Proof. We have to show that

\begin{equation*}m(a_1\!+\!a_2, b_1\!+\!b_2, *, *, 1)\ge m(a_1, b_1, *, *, 1)\cdot m(a_2, b_2, *, *, 1).\end{equation*}

Consider $t=m(a_2, b_2, *, *, 1)$ pairwise disjoint ground sets $V_1,\dots,V_t$ and for all $i\in [t]$ a copy $({\mathcal{A}}_i,{\mathcal{B}}_i)$ of a construction giving an $(a_1,b_1)$ -bounded $1$ -cross-intersecting SPS of size $s$ such that ${\mathcal{A}}_i=\{A_{i,1},\dots,A_{i,s}\},\;{\mathcal{B}}_i=\{B_{i,1},\dots,B_{i,s}\}$ , where $s=m(a_1, b_1, *, *, 1)$ . Let $({\mathcal{A}},{\mathcal{B}})$ be a copy of an $(a_2,b_2)$ -bounded $1$ -cross-intersecting SPS of size $t$ on the ground set $V$ such that ${\mathcal{A}}=\{A_1,\dots,A_t\},\;{\mathcal{B}}=\{B_1,\dots,B_t\}$ , where $V$ is disjoint from all $V_i$ -s. For any $1\le i\le t,\; 1\le j\le s$ define

\begin{equation*}{A_{i,j}^{\prime}}=A_{i,j}\cup A_i, \; {B_{i,j}^{\prime}}=B_{i,j}\cup B_i.\end{equation*}

The pairs $({A_{i,j}^{\prime}},{B_{i,j}^{\prime}})$ form a $1$ -cross-intersecting SPS such that $|{A_{i,j}^{\prime}}|\le a_1+a_2$ and $|{B_{i,j}^{\prime}}|\le$ $b_1+b_2$ .

Proof. Let ${\textbf{a}}_i$ (resp. ${\textbf{b}}_i$ ) denote the characteristic vector of $A_i$ (resp. $B_i$ ), i.e. ${\textbf{a}}_i(v)=1$ for $v\in V$ if and only if $v\in A_i$ . Otherwise the coordinates are $0$ . Suppose that

\begin{equation*} \sum _{i=1}^m \lambda _i {\textbf {a}}_i = {\boldsymbol {0}}.\end{equation*}

Take the dot product of both sides of this equation with ${\textbf{b}}_j$ . Since $|A_i\cap B_j|=1$ for $i\ne j$ and $|A_i\cap B_j|=0$ for $i=j$ , we get that

\begin{equation*}\left (\sum _{i=1}^m \lambda _i\right )-\lambda _j=0.\end{equation*}

Adding these for all $j$ yields $(m-1) (\sum _{i=1}^m \lambda _i )=0$ . Consequently (using $m\gt 1$ ) $\sum _{i=1}^m \lambda _i=0$ . Thus $\lambda _j=0$ for all $j$ .

Proof. Let $(\mathcal{A},\mathcal{B})$ be a $(2,n)$ -bounded $1$ -cross-intersecting SPS of size $m$ . It is convenient to assume that $\mathcal{A}$ is two-uniform (a graph without multiple edges) and $\mathcal{B}$ is an $n$ -uniform hypergraph. (For smaller sets dummy vertices can be added).

Consider the simple graph $\mathcal{A}$ .

Assume next that $\mathcal{A}$ is an acyclic graph.

Lemma 3.2. Assume that $T\subseteq{\mathcal{A}}$ is a non-star tree component with $t$ edges. Then

\begin{equation*}\max _{A_i\in T} |B_i\cap V(T)|\ge \left \lceil {t\over 2}\right \rceil .\end{equation*}

Proof. Let $P=x,y,z,z_2, \dots$ be a maximal path of $T$ , set $A_1=\{x,y\},A_2=\{y,z\}$ . Let $S\subseteq V(T)$ the set of leaves connected to $y$ . Note that $t\geq 3$ , $|V(T)|=t+1$ , $\;N_T(y)=S\cup \{ z\}$ and $x\in S$ . Then $B_1\cap V(T)$ is the set $X$ of vertices with odd distance from $y$ in the tree $T-x$ . On the other hand, $B_2\cap V(T)$ is the set $X'=S\cup D$ where $D$ is the set of vertices with odd distance from $z$ in the tree $T-(S\cup \{ y\})$ . Then $|X|+|X'|=t+|S|-1\ge t$ . Therefore

\begin{equation*} \max \{|B_1\cap V(T)|, |B_2\cap V(T)|\} =\max \{ |X|, |X'|\} \ge \left \lceil {t\over 2}\right \rceil . \end{equation*}

Assume that there is a non-star tree component $T$ in $\mathcal{A}$ with $t$ edges, $A_1,\dots,A_t$ , $(t\ge 3)$ . We define another $(2,n)$ -bounded $1$ -cross-intersecting SPS $(\mathcal{A'},\mathcal{B'})$ of size $m$ . Let ${\mathcal{A}}'$ be the graph defined by replacing $T$ with $S$ , where $S$ is the union of two vertex disjoint stars $S_1$ and $S_2$ with centres $s_1,s_2$ having $ \lceil{t\over 2} \rceil$ and $ \lfloor{t\over 2} \rfloor$ edges, respectively. We keep all edges of the other components of $\mathcal{A}$ , i.e., ${\mathcal{A}}'= ({\mathcal{A}}\setminus E(T))\cup E(S)$ .

For $i=1,\dots,t$ in case of ${A_{i}^{\prime}}\in E(S_\alpha )$ let $C_i$ be the complement of ${A_{i}^{\prime}}$ in the star $S_\alpha$ together with the centre of the other star of $S$ , i.e., $C_i= (V(S_\alpha )\setminus {A_{i}^{\prime}} )\cup \{ s_{3-\alpha }\}$ . Note that $|C_i|$ is either $\lfloor{t\over 2}\rfloor$ or $\lceil{t\over 2}\rceil$ . According to Lemma 3.2 there is a hyperedge, say $B_1$ , with $|B_1\cap V(T)|\ge \lceil{t\over 2} \rceil$ . Define ${\mathcal{B}}'$ as follows.

\begin{equation*}B_i^{\prime}\,:\!=\,\left \{\begin {array}{ll} C_i\cup (B_1\setminus V(T)) &\qquad \text {for } 1\leq i\leq t, \\ \{ s_1, s_2\}\cup (B_i\setminus V(T))& \qquad \text {for } i\gt t. \end {array}\right . \end{equation*}

Proof. It is clear that $({{\mathcal{A}}'},{{\mathcal{B}}'})$ is a $1$ -cross-intersecting SPS of size $m$ . To prove that it is $(2,n)$ - bounded, assume first that $1\le i\le t$ . Then

\begin{equation*} |B_i^{\prime}|= |C_i|+|B_1\setminus V(T))| \le \left \lceil {t/2}\right \rceil +\left ( |B_1|-\left \lceil {t/2}\right \rceil \right )=|B_1|\leq n. \end{equation*}

If $i\gt t$ , we have

\begin{equation*}|B_i^{\prime}|= 2+|B_i\setminus V(T)|\le |B_i\cap V(T)|+|B_i\setminus V(T)|\leq n,\end{equation*}

where the inequality $2\le |B_i\cap V(T)|$ holds because $T$ is not a star.

Applying Claim 3.3 repeatedly, we may assume that all components of $\mathcal{A}$ are stars, $S_1,\dots,S_k$ , where $S_i$ has $t_i\ge 1$ edges. For any edge $A_j\in S_i$ , $n\ge |B_j|=t_i-1+k-1$ . Adding these inequalities for $i=1,\dots,k$ , we obtain that $kn\ge m-2k+k^2$ which leads to $k(n+2-k)\ge m$ . Hence

\begin{equation*}m\leq k(n+2-k)\leq \Bigl (\left \lfloor {n\over 2}\right \rfloor +1\Bigr )\Bigl (\left \lceil {n\over 2}\right \rceil +1\Bigr ). \end{equation*}

Taking together the bounds for odd cycles and acyclic graphs, we get that

\begin{equation*}m\le \max \left \{2n+1,\; \Bigl (\left \lfloor {n\over 2}\right \rfloor +1\Bigr )\Bigl (\left \lceil {n\over 2}\right \rceil +1\Bigr )\right \}.\end{equation*}

For $n=2,3$ the first term is larger, for $n=4$ they are equal, and for $n\ge 5$ the second term takes over. This proves the upper bound for $m$ .

The matching lower bound for $n\ge 4$ comes from Proposition 1.1 applied to the standard construction with values $(1,\lceil{n\over 2}\rceil )$ and $(1,\lfloor{n\over 2}\rfloor )$ . For $n=2$ the hypergraph ${\mathcal H}(2,2)$ works (defined in Subsection 1.1). For $n=3$ we can define ${\mathcal H}(2,3)$ as the pairs $ (\{i, i\!+\!1\}, \{i\!+\!2, i\!+\!4, i\!+\!6\} )$ taken modulo $7$ .

Proof. Our first observation here is the following.

Proof. Suppose $v\in A_1\cap \ldots \cap A_{n+2}$ . Then $v\not \in B_i$ for $i\le n+2$ and in $\bigcup _{i=1}^{n+2}A_i\setminus \{v\}$ the sets $A_i^{\prime}=A_i\setminus \{v\}$ are pairwise disjoint. The set $B_{n+2}$ must intersect each ${A_{1}^{\prime}},\ldots,{A_{n+1}^{\prime}}$ which is impossible.

Proof. Suppose that $({\mathcal{A}},{\mathcal{B}})$ is an $(n,n)$ -bounded $1$ -cross-intersecting SPS of size $m$ such that both $\mathcal{A}$ and $\mathcal{B}$ are linear hypergraphs. We have $m_2({01{\textrm{-int}}},{01{\textrm{-int}}}, 1)\leq 5$ by Theorem 1.4 so we may suppose that $n\geq 3$ . If $m\leq 2n+2$ then there is nothing to prove, so from now on, we may suppose that $m\geq 2n+3$ .

We claim that for every $v\in V$ , $d_{\mathcal{A}}(v)$ , $d_{\mathcal{B}}(v)\le n$ . Indeed, $d_{\mathcal{A}}(v)\leq n+1$ (and in the same way $d_{\mathcal{B}}(v)\le n+1$ ) is obvious from Claim 4.1. Suppose $d_{\mathcal{A}}(v)= n+1$ , say $v\in A_1\cap \dots \cap A_{n+1}$ then $m\gt 2n+2\geq d_{\mathcal{A}}(v)+d_{\mathcal{B}}(v)$ so there is a pair $A_i,B_i$ with $i\gt n+1$ such that $v\notin A_i\cup B_i$ . Thus $B_i$ cannot intersect all $A_j$ -s containing $v$ , proving the claim.

Since $(\mathcal{A},\mathcal{B})$ is $1$ -cross-intersecting we have $\sum _{v\in B_i} d_{\mathcal{A}}(v)=m-1$ for each $B_i$ . Adding up these $m$ equations we get

(4) \begin{equation} \sum _v d_{\mathcal{A}}(v)d_{\mathcal{B}}(v)=m^2-m. \end{equation}

Let ${\mathcal{A}}_i$ be the set of $A_j$ -s that intersect $A_i$ and different from $A_i$ . Our crucial observation is that if $A_i$ and $A_j$ do not intersect then

(5) \begin{equation} |{\mathcal{A}}_i|+|{\mathcal{A}}_j|\leq n^2. \end{equation}

Indeed, the left-hand side of (5) equals to $\sum _{\ell : \ell \neq i,j} |A_\ell \cap (A_i\cup A_j)|$ . For two disjoint sets $X,Y$ we say that a pair $(x,y)$ joins $X,Y$ if $x\in X,y\in Y$ . For $\ell \neq i,j$ we have $|A_\ell \cap (A_i\cup A_j)|\leq 2$ . In case of $|A_\ell \cap (A_i\cup A_j)|= 2$ , we select two pairs $(x,y),(x',y')$ joining $A_i,A_j$ , namely $(x,y)=A_\ell \cap (A_i\cup A_j)$ and $(x',y')=B_\ell \cap (A_i\cup A_j)$ . In case of $|A_\ell \cap (A_i\cup A_j)|= 1$ we select one pair $(x,y)$ joining $A_i,A_j$ , namely $(x,y)=B_\ell \cap (A_i\cup A_j)$ . These pairs are distinct because

\begin{equation*}|A_{\ell }\cap B_{\ell '}|\le 1,|A_{\ell }\cap A_{\ell '}|\le 1, |B_{\ell }\cap B_{\ell '}|\le 1. \end{equation*}

Since there are $n^2$ pairs between $A_i$ and $A_j$ we obtain that $\sum _{\ell : \ell \neq i,j} |A_\ell \cap (A_i\cup A_j)|\leq n^2$ , completing the proof of (5).

If $A_i\cap A_j=\{v\}$ then we will prove that

(6) \begin{equation} |{\mathcal{A}}_i|+|{\mathcal{A}}_j|\leq (n-1)^2+d_{\mathcal{A}}(v)+d_{\mathcal{B}}(v) \leq n^2+1. \end{equation}

Indeed, as before,

\begin{equation*} |{\mathcal {A}}_i|+|{\mathcal {A}}_j|=\sum _{\ell : \ell \neq i} |A_\ell \cap A_i| +\sum _{\ell : \ell \neq j} |A_\ell \cap A_j|. \end{equation*}

For every $\ell \neq i,j$ we select (at most) two pairs joining $A_i\setminus \{ v\}$ to $A_j\setminus \{ v\}$ , namely $A_\ell \cap ((A_i\setminus \{ v\})\cup (A_j\setminus \{ v\}))$ and $B_\ell \cap ((A_i\setminus \{ v\})\cup (A_j\setminus \{ v\}))$ . In this way we selected at least $ |A_\ell \cap A_i|+|A_\ell \cap A_j|$ distinct pairs except if $v\in A_\ell \cup B_\ell$ . In the latter case we still have selected at least $ |A_\ell \cap A_i|+|A_\ell \cap A_j|-1$ pairs. So the left-hand side of (6) is at most the number of pairs joining $A_i\setminus \{ v\}$ to $A_j\setminus \{ v\}$ plus $d_{\mathcal{A}}(v)+d_{\mathcal{B}}(v)$ . This completes the proof of (6).

Next we prove that

(7) \begin{equation} \sum _{v\in V} d_{\mathcal{A}}(v)^2 \leq m\left (\frac{1}{2}n^2 + n + \frac{1}{2}\right ). \end{equation}

Add up inequalities (5) and (6) for all $1\leq i\lt j\leq m$

\begin{equation*} \frac {1}{m-1}\sum _{1\leq i\lt j\leq m} |{\mathcal {A}}_i|+|{\mathcal {A}}_j| \leq \frac {1}{m-1}\binom{m}{2} (n^2+1)= m\left (\frac {1}{2}n^2 + \frac {1}{2}\right ). \end{equation*}

Here, the left-hand side is

\begin{equation*} \sum _{1\leq i\leq m} |{\mathcal {A}}_i|= \sum _{1\leq i\leq m}\left ( \sum _{v\in A_i} (d_{\mathcal {A}}(v)\!-\!1)\right ) =\sum _{v\in V} \left (d_{\mathcal {A}}(v)^2-d_{\mathcal {A}}(v)\right )= \left (\sum _{v\in V} d_{\mathcal {A}}(v)^2\right ) -mn. \end{equation*}

The last two displayed formulas yield (7) and equality can hold only if (5) was not used. Note that similar upper bound must hold for $ \sum _{v\in V} d_{\mathcal{B}}(v)^2$ , too.

Apply (7) to $\mathcal{A}$ and to $\mathcal{B}$ and subtract the double of (4). We obtain

\begin{align*} 0\leq \sum _{v\in V} (d_{\mathcal{A}}(v)-d_{\mathcal{B}}(v))^2 & = \sum _{v} d_{\mathcal{A}}(v)^2 + \sum _{v} d_{\mathcal{A}}(v)^2 -2\sum _v d_{\mathcal{A}}(v)d_{\mathcal{B}}(v) \\ &\leq 2m\left ( \frac{1}{2}n^2 + n + \frac{1}{2}\right ) -2m(m-1)= 2m \left ( \frac{1}{2}n^2 + n + \frac{3}{2}-m\right ). \end{align*}

This implies $m\leq \frac{1}{2}n^2 + n + \frac{3}{2}$ . As a last step, we show that this inequality is strict completing the proof of the upper bound on $m$ . Indeed, equality can hold only if (5) was never used to $\mathcal{A}$ neither to $\mathcal{B}$ . This implies that $\mathcal{A}$ and $\mathcal{B}$ are 1-intersecting and because of (6) there exists a $v$ with $d_{\mathcal{A}}(v)=d_{\mathcal{B}}(v)=n$ . Suppose

\begin{equation*}v\in A_1\cap \dots \cap A_n\cap B_{n+1}\cap \dots \cap B_{2n}.\end{equation*}

Then $A_{n+1}\cap B_{n+2}=\emptyset$ because $A_{n+1}\cap B_i$ , $B_{n+2}\cap A_i$ are nonempty for $i=1,\dots,n$ . This contradicts the $1$ -intersection property.

Proof. Recall that ${\mathcal{H}}={\mathcal{A}}\cup{\mathcal{B}}$ . First, consider the case when there exists a vertex $v$ with $d_{\mathcal{H}}(v)\geq n+1$ , say $v\in A_i\cup B_i$ for $i\in \{1, 2, \dots, n+1 \}$ . Then one of the members of $\{ A_{n+2}, B_{n+2}\}$ does not cover $v$ , say, $v\notin A_{n+2}$ . Then, $A_{n+2}$ cannot intersect all members of $\{ A_i, B_i\}_{1\leq i\leq n+1}$ containing $v$ , a contradiction. So in this case $m=n+1$ and we are done.

From now on, we may suppose that $m\gt n+1$ , and $d_{\mathcal{H}}(v)\leq n$ for all $v\in V$ . Since only $B_1$ is disjoint from $A_1$ we get

\begin{equation*} 2m=|{\mathcal {H}}|= 2+ \sum _{v\in A_1} (d_{\mathcal {H}}(v)-1)\leq 2+ n(n-1). \end{equation*}

and we conclude that $m\le\binom{n}{2}+1$ . If $n\ge 4$ and equality holds, then all vertices of $A_1$ (and of all other hyperedges) must have degree $n$ .