Proof of Theorem 1.3

Let T be a maximum $3$ -wise intersecting family. By Lemma 1.1, $|T|\leq 2^{m-1}$ . However, a star has size $2^{m-1}$ and, hence, $|T|=2^{m-1}$ . Assume for contradiction that T is not a star.

For each k with $0\leq k\leq m$ , we can partition T according to the first k bits. More precisely, given $\boldsymbol {\delta }=(\delta _1, \ldots , \delta _k) \in \{0,1\}^k$ , we define

$$ \begin{align*} T_{\boldsymbol{\delta}} = \{\mathbf{x}\in T \ | \ x_i = \delta_i \text{ for } 1\leq i\leq k \}. \end{align*} $$

When $k=0$ , we define $T_{\boldsymbol {\delta }}=T$ . Then, T is partitioned as

$$ \begin{align*} T = \bigcup_{\boldsymbol{\delta}\in\{0,1\}^{k}} T_{\boldsymbol{\delta}}. \end{align*} $$

Our key observation is the following claim.

Claim 1. Assume that T is $3$ -wise intersecting and not a star. For each k with $0\leq k\leq m-1$ and each $\boldsymbol {\delta } \in \{0,1\}^k$ ,

$$ \begin{align*} |T_{\boldsymbol{\delta}}| = 2^{m-k-1}. \end{align*} $$

Equivalently, among the words of T, each prefix of length k appears exactly $2^{m-k-1}$ times.

Proof of the Claim

We prove the claim by induction on k. When $k=0$ , the assertion is trivial.

Fix $k\geq 1$ . First, we establish that $T_{\boldsymbol {\delta }}\neq \emptyset $ for each $\boldsymbol {\delta } \in \{0,1\}^k$ . Suppose, to the contrary, that $T_{\boldsymbol {\alpha }}=\emptyset $ for some $\boldsymbol {\alpha }=(\alpha _1, \ldots , \alpha _k)$ . By induction,

$$ \begin{align*} |T_{(\alpha_1, \ldots, \alpha_{k-1})}|= 2^{m-(k-1)-1} = 2^{m-k}. \end{align*} $$

After decomposing the set above into a disjoint union

$$ \begin{align*} T_{(\alpha_1, \ldots, \alpha_{k-1})} = T_{(\alpha_1, \ldots, \alpha_{k-1}, \alpha_k)} \cup T_{(\alpha_1, \ldots, \alpha_{k-1}, 1-\alpha_k)} \end{align*} $$

and using $T_{\boldsymbol {\alpha }}=\emptyset $ , we see that $|T_{(\alpha _1, \ldots , \alpha _{k-1}, 1-\alpha _k)}| = 2^{m-k}$ and, thus,

$$ \begin{align*} T_{(\alpha_1, \ldots, \alpha_{k-1}, 1-\alpha_k)}=\{(\alpha_1, \ldots, \alpha_{k-1}, 1-\alpha_k, x_{k+1}, \ldots, x_m) \mid x_j \in \{0,1\} \text{ for } k+1\leq j \leq m\}. \end{align*} $$

Suppose

$$ \begin{align*} T_{(1-\alpha_1, 1-\alpha_2, \ldots, 1-\alpha_{k-1}, \alpha_k)}\neq \emptyset. \end{align*} $$

Pick $(1-\alpha _1, 1-\alpha _2, \ldots , 1-\alpha _{k-1}, \alpha _k, y_{k+1}, \ldots , y_m) \in T_{(1-\alpha _1, 1-\alpha _2, \ldots , 1-\alpha _{k-1}, \alpha _k)}$ . Then,

$$ \begin{align*}(\alpha_1, \alpha_2, \ldots, \alpha_{k-1}, 1-\alpha_k, 1-y_{k+1}, \ldots, 1-y_m) \in T_{(\alpha_1, \ldots, \alpha_{k-1}, 1-\alpha_k)},\end{align*} $$

but

$$ \begin{align*}(1-\alpha_1, 1-\alpha_2, \ldots, 1-\alpha_{k-1}, \alpha_k, y_{k+1}, \ldots, y_m)\end{align*} $$

and

$$ \begin{align*}(\alpha_1, \alpha_2, \ldots, \alpha_{k-1}, 1-\alpha_k, 1-y_{k+1}, \ldots, 1-y_m)\end{align*} $$

do not intersect, which gives a contradiction. Thus,

$$ \begin{align*} T_{(1-\alpha_1, 1-\alpha_2, \ldots, 1-\alpha_{k-1}, \alpha_k)}= \emptyset. \end{align*} $$

By a similar argument as above,

$$ \begin{align*} T_{(1-\alpha_1, 1-\alpha_2, \ldots, 1-\alpha_k)}=\{(1-\alpha_1, \ldots, 1-\alpha_k, x_{k+1}, \ldots, x_m) \mid x_j \in \{0,1\} \text{ for } k+1\leq j \leq m\}. \end{align*} $$

Since T is not a star, it is not the family of all binary vectors whose kth bit is $1-\alpha _k$ . Therefore, we can choose $\boldsymbol {\rho } = (\rho _1, \ldots , \rho _k, \ldots , \rho _m)\in T$ , where $\rho _k=\alpha _k$ . However, the three elements,

$$ \begin{align*} & (\rho_1, \ldots, \rho_{k-1}, \alpha_k, \rho_{k+1}, \ldots, \rho_m), \\ & (\alpha_1, \ldots, \alpha_{k-1}, 1-\alpha_k, 1-\rho_{k+1}, \ldots, 1-\rho_m)\in T_{(\alpha_1, \ldots, \alpha_{k-1}, 1-\alpha_k)},\\ & (1-\alpha_1, 1-\alpha_2, \ldots, 1-\alpha_{k-1}, 1-\alpha_{k}, 1-\rho_{k+1}, \ldots, 1-\rho_m) \in T_{(1-\alpha_1, 1-\alpha_2, \ldots, 1-\alpha_k)}, \end{align*} $$

are in T, but there is no coordinate in which all three words agree, which contradicts the hypothesis that T is a $3$ -wise intersecting family. This completes the proof that $T_{\boldsymbol {\delta }}\neq \emptyset $ for each $\boldsymbol {\delta }\in \{0,1\}^k$ .

Second, we show that for each $\boldsymbol {\delta }=(\delta _1, \ldots , \delta _k)\in \{0,1\}^k$ ,

$$ \begin{align*} \widetilde{T_{\boldsymbol{\delta}}}=\{\widetilde{\mathbf{x}}=(x_{k+1}, \ldots, x_m)\in \{0,1\}^{m-k} \mid (\delta_1, \delta_2, \ldots, \delta_k, x_{k+1}, \ldots, x_m)\in T_{\boldsymbol{\delta}}\} \end{align*} $$

is intersecting. Given $\widetilde {\mathbf {x}}=(x_{k+1}, \ldots , x_{m}), \widetilde {\mathbf {y}} = (y_{k+1}, \ldots , y_{m}) \in \widetilde {T_{\boldsymbol {\delta }}}$ , pick an arbitrary element

$$ \begin{align*} \mathbf{z} = (1-\delta_1, 1-\delta_2, \ldots, 1-\delta_k, z_{k+1},\ldots, z_m ) \in T_{(1-\delta_1, 1-\delta_2, \ldots, 1-\delta_{k})}. \end{align*} $$

Since T is $3$ -wise intersecting, the three binary words $(\boldsymbol {\delta },\widetilde {\mathbf {x}}), (\boldsymbol {\delta }, \widetilde {\mathbf {y}})$ and $\mathbf {z}$ intersect, that is, they agree in some coordinate. This shared bit must be in a position indexed by some $j\in \{k+1, \ldots , m\}$ . In particular, $\widetilde {\mathbf {x}}$ and $\widetilde {\mathbf {y}}$ intersect; as a result, $\widetilde {T_{\boldsymbol {\delta }}}\subseteq \{0,1\}^{m-k}$ is intersecting. By Lemma 1.1,

(2.1) $$ \begin{align} |T_{\boldsymbol{\delta}}|=|\widetilde{T_{\boldsymbol{\delta}}}|\leq 2^{m-k-1}. \end{align} $$

Partition

(2.2) $$ \begin{align} T_{(\delta_1, \delta_2, \ldots, \delta_{k-1})}=T_{(\delta_1, \delta_2, \ldots, \delta_{k-1},\delta_k)} \cup T_{(\delta_1, \delta_2, \ldots, \delta_{k-1},1-\delta_k)}. \end{align} $$

By the inductive hypothesis, $|T_{(\delta _1, \delta _2, \ldots , \delta _{k-1})}|= 2^{m-k}$ . Combining (2.1) and (2.2) finishes the proof of the claim.

We now apply Claim 1 (which is valid since T is assumed not to be a star) to complete the proof by contradiction. Setting $k=m-1$ and $\boldsymbol {\delta }=(0,0,\ldots ,0)\in \{0,1\}^{m-1}$ , we get $|T_{\boldsymbol {\delta }}|=1$ , so T contains a unique word of the form $\mathbf {x}=(0,0,\ldots ,0,u)$ for some $u\in \{0,1\}$ . Setting $\boldsymbol {\delta }=(1,1,\ldots ,1)$ , we similarly find a unique word $\mathbf {y}=(1,1,\ldots ,1,v)\in T$ . Since $\mathbf {x}$ and $\mathbf {y}$ agree only at position m, and T is intersecting, we must have $u=v$ .

Suppose some $\mathbf {z}\in T$ has last coordinate $1-u$ . Then, $\mathbf {x}$ , $\mathbf {y}$ and $\mathbf {z}$ share no common coordinate: any common coordinate would have to be position m (the only position where $\mathbf {x}$ and $\mathbf {y}$ agree), but $\mathbf {z}$ takes the value $1-u\neq u$ there. This contradicts the $3$ -wise intersecting hypothesis. As a result, every word in T has last coordinate u. Since the family of all binary words with last coordinate u is a star of size $2^{m-1}=|T|$ , it follows that T equals this star; this contradicts our assumption that T is not a star.

Proof of Theorem 1.2

We proceed by induction on m. The base case $m=1$ is trivial. We may assume, without loss of generality, that the alphabet X is given by $\mathbb {Z}_q := \mathbb {Z}/q\mathbb {Z}$ .

Since T is a maximum intersecting family, $|T|=q^{m-1}$ by Lemma 1.1. For each i with $1\leq i \leq q$ , we partition the words in T according to the first letter by defining

$$ \begin{align*} T_i=\{\mathbf{x}=(x_1, \ldots, x_m)\in T \mid x_1=i\} \quad \text{and} \quad \widetilde{T}_i=\{\widetilde{\mathbf{x}}=(x_2, \ldots, x_m) \mid (i,\widetilde{\mathbf{x}})\in T_i\}. \end{align*} $$

Given $\boldsymbol {\delta }\in \mathbb {Z}_q^{m-1}$ , consider the set of q words $S(\boldsymbol {\delta })=\{\boldsymbol {\delta }, \boldsymbol {\delta }+\mathbf {1}, \ldots , \boldsymbol {\delta }+(\mathbf {q-1})\}$ , where $\mathbf {c}=(c, \ldots , c)$ represents a constant vector in $\mathbb {Z}_q^{m-1}$ . There are $q^{m-1}$ choices of $\boldsymbol {\delta }\in \mathbb {Z}_q^{m-1}$ . Since each set $S(\boldsymbol {\delta })$ has q elements, this yields $q^{m-2}$ distinct sets of the form $S(\boldsymbol {\delta })$ . Our key observation is the following claim.

Claim 2. For each such $S(\boldsymbol {\delta })$ ,

$$ \begin{align*} \sum_{i=1}^{q} |\widetilde{T}_i\cap S(\boldsymbol{\delta})| \leq q. \end{align*} $$

Moreover, the equality holds if and only if $S(\boldsymbol {\delta }) \subseteq \widetilde {T}_i$ for a unique i with $1\leq i \leq q$ , or there exists $\mathbf {k}=(k, \ldots , k)$ such that $S(\boldsymbol {\delta }) \cap \widetilde {T}_i=\{\boldsymbol {\delta }+\mathbf {k}\}$ for all i with $1\leq i \leq q$ .

Proof of the Claim

Fix $\boldsymbol {\delta }\in \mathbb {Z}_q^{m-1}$ . We consider two cases.

Case 1. There exists some i with $1\leq i \leq q$ such that $|S(\boldsymbol {\delta })\cap \widetilde {T}_i|\geq 2$ .

Take $\boldsymbol {\delta }+\mathbf {c_1}, \boldsymbol {\delta }+\mathbf {c_2}\in S(\boldsymbol {\delta })\cap \widetilde {T}_i$ with $c_1\neq c_2$ . If $\boldsymbol {\delta }+\mathbf {c_3}\in S(\boldsymbol {\delta })\cap \widetilde {T}_j$ for $j\neq i$ , then $\mathbf {c_3}\neq \mathbf {c_1}$ or $\mathbf {c_3}\neq \mathbf {c_2}$ ; however, $(i,\boldsymbol {\delta }+\mathbf {c_1})$ or $(i,\boldsymbol {\delta }+\mathbf {c_2})$ does not intersect $(j,\boldsymbol {\delta }+\mathbf {c_3})$ . This shows $S(\boldsymbol {\delta })\cap \widetilde {T}_j=\emptyset $ for each $j\neq i$ . Therefore,

$$ \begin{align*} \sum_{j=1}^{q} |\widetilde{T}_j\cap S(\boldsymbol{\delta})| = |\widetilde{T}_i\cap S(\boldsymbol{\delta})| \leq |S(\boldsymbol{\delta})| = q, \end{align*} $$

with the equality holding if and only if $S(\boldsymbol {\delta })\subseteq \widetilde {T}_i$ .

Case 2. For all i with $1\leq i\leq q$ , we have $|S(\boldsymbol {\delta })\cap \widetilde {T}_i|\leq 1$ .

In this case, we have $\sum _{i=1}^{q}|\widetilde {T}_i\cap S(\boldsymbol {\delta })|\leq q$ , with the equality holding if and only if $|S(\boldsymbol {\delta }){\kern-1pt}\cap{\kern-1pt} \widetilde {T}_i|{\kern-1pt}={\kern-1pt}1$ for $1{\kern-1pt}\leq{\kern-1pt} i {\kern-1pt}\leq{\kern-1pt} q$ . Suppose the equality holds. If these singleton sets ${S(\boldsymbol {\delta }){\kern-1pt}\cap{\kern-1pt} \widetilde {T}_i}$ do not all coincide, then a similar argument as in the previous paragraph leads to a contradiction. Hence, we can find the same constant $\mathbf {k}$ for which $S(\boldsymbol {\delta }) \cap \widetilde {T}_i=\{\boldsymbol {\delta }+\mathbf {k}\}$ for $1\leq i\leq q$ .

We count triples $(\boldsymbol {\delta },j,\mathbf {w})$ with $\mathbf {w}\in \widetilde {T}_j\cap S(\boldsymbol {\delta })$ in two different ways. First, summing $|\widetilde {T}_j\cap S(\boldsymbol {\delta })|$ over all $\boldsymbol {\delta }\in \mathbb {Z}_q^{m-1}$ and all $1\leq j\leq q$ counts these triples by fixing $\boldsymbol {\delta }$ and j, and then counting the admissible choices of $\mathbf {w}$ . Second, for each fixed $\mathbf {w}\in \widetilde {T}_j$ , there are exactly q choices of $\boldsymbol {\delta }\in \mathbb {Z}_q^{m-1}$ for which $\mathbf {w}\in S(\boldsymbol {\delta })$ , namely $\boldsymbol {\delta }=\mathbf {w}-\mathbf {c}$ with $c\in \mathbb {Z}_q$ . Since $\sum _{j=1}^q |\widetilde {T}_j|=|T|$ , it follows that

$$ \begin{align*} \sum_{\boldsymbol{\delta}\in \mathbb{Z}_q^{m-1}} \sum_{j=1}^{q} |\widetilde{T}_{j}\cap S(\boldsymbol{\delta})| = q|T| = q^{m}. \end{align*} $$

The outer sum has $q^{m-1}$ terms and each inner summand is at most q by Claim 2. Thus, for equality to hold, it must be the case that

$$ \begin{align*} \sum_{i=1}^{q} |\widetilde{T}_i\cap S(\boldsymbol{\delta})| = q \end{align*} $$

for each $\boldsymbol {\delta }\in \mathbb {Z}_q^{m-1}$ . Therefore, by the ‘moreover’ part of Claim 2, for each $\boldsymbol {\delta }\in \mathbb {Z}_q^{m-1}$ , either $S(\boldsymbol {\delta })\subseteq \widetilde {T}_{i}$ for a unique choice of $i\in \{1, 2, \ldots , q\}$ or there exists k (which might depend on $\boldsymbol {\delta }$ ) such that $S(\boldsymbol {\delta })\cap \widetilde {T}_{i}$ consists of the same single element $\boldsymbol {\delta }+\mathbf {k}$ for all i.

Next, we consider these two cases separately.

Case 1: there exist i with $1\leq i \leq q$ and $\boldsymbol {\delta } \in \mathbb {Z}_q^{m-1}$ such that $S(\boldsymbol {\delta }) \subseteq \widetilde {T}_i$ .

We claim that $\widetilde {T}_i = \mathbb {Z}_q^{m-1}$ so that T is a star consisting of all words whose first coordinate is i.

Let $\mathbf {u}\in \mathbb {Z}_q^{m-1}$ . Since $q\geq 3$ , we can find $\mathbf {v}, \mathbf {v'}\in (\mathbb {Z}_q\setminus \{0\})^{m-1}$ such that $\mathbf {u}=\boldsymbol {\delta }+\mathbf {v}+\mathbf {v'}$ . Indeed, any element in $\mathbb {Z}_q=\{0, 1, \ldots , q-1\}$ can be expressed as a sum of two nonzero elements: $j=1+(j-1)$ for $j\neq 1$ and $1=2+(q-1)$ .

Observe that for each $\mathbf {w}\in S(\boldsymbol {\delta })$ and each $j\neq i$ with $1\leq j \leq q$ , we have $\mathbf {w}+\mathbf {v} \notin \widetilde {T}_j$ ; for otherwise, $(i,\mathbf {w})$ and $(j, \mathbf {w}+\mathbf {v})$ are both in T but do not intersect. This shows that $(S(\boldsymbol {\delta })+\mathbf {v})\cap \widetilde {T}_{j}=\emptyset $ for each $j\neq i$ and, thus, Claim 2 forces $S(\boldsymbol {\delta })+\mathbf {v} \subseteq \widetilde {T}_i$ . Consequently, $S(\boldsymbol {\delta }+\mathbf {v})\subseteq \widetilde {T}_i$ . Now, we apply the same argument with $\boldsymbol {\delta }$ replaced with $\widetilde {\boldsymbol {\delta }}=\boldsymbol {\delta }+\mathbf {v}$ to deduce that $S(\widetilde {\boldsymbol {\delta }}+\mathbf {v'})\subseteq \widetilde {T}_i$ . In particular, $\mathbf {u}=\boldsymbol {\delta }+\mathbf {v}+\mathbf {v'}=\widetilde {\boldsymbol {\delta }}+\mathbf {v'}\in \widetilde {T}_i$ . Since $\mathbf{u}\in \mathbb {Z}_q^{m-1}$ is arbitrary, we have shown that $\widetilde {T}_i = \mathbb {Z}_q^{m-1}$ , as desired.

Case 2: for each $\boldsymbol {\delta }\in \mathbb {Z}_q^{m-1}$ , there is $k=k(\boldsymbol {\delta })$ such that

$$ \begin{align*} S(\boldsymbol{\delta}) \cap \widetilde{T}_1 = S(\boldsymbol{\delta}) \cap \widetilde{T}_2 =\cdots = S(\boldsymbol{\delta}) \cap \widetilde{T}_q = \{\boldsymbol{\delta} + \mathbf{k}\}. \end{align*} $$

It follows that $\widetilde {T}_1=\widetilde {T}_2=\cdots =\widetilde {T}_q$ ; write this common family as U. We claim that U is intersecting. Indeed, if $\mathbf {u},\mathbf {v}\in U$ , then $(1,\mathbf {u})$ and $(2,\mathbf {v})$ are both in T, so since T is intersecting and these two words differ in the first coordinate, they must agree in one of the last $m-1$ coordinates. Hence, $\mathbf {u}$ and $\mathbf {v}$ intersect. Moreover, $|U|=q^{m-2}$ . By the inductive hypothesis, U is a star in $\mathbb {Z}_q^{m-1}$ and, therefore, T is a star in $\mathbb {Z}_q^m$ .

In both cases, T is a star. This completes the proof.