F\in{\mathcal{F}}$ is $k$ -wise covered. Consider the bipartite graph with partite sets $\mathcal{F}$ , $\partial{\mathcal{F}}$ and an edge between $F$ and $G$ iff $G\subset F$ . It is clear that each $F\in{\mathcal{F}}$ has degree $k$ . On the other hand, the condition implies that each $G\in \partial{\mathcal{F}}$ has degree at least $k$ . Consequently, $|{\mathcal{F}}|\geq |\partial{\mathcal{F}}|$ . In view of (7), we see $|{\mathcal{F}}|= |\partial{\mathcal{F}}|$ and equality holds iff ${\mathcal{F}}=\binom{X}{k}$ with $|X|=2k-1$ . The same argument implies $f(n,k,r)=0$ for $r\geq k+1$ .

Proof. Take an arbitrary set $F$ from ${\mathcal{F}}^{i}$ . Since $|F|\leq k$ , we have $F\cap (F_j\setminus C_i)=\emptyset$ for some $j$ , $0\leq j\leq k$ . Then $F\cap C_i=F\cap F_j\neq \emptyset$ .

Proof of Theorem 1.5. By proposition 4.1, we may assume $3\leq r\leq k$ . Let $\mathcal{F}$ be an $r$ -complete intersecting family of maximal size and let ${\mathcal{B}}={\mathcal{B}}({\mathcal{F}})$ be its basis. Let $X=\mathop{\cup }\limits _{B\in{\mathcal{B}}} B$ . By (21) we have

\begin{equation*} |X|\leq \sum _{1\leq \ell \leq k} \ell !k^\ell \leq 2k!k^k\leq k^{2k}. \end{equation*}

By the definition of $\mathcal{B}$ , for any $F\in{\mathcal{F}}$ there exists $B\in{\mathcal{B}}$ such that $B\subset F$ . Then for $F,F'\in{\mathcal{F}}$ , there exist $B,B'\in{\mathcal{B}}$ such that $B\subset F$ and $B'\subset F'$ . Since $\mathcal{B}$ is intersecting, $\emptyset \neq B\cap B'\subset F\cap F'\cap X$ . Thus, for all $F,F'\in{\mathcal{F}}$ , $F\cap F'\cap X \neq \emptyset$ .

Let us define $p=\min \left \{|F\cap X|\colon F\in{\mathcal{F}}\right \}$ and choose an arbitrary pair $(F,P_0)$ , $P_0\in \binom{X}{p}$ , $F\cap X=P_0$ . Set $H=F\setminus P_0$ and define

\begin{equation*} {\mathcal {P}}(H) =\left \{P\in \binom {X}{p}\colon H\cup P\in {\mathcal {F}}\right \}. \end{equation*}

Note that $P_0\in{\mathcal{P}}(H)$ .

If $p\lt r$ , by Claim 4 and Proposition 4.1 we have $1\leq |{\mathcal{P}}(H)|\leq f(|X|,p,r)=0$ , a contradiction. Thus $p\geq r$ . Define

\begin{equation*} {\mathcal {F}}_0 =\left \{F\in {\mathcal {F}}\colon |F\cap X|\geq r+1\right \}. \end{equation*}

Then

\begin{equation*} |{\mathcal {F}}_0| \leq \sum _{r+1\leq i\leq k} \binom {|X|}{i}\binom {n-|X|}{k-i}\leq \sum _{r+1\leq i\leq k} \binom {k^{2k}}{i}\binom {n-k^{2k}}{k-i} \lt 2\binom {k^{2k}}{r+1}\binom {n-2r}{k-r-1}. \end{equation*}

If $p\geq r+1$ , then

\begin{equation*} |{\mathcal {F}}|= |{\mathcal {F}}_0| \leq 2\binom {k^{2k}}{r+1}\binom {n-2r}{k-r-1}\leq \binom {2r-1}{r}\binom {n-2r+1}{k-r}\lt |{\mathcal {L}}(n,k,r)|. \end{equation*}

Thus we assume $p=r$ .

If $|{\mathcal{P}}(H)|\leq \binom{2r-1}{r}-1$ holds for all $H\in \binom{[n]\setminus X}{k-r}$ , then

\begin{align*} |{\mathcal{F}}|&\leq \sum _{H\in \binom{[n]\setminus X}{k-r}}|{\mathcal{P}}(H)|+|{\mathcal{F}}_0|\\[5pt] &\leq \left (\binom{2r-1}{r}-1\right )\binom{n-|X|}{k-r}+2\binom{k^{2k}}{r+1}\binom{n-2r}{k-r-1}\\[5pt] &\leq \binom{2r-1}{r}\binom{n-2r+1}{k-r}\ (\mbox{for } n\geq n_0(k,r))\\[5pt] &\lt |{\mathcal{L}}(n,k,r)|. \end{align*}

Assume now that for some $H\in \binom{[n]\setminus X}{k-r}$ , $|{\mathcal{P}}(H)|=\binom{2r-1}{r}$ . By Proposition 4.1 we may assume that ${\mathcal{P}}(H)=\binom{Y}{r}$ , $Y\in \binom{X}{2r-1}$ . We claim that $|F\cap Y|\geq r$ for all $F\in{\mathcal{F}}$ . Indeed the opposite would mean that $F\cap P=\emptyset$ for some $P\in \binom{Y}{r}$ . Then $F\cap (H\cup P)\cap X=F\cap P=\emptyset$ , a contradiction. Consequently ${\mathcal{F}}\subset \{F\in \binom{[n]}{k}\colon |F\cap Y|\geq r\}$ , i.e., $\mathcal{F}$ is contained in an isomorphic copy of ${\mathcal{L}}(n,k,r)$ .

Proof. Let ${\mathcal{B}}={\mathcal{B}}({\mathcal{F}})$ . For the proof we use a branching process. During the proof a sequence $S=(x_1,x_2,\ldots,x_\ell )$ is an ordered sequence of distinct elements of $[n]$ and we use $\widehat{S}$ to denote the underlying unordered set $\{x_1,x_2,\ldots,x_\ell \}$ . At the beginning, we assign weight 1 to the empty sequence $S_{\emptyset }$ . At the first stage, we choose $B_1\in{\mathcal{B}}$ with $|B_1|$ minimal. For any vertex $x\in B_1$ , define one sequence $(x)$ and assign the weight $|B_1|^{-1}$ to it.

In each subsequent stage, we pick a sequence $S=(x_1,\ldots,x_p)$ and denote its weight by $w(S)$ . If $\widehat{S}\cap B\neq \emptyset$ holds for all $B\in{\mathcal{B}}$ then we do nothing. Otherwise we pick $B\in{\mathcal{B}}$ satisfying $\widehat{S}\cap B= \emptyset$ and replace $S$ by the $|B|$ sequences $(x_1,\ldots,x_p,y)$ with $y\in B$ and assign weight $\frac{w(S)}{|B|}$ to each of them. Clearly, the total weight is always 1.

We continue until $\widehat{S}\cap B\neq \emptyset$ for all sequences $S$ and all $B\in{\mathcal{B}}$ . Since $[n]$ is finite, each sequence has length at most $n$ and eventually the process stops. Let $\mathcal{S}$ be the collection of sequences that survived in the end of the branching process and let ${\mathcal{S}}^{(\ell )}$ be the collection of sequences in $\mathcal{S}$ with length $\ell$ .

By Claim 5, we see that $|{\mathcal{B}}^{(\ell )}|\leq |{\mathcal{S}}^{(\ell )}|$ for all $\ell \geq 1$ . Let $S=(x_1,\ldots,x_\ell )\in{\mathcal{S}}^{(\ell )}$ and let $S_i=(x_1,\ldots,x_i)$ for $i=1,\ldots,\ell$ . At the first stage, $w(S_1)=1/|B_1|$ . Assume that $B_{i}$ is the selected set when replacing $S_{i-1}$ in the branching process for $i=2,\ldots,\ell$ . Then

\begin{equation*} w(S)= \prod _{i=1}^\ell \frac {1}{|B_{i}|} \geq k^{-\ell }. \end{equation*}

It follows that

\begin{equation*} k^{-k}\sum _{1\leq \ell \leq k}|{\mathcal {B}}^{(\ell )}| \leq \sum _{1\leq \ell \leq k}k^{-\ell }|{\mathcal {B}}^{(\ell )}|\leq \sum _{1\leq \ell \leq k}k^{-\ell }|{\mathcal {S}}^{(\ell )}|\leq \sum _{1\leq \ell \leq k}\sum _{S\in {\mathcal {S}}^{(\ell )}}w(S)\leq \sum _{S\in {\mathcal {S}}}w(S)=1. \end{equation*}

Thus $|{\mathcal{B}}|=\sum \limits _{1\leq \ell \leq k}|{\mathcal{B}}^{(\ell )}| \leq k^k$ .

Proof. Suppose that $|F\cap F'|\leq 2$ and $F,F'\in{\mathcal{F}}$ . If $F\cap F'=\{x,x'\}$ then $F\setminus \{x\}, F'\setminus \{x'\}\in \partial{\mathcal{F}}$ and they are disjoint, a contradiction. The case $F\cap F'=\{x\}$ is even easier.

Proof. Let $r=3$ . Suppose that there exist two edges intersecting in one vertex, say $(x_1,x_2,z), (y_1,y_2,z)\in{\mathcal{F}}^*$ , since $(x_1,x_2)$ is 3-fold covered and $\mathcal{F}$ is intersecting, we have $(x_1,x_2,y_i)\in{\mathcal{F}}$ , $i=1,2$ . Similarly, $(y_1,y_2,x_i)\in{\mathcal{F}}$ , $i=1,2$ . Since $(x_1,x_2,y_2)\in{\mathcal{F}}$ and $(z,y_1)$ is 3-fold covered, $(z,y_1,x_1),(z,y_1,x_2)\in{\mathcal{F}}$ . Similarly, $(z,y_2,x_1),(z,y_2,x_2)\in{\mathcal{F}}$ . Hence $\{x_1,x_2,y_1,y_2,z\}$ spans a complete 3-graph in $\mathcal{F}$ . Since $\mathcal{F}$ is intersecting, we conclude that ${\mathcal{F}}=\binom{[5]}{3}$ up to isomorphism and $|{\mathcal{F}}^*|=10$ . Suppose next that there are two edges intersecting in two vertices say $(x,z_1,z_2),(y,z_1,z_2)\in{\mathcal{F}}^*$ , since $(x,z_1),(y,z_2)$ are 3-fold covered and $\mathcal{F}$ is intersecting, there exists $w$ such that $(x,z_1,y),(y,z_2,x),(x,z_1,w),(y,z_2,w)\in{\mathcal{F}}$ . Arguing with $(x,z_2)$ and $(y,z_1)$ , we infer that $(x,z_2,y),(y,z_1,x),(x,z_2,w),(y,z_1,w)\in{\mathcal{F}}$ . Hence $\{x,y,z_1,z_2,w\}$ spans a complete 3-graph in $\mathcal{F}$ . Since $\mathcal{F}$ is intersecting, we conclude that ${\mathcal{F}}=\binom{[5]}{3}$ up to isomorphism and $|{\mathcal{F}}^*|=10$ . If ${\mathcal{F}}^*$ is 3-intersecting, then $|{\mathcal{F}}^*|\leq 1$ holds trivially.

For $r\geq 4$ , we claim that each member in $\partial{{\mathcal{F}}^*}$ is a transversal of $\mathcal{F}$ . Otherwise, let $G\in \partial{{\mathcal{F}}^*}$ be a $2$ -set that is not a transversal. Then there exists $F\subset{\mathcal{F}}$ such that $F\cap G=\emptyset$ . Since $G\in \partial{\mathcal{F}}^*$ and $r\geq 4\gt |F|$ , there exists $x$ such that $G\cup \{x\}\in{\mathcal{F}}$ and $F\cap (G\cup \{x\})=\emptyset$ , a contradiction.

Thus $\partial{{\mathcal{F}}^*}\subset{\mathcal{T}}({\mathcal{F}})$ . By Lemma 5.1 (i) $\partial{{\mathcal{F}}^*}$ is intersecting. In view of Proposition 5.3, ${\mathcal{F}}^*$ is 3-intersecting. For $n\geq 6$ , by Proposition 5.3 and (1) we have $|{\mathcal{F}}^*|\leq \binom{n-3}{3-3}=1$ . In the case of equality, by symmetry we may assume that ${\mathcal{F}}^*=[3]$ .

Then we claim $|F\cap [3]|\geq 2$ for all $F\in{\mathcal{F}}$ . Indeed, otherwise $|F\cap [3]|=1$ for some $F\in{\mathcal{F}}$ , without loss of generality assume $F\cap [3]=\{1\}$ , then we can find an $F'\in{\mathcal{F}}$ disjoint to $F$ since $(2,3)$ is $r$ -fold covered with $r\geq 4\gt |F|$ , a contradiction. Thus $|F\cap [3]|\geq 2$ for all $F\in{\mathcal{F}}$ . Using that $\mathcal{F}$ is saturated, we conclude that ${\mathcal{F}}={\mathcal{L}}(n,3,2)$ up to isomorphism. For $n\leq 5$ , clearly ${\mathcal{F}}\subset \binom{[5]}{3}$ . Since no 2-set is contained in 4 or more 3-sets in $\binom{[5]}{3}$ , we have $|{\mathcal{F}}^*|=0$ .

Proof. Let ${\mathcal{F}}\subset \binom{[n]}{k}$ be a saturated intersecting family and let ${\mathcal{F}}^*$ be the family of $r$ -complete sets in $\mathcal{F}$ . Let ${\mathcal{B}}={\mathcal{B}}({\mathcal{F}})$ and $X=\mathop{\cup }\limits _{B\in{\mathcal{B}}} B$ . By Lemma 5.2, we have $|X|\leq k\cdot k^k=k^{k+1}$ . Let

\begin{equation*} p=\min \left \{|F\cap X|\colon F\in {\mathcal {F}}^*\right \}. \end{equation*}

If $p\geq 4$ , then for $n\geq n_0(k,r)$ we have

\begin{equation*} |{\mathcal {F}}^*|\leq \sum _{4\leq i\leq k} \binom {|X|}{i}\binom {n-|X|}{k-i}\leq \sum _{4\leq i\leq k} \binom {k^{k+1}}{i}\binom {n-k^{k+1}}{k-i} \leq 2\binom {k^{k+1}}{4}\binom {n-6}{k-4}\lt \binom {n-3}{k-3} \end{equation*}

and we are done. If $p=1$ , then there exists $F^*\in{\mathcal{F}}^*$ such that $F^*\cap X=\{x\}$ . It follows that $\{x\}\in{\mathcal{B}}$ . By saturatedness we have ${\mathcal{F}}={\mathcal{S}}_x$ and $|{\mathcal{F}}^*|=0$ . If $p=2$ then for some $F^*\in{\mathcal{F}}^*$ , $F^*\cap X=\{x,y\}\in{\mathcal{B}}$ . Using $r$ -completeness we find $F_x\in{\mathcal{F}}$ , $F_x\supset F^*\setminus \{y\}$ and $F_y\in{\mathcal{F}}$ , $F_y\supset F^*\setminus \{x\}$ and $F_x\cap F_y\cap X=\emptyset$ , contradicting (22). Thus we may assume $p=3$ .

Define the 3-graph

\begin{equation*} {\mathcal {T}}^* =\left \{T\in \binom {X}{3}\colon \exists F^*\in {\mathcal {F}}^*, F^*\cap X=T\right \},\ {\mathcal {T}} =\left \{T\in \binom {X}{3}\colon \exists F\in {\mathcal {F}}, F\cap X=T\right \}. \end{equation*}

Note that $p=3$ implies ${\mathcal{T}}\neq \emptyset$ . By (22) we infer that $\mathcal{T}$ is intersecting. We distinguish two cases.

Case 1. $r=3$ .

If there exist $(x_1,y_1,z), (x_2,y_2,z)\in{\mathcal{T}}^*$ , let $H_i\in \binom{[n]\setminus X}{k-3}$ such that $H_i\cup \{x_i,y_i,z\}\in{\mathcal{F}}^*$ , $i=1,2$ . By 3-completeness and (22), we have

\begin{equation*} (x_1,y_1,x_2),(x_1,y_1,y_2),(x_2,y_2,x_1),(x_2,y_2,y_1)\in {\mathcal {T}}. \end{equation*}

Then there exists $H_3\in \binom{[n]\setminus X}{k-3}$ such that $H_3\cup \{x_1,y_1,x_2\}\in{\mathcal{F}}$ . Since $H_2\cup \{y_2,z\}$ is covered by at least 3 members of $\mathcal{F}$ , by (22) we infer $H_2\cup (x_1,y_2,z),H_2\cup (y_1,y_2,z)\in{\mathcal{F}}$ . Similarly, we have $H_2\cup (x_2,y_1,z),H_2\cup (x_1,x_2,z)\in{\mathcal{F}}$ . Hence $\{x_1,x_2,y_1,y_2,z\}$ spans a complete 3-graph in $\mathcal{T}$ . We claim that $|F\cap \{x_1,x_2,y_1,y_2,z\}|\geq 3$ for all $F\in{\mathcal{F}}$ . Indeed, otherwise suppose that there is $F\in{\mathcal{F}}$ with $|F\cap \{x_1,x_2,y_1,y_2,z\}|\leq 2$ . Without loss of generality assume that $F\cap \{y_1,y_2,z\}=\emptyset$ . Since $\{y_1,y_2,z\}\in{\mathcal{T}}$ , there exists $H\in \binom{[n]\setminus X}{k-3}$ such that $H\cup \{y_1,y_2,z\}=:F'\in{\mathcal{F}}$ . But then $F\cap F'\cap X=\emptyset$ , contradicting (22). By saturatedness, we conclude that ${\mathcal{F}}={\mathcal{L}}(n,k,3)$ up to isomorphism and $|{\mathcal{F}}^*|= |{\mathcal{L}}(n,k,3)|$ .

If there exist $(x_1,y,z), (x_2,y,z)\in{\mathcal{T}}^*$ , let $H_i\in \binom{[n]\setminus X}{k-3}$ such that $H_i\cup \{x_i,y,z\}\in{\mathcal{F}}^*$ , $i=1,2$ . Since $H_1\cup \{x_1,y\},H_2\cup \{x_2,z\}$ are 3-fold covered, by (22) there exists $w\in X$ such that $H_1\cup \{x_1,y,x_2\},H_1\cup \{x_1,y,w\},H_2\cup \{x_2,z,x_1\},H_2\cup \{x_2,z,w\}\in{\mathcal{F}}$ . Similarly, $H_1\cup \{x_1,z,x_2\},H_1\cup \{x_1,z,w\},H_2\cup \{x_2,y,x_1\},H_2\cup \{x_2,y,w\}\in{\mathcal{F}}$ . Then $\{x_1,x_2,y,z,w\}$ spans a complete 3-graph in $\mathcal{T}$ . By the same argument and saturatedness, we conclude that ${\mathcal{F}}={\mathcal{L}}(n,k,3)$ up to isomorphism and $|{\mathcal{F}}^*|= |{\mathcal{L}}(n,k,3)|$ .

Now we may assume that ${\mathcal{T}}^*$ is 3-intersecting. Since ${\mathcal{T}}^*$ is a 3-graph, we trivially have $|{\mathcal{T}}^*|\leq 1$ . Then for $n\geq n_0(k)$ we obtain that

\begin{equation*} |{\mathcal {F}}^*| \leq \binom {n-|X|}{k-3}+\sum _{4\leq i\leq k} \binom {|X|}{i}\binom {n-|X|}{k-i} \leq 10\binom {n-5}{k-3} \leq |{\mathcal {L}}(n,k,3)|. \end{equation*}

Case 2. $r\geq 4$ .

Claim 6. For all $F\in{\mathcal{F}}^*$ and $T\in{\mathcal{T}}^*$ ,

(23) \begin{align} |F\cap T|\geq 2. \end{align}

Proof. Otherwise using Claim 6, without loss of generality, $\{1,2,3\},\{1,2,4\}\in{\mathcal{T}}^*$ . Let $H_i\in \binom{[n]\setminus X}{k-3}$ such that $H_i\cup \{1,2,i\}\in{\mathcal{F}}^*$ , $i=3,4$ . Let $x_1,\ldots,x_r$ be such that $H_3\cup \{1,3,x_j\}\in{\mathcal{F}}$ , $j=1,\ldots,r$ . Let $y_1,\ldots,y_r$ be such that $H_4\cup \{2,4,y_j\}\in{\mathcal{F}}$ , $j=1,\ldots,r$ . By $r\geq 4$ , without loss of generality assume $x_1\notin \{2,4\}$ and $y_1\notin \{1,3,x_1\}$ . Then

\begin{equation*} (H_3\cup \{1,3,x_1\})\cap (H_4\cup \{2,4,y_1\})\cap X=\emptyset, \end{equation*}

contradicting (22).

By Claim 7, we may assume that ${\mathcal{T}}^*=\{(1,2,3)\}$ . Define

\begin{equation*} {\mathcal {F}}_i^* =\{F\in {\mathcal {F}}^*\colon F\cap [3]=[3]\setminus \{i\}\},\ i=1,2,3. \end{equation*}

Proof. By symmetry assume $i=1$ and set $S=F\cap X$ . Suppose indirectly $|S|\lt r$ . Let $\tilde{F}\in{\mathcal{F}}^*$ with $\tilde{F}\cap X=[3]$ . By $r$ -completeness there are $x_1,x_2,\ldots,x_r$ distinct elements with $(\tilde{F}\setminus \{3\})\cup \{x_j\}\in{\mathcal{F}}$ , $1\leq j\leq r$ . Also $F\setminus \{2\}$ is contained in $r\gt 3$ members of $\mathcal{F}$ . Let $\hat{F}$ be one of them with $\hat{F}\cap [3] = \{3\}$ and let $\hat{S}= \hat{F}\cap X$ . Clearly, $|\hat{S}|\leq |S|\lt r$ . Consequently we can choose $x_j\notin \hat{S}$ . Then

\begin{equation*} \left (\left (\tilde {F}\setminus \{3\}\right )\cup \{x_j\}\right )\cap \hat {F}\cap X=\emptyset, \end{equation*}

contradicting (22).

By Claims 6, 7, 8, we have

\begin{align*} |{\mathcal{F}}^*|&\leq |{\mathcal{T}}^*|\binom{n-|X|}{k-3}+\sum _{1\leq i\leq 3}|{\mathcal{F}}_i^*|\\[5pt] &\leq \binom{n-|X|}{k-3}+\sum _{r\leq i\leq k} \binom{3}{2}\binom{|X|}{i-2}\binom{n-|X|}{k-i}\\[5pt] &=\binom{n-3}{k-3}+O(n^{k-r}) \end{align*}

and the proposition is proven.

Proof. Suppose that $F_1,F_2,\ldots,F_{k+1}$ is a sunflower in ${\mathcal{F}}^*$ with centre $[3]$ and let $G_i=F_i\setminus [3]$ , $i=1,\ldots,k+1$ .

If there exists $F\in{\mathcal{F}}^*$ with $|F\cap [3]|\leq 1$ , pick $G\in \binom{F}{k-1}$ with $G\cap [3]=\emptyset$ . Then $G\cap F_i\neq \emptyset$ can hold for at most $k-1$ values of $i$ . Pick $F_p,F_r$ disjoint to $G$ . Now $k$ -completeness and $k\geq 4$ imply that we can choose $z\notin [3]$ , $G\cup \{z\}\in{\mathcal{F}}$ . Then either $F_p$ or $F_r$ is disjoint to $G\cup \{z\}$ , a contradiction.

If there exists $F\in{\mathcal{F}}^*$ with $|F\cap [3]|= 2$ , without loss of generality, assume that $F\cap [3]= \{1,2\}$ and let $G=F\setminus \{1,2\}$ . Pick $F_p,F_q,F_r$ disjoint to $G$ . Since $k\geq 4$ , we can choose $z,w\notin [3]$ such that $G\cup \{1,z\},G\cup \{1,w\}\in{\mathcal{F}}$ . Then one of $F_p, F_q, F_r$ , without loss of generality say $F_p$ , is disjoint to both $G\cup \{z\}$ and $G\cup \{w\}$ . Since $F_p\setminus \{1\}$ is covered by at least $k$ members of $\mathcal{F}$ , we can find $u\notin G\cup \{1\}$ such that $(F_p\setminus \{1\})\cup \{u\}\in{\mathcal{F}}$ . Then either $z\neq u$ or $w\neq u$ holds. Without loss of generality, assume that $z\neq u$ , then $(F_p\setminus \{1\})\cup \{u\}$ and $G\cup \{1,z\}$ are disjoint, a contradiction. Thus, $[3]\subset F$ for all $F\in{\mathcal{F}}^*$ and the lemma follows.

Proof. Let ${\mathcal{F}}\subset \binom{[n]}{k}$ be a saturated intersecting family. Let ${\mathcal{F}}^*$ be the family of $r$ -complete sets in $\mathcal{F}$ and let ${\mathcal{G}}=\partial{{\mathcal{F}}^*}$ . We claim that each member in $\mathcal{G}$ is a transversal of $\mathcal{F}$ . Otherwise, let $G\in{\mathcal{G}}$ be a $(k-1)$ -set that is not a transversal. Then there exists $F\subset{\mathcal{F}}$ such that $F\cap G=\emptyset$ . Since $G\in \partial{\mathcal{F}}^*$ and $r\gt |F|$ , there exists $x$ such that $G\cup \{x\}\in{\mathcal{F}}$ and $F\cap (G\cup \{x\})=\emptyset$ , a contradiction. Thus ${\mathcal{G}}\subset{\mathcal{T}}({\mathcal{F}})$ .

Since $\mathcal{F}$ is saturated, all $k$ -element supersets of any $G\in{\mathcal{T}}({\mathcal{F}})$ are members of $\mathcal{F}$ . By Lemma 5.1 (i) we see that $\mathcal{G}$ is intersecting. In view of Proposition 5.3, ${\mathcal{F}}^*$ is 3-intersecting. Since $n\geq 4(k-2)$ , by (1) we have $|{\mathcal{F}}^*|\leq \binom{n-3}{k-3}$ .

Proof. Let ${\mathcal{F}}\subset \binom{[n]}{k}$ be a saturated intersecting family. Let ${\mathcal{F}}^*$ be the family of $k$ -complete sets in $\mathcal{F}$ and let ${\mathcal{G}}=\partial{{\mathcal{F}}^*}$ . Define

\begin{equation*} {\mathcal {G}}'=\{G\in {\mathcal {G}}\colon G\notin {\mathcal {T}}({\mathcal {F}})\} \mbox { and } {\mathcal {E}}=\left \{E\in {\mathcal {F}}^*\colon \binom {E}{k-1}\cap {\mathcal {G}}'\neq \emptyset \right \}. \end{equation*}

From Claim 9 it is clear that $H(G)\neq H(G')$ imply $G\cap H(G')\neq \emptyset$ . Define

\begin{equation*} {\mathcal {H}}=\{H(G)\colon G\in {\mathcal {G}}'\}. \end{equation*}

Let ${\mathcal{H}}=\{H_1,\ldots,H_m\}$ . To each $H_i\in{\mathcal{H}}$ , fix $G_i\in{\mathcal{G}}'$ satisfying $H(G_i)=H_i$ . Now $H_i\cap G_j=\emptyset$ iff $i=j$ , By (9), we obtain

(24) \begin{align} |{\mathcal{H}}|=m\leq \binom{2k-1}{k}. \end{align}

For each $H\in{\mathcal{H}}$ , let

\begin{equation*} {\mathcal {G}}'(H)=\{G\in {\mathcal {G}}'\colon H(G)=H\}. \end{equation*}

Now we distinguish two cases.

Case 1. There exists $H\in{\mathcal{H}}$ such that $|{\mathcal{G}}'(H)|\gt k(k-1)\binom{n-4}{k-4}$ .

Since ${\mathcal{G}}'(H)$ is a 2-intersecting $(k-1)$ -graph, by (11) ${\mathcal{G}}'(H)$ is a 2-star. So let ${\mathcal{G}}'(H)$ be a star with centre $\{1,2\}$ . Let $H=\{x_1,\ldots,x_k\}$ . Define

\begin{equation*} {\mathcal {G}}_i^H = \left \{G\in {\mathcal {G}}'(H)\colon \ G\cup \{x_i\}\in {\mathcal {F}}^*\right \}. \end{equation*}

Note that ${\mathcal{G}}'(H)={\mathcal{G}}_1^H \cup \ldots \cup{\mathcal{G}}_k^H$ . Without loss of generality, we may assume that $|{\mathcal{G}}_1^H| \geq \ldots \geq |{\mathcal{G}}_k^H|$ .

Proof. Suppose for contradiction that ${\mathcal{G}}_2^H \neq \emptyset$ and let $ R_2\in{\mathcal{G}}_2^H([2])$ . Since

\begin{equation*} |{\mathcal {G}}_1^H([2])|\geq \frac {1}{k} |{\mathcal {G}}'(H)|\geq (k-1)\binom {n-4}{k-4}, \end{equation*}

we have

\begin{align*} \left |{\mathcal{G}}_1^H([2])\cap \binom{[n]\setminus R_2}{k-3}\right |&\geq |{\mathcal{G}}_1^H([2])|-|R_2|\binom{n-k-3}{k-4}\\[5pt] &\geq (k-1)\binom{n-4}{k-4} -(k-3)\binom{n-4}{k-4}\\[5pt] &=2\binom{n-4}{k-4}. \end{align*}

By (10) we have $\nu ({\mathcal{G}}_1^H([2])\cap \binom{[n]\setminus R_2}{k-3})\geq 2$ . It follows that there are $R_0, R_1\in{\mathcal{G}}_1^H([2])$ such that $R_0, R_1,R_2$ are pairwise disjoint sets. Set $G_i=R_i\cup [2]$ for $i=0,1,2$ . Since $G_1\cup \{x_1\},\ G_2\cup \{x_2\}\in{\mathcal{F}}^*$ , we know that $E_1=R_1\cup \{1,x_1\}$ , $E_2=R_2\cup \{2,x_2\}$ are both covered by $k$ members of $\mathcal{F}$ . To avoid disjointness, we have $E_1\cup \{y_2\}\in{\mathcal{F}}$ for each $y_2\in E_2$ and $E_2\cup \{y_1\}\in{\mathcal{F}}$ for each $y_1\in E_1$ . Moreover, there is an extra element $z$ such that $E_1\cup \{z\},\ E_2\cup \{z\} \in{\mathcal{F}}$ .

Let $E_0=R_0\cup \{1,x_1\}$ and clearly $E_0\cap E_2=\emptyset$ . Since $E_0\subset G_0\cup \{x_1\}\in{\mathcal{F}}^*$ , $E_0$ is covered by $k$ members of $\mathcal{F}$ , we can find $w\notin E_2$ such that $E_0\cup \{w\}\in{\mathcal{F}}$ . For $k\geq 5$ we may choose $u\in R_1$ , $u\neq w$ . Then $E_2\cup \{u\}\in{\mathcal{F}}$ and $E_0\cup \{w\},\ E_2\cup \{u\}$ are disjoint, a contradiction.

Claim 11 implies ${\mathcal{G}}'(H) ={\mathcal{G}}_1^H$ and $G\cup \{x_1\}\in{\mathcal{F}}^*$ for all $G\in{\mathcal{G}}'(H)$ . Then by Lemma 6.1 we may assume that $\nu ({\mathcal{G}}_1^H([2]))\leq k$ . By (10),

\begin{equation*} |{\mathcal {G}}'(H)|\leq k\binom {n-4}{k-4}, \end{equation*}

contradicting our assumption.

Case 2. For each $H\in{\mathcal{H}}$ , $|{\mathcal{G}}'(H)|\leq k(k-1)\binom{n-4}{k-4}$ .

By the definition of $\mathcal{E}$ , we infer that each $E$ in $\mathcal{E}$ contains a $(k-1)$ -set $G\in{\mathcal{G}}'$ , implying $|{\mathcal{E}}|\leq |{\mathcal{G}}'|$ . By Claim 9 and (24),

\begin{equation*} |{\mathcal {E}}|\leq |{\mathcal {G}}'| \leq \sum _{H\in {\mathcal {H}}} |{\mathcal {G}}'(H)| \leq \binom {2k-1}{k}k(k-1)\binom {n-4}{k-4}. \end{equation*}

Define ${\mathcal{F}}_1={\mathcal{F}}^*\setminus{\mathcal{E}}$ . Note that each member of $\partial{\mathcal{F}}_1$ is a transversal of $\mathcal{F}$ . Since $\mathcal{F}$ is saturated, by Lemma 5.1 (i) the family $\partial{\mathcal{F}}_1$ is intersecting. By Proposition 5.3, ${\mathcal{F}}_1$ is 3-intersecting. If $|{\mathcal{F}}_1|\leq k\binom{n-4}{k-4}$ , then

\begin{equation*} |{\mathcal {F}}^*| =|{\mathcal {E}}|+|{\mathcal {F}}_1|\leq \left (\binom {2k-1}{k}k(k-1)+k\right ) \binom {n-4}{k-4}\leq \binom {n-3}{k-3}. \end{equation*}

Otherwise, by (11) we have $[3]\subset F$ for all $F\in{\mathcal{F}}_1$ . Then by Lemma 6.1, we may assume $\nu ({\mathcal{F}}_1([3]))\leq k$ and (10) implies

\begin{equation*} |{\mathcal {F}}_1|\leq k\binom {n-4}{k-4}, \end{equation*}

which contradicts the assumption and the proposition is proven.

Proof. Let ${\mathcal{F}}\subset \binom{[n]}{4}$ be a saturated intersecting family. Let ${\mathcal{F}}^*$ be the family of $4$ -complete sets in $\mathcal{F}$ .

By Claim 12 and $n\geq 38$ , we may assume that ${\mathcal{F}}^*$ is 2-intersecting.

Proof. Assume that $(1,2,3,4), (1,2,5,6),(1,2,7,8)$ form such a sunflower in ${\mathcal{F}}^*$ . Since $(2,3,4),(1,5,6)$ are 4-fold covered in $\mathcal{F}$ , we may assume that

\begin{equation*} (2,3,4,a),(2,3,4,b),(2,3,4,c),(1,5,6,p),(1,5,6,q),(1,5,6,r)\in {\mathcal {F}}. \end{equation*}

The intersecting property of $\mathcal{F}$ implies that (by symmetry) $(a,b)=(5,6)$ , $(p,q)=(3,4)$ , $c=r$ . Since $(1,7,8)$ is 4-fold covered in $\mathcal{F}$ , let $(1,7,8,u),(1,7,8,v),(1,7,8,w)\in{\mathcal{F}}$ . Then one of $u,v,w$ is not in $\{5,6\}$ , by symmetry assume $w\notin \{5,6\}$ , it implies that $(2,3,4,a),(1,7,8,w)$ are disjoint, a contradiction.

Let $F\in{\mathcal{F}}^*$ . For any $P\in \binom{F}{2}$ , define

\begin{equation*} {\mathcal {G}}(P) =\left \{F'\setminus P\colon P\subset F'\in {\mathcal {F}}^*\right \}. \end{equation*}

By Claim 13, $\nu ({\mathcal{G}}(P))\leq 2$ . By Lemma 6.1, we may assume that ${\mathcal{F}}^*$ contains no sunflower with $5$ petals and a centre of size $3$ . It follows that the maximum degree of ${\mathcal{G}}(P)$ is at most 4. Then (25) implies $|{\mathcal{G}}(P)|\leq 2*4+2=10$ . Since ${\mathcal{F}}^*$ is 2-intersecting, we conclude that for $n\geq 63$ ,

\begin{equation*} |{\mathcal {F}}^*| \leq \sum _{P\in \binom {F}{2}} |{\mathcal {G}}(P)| \leq \binom {4}{2} *10 =60\leq n-3 \end{equation*}

and the proposition is proven.