Proof. Suppose that there is $v \in G$ such that $L(v)$ is not disperse, and let $X = \{v_1, \dots , v_{\ell }\} \subseteq V \setminus \{v\}$ witness this, namely, $2 \leq |e_{L(v)}(X)| \leq \ell - 2$ . Then $2 \leq e_G(X \cup \{v\}) \leq \ell -1$ , contradicting that $G$ is disperse.

Proof. If $v \in g$ then the assertion clearly holds, because $g \setminus \{v\}$ is a (one-edge) tight walk between $A$ and $B$ in $L(v)$ . So suppose that $v \notin g$ . Then we can write $f = \{v,v_1,\dots ,v_{\ell -1}\}, g = \{v_1,\dots ,v_{\ell -1},w\}$ with $w \neq v$ . As $G$ is disperse, the $(\ell +1)$ -set $f \cup g$ must miss at most one edge. This means that there are $e_1,e_2 \in E(G)$ with $\{v\} \cup A \subseteq e_1$ and $\{v\} \cup B \subseteq e_2$ . Now $e_1 \! \setminus \! \{v\},e_2 \! \setminus \! \{v\}$ is a tight walk between $A$ and $B$ in $L(v)$ .

Proof. We first consider some easy degenerate cases. If $v \in f$ or $v \in g$ then we can take $(f',g') = (f,g)$ or $(f',g') = (g,g)$ , respectively. If $B \subseteq f$ then take $(f',g') = (e,f)$ , and if $|e \cap g| \geq \ell -1$ then take $(f',g') = (e,g)$ . Assuming that none of the above holds, without loss of generality we may write $e = \{v, w_1, \dots , w_{\ell - 1}\}, f = \{w_1, \dots , w_{\ell }\}$ and $g = \{w_2, \dots , w_{\ell + 1}\}$ with the vertices $v,w_1,\dots ,w_{\ell +1}$ being distinct, and $B = \{w_2, \dots , w_{\ell + 1}\} \backslash \{w_i\}$ for some $i \neq \ell + 1$ . Note that $|f \cap B| = \ell -2$ .

Suppose, for the sake of contradiction, that there is no tight walk $f',g'$ as in the lemma. We proceed via a series of claims which will eventually give a contradiction.

Now, we use the above claims to derive a contradiction and hence prove Lemma 2.3. Fix an arbitrary $x \in f \cap B$ (this is possible because $|f \cap B| = \ell -2 \geq 1$ ). Set $C = g \backslash \{x\}$ , and note that $|C| = \ell -1$ ; $C \neq B$ and hence $|C \cap B| = \ell -2$ ; $w_{\ell +1} \in C$ (as $w_{\ell +1} \notin f$ and so $x \neq w_{\ell +1}$ ); and $C \cap B \neq f \cap B$ (as $x \in f \cap B$ while $x \notin C$ ). Consider the set $X = \{v, w_1\} \cup C$ , so $|X| = \ell +1$ . We will show that there exist two edges and two non-edges on $X$ . Indeed, by Claim 2.4, $\{v, w_1\} \cup (C \backslash \{w_{\ell + 1}\}) \in E(G)$ , and by Claim 2.6, $\{w_1\} \cup C \in E(G)$ . On the other hand, by Claim 2.5, $\{v\} \cup C \not \in E(G)$ , and by Claim 2.7, $\{v, w_1\} \cup (C \cap B) \not \in E(G)$ . This contradicts the assumption that $G$ is disperse, as desired.

Proof. We prove this by induction on $i$ . For $i = 1$ the claim is trivial by letting $f = g = e_1$ . For the inductive step, let $f, g$ be a tight walk with $v \in f$ and $e_{i - 1} \cap e_i \subseteq g$ . Note that $f, g, e_i$ is a tight walk as $f,g$ is a tight walk and $|g \cap e_i| \geq |e_{i - 1} \cap e_i| = \ell - 1$ . Moreover, $v \in f$ and $B := e_i \cap e_{i + 1} \subseteq e_i$ , so we may apply Lemma 2.3 to obtain a tight walk $f',g'$ with $v \in f'$ and $e_i \cap e_{i + 1} \subseteq g'$ , as desired.

Proof. Let $e_1,\dots ,e_m$ be a tight walk in $G$ between $A \cup \{v\}$ and $B \cup \{v\}$ . We will prove the lemma by induction on $m$ . For $m \leq 2$ the lemma is trivial, by simply observing that $e_1 \backslash \{v\}, e_m \backslash \{v\}$ is a tight walk between $A$ and $B$ in $L(v)$ . So, suppose $m \geq 3$ . First, consider the case where $v \in e_i$ for some $1 \lt i \lt m$ . Fix any $C \subseteq e_i \! \setminus \! \{v\}$ of size $\ell - 2$ . By the inductive hypothesis, $A$ and $C$ are connected in $L(v)$ , because $e_1,\dots ,e_i$ is a tight walk between $A \cup \{v\}$ and $C \cup \{v\}$ in $G$ . Similarly, $C$ and $B$ are connected in $L(v)$ , by applying the inductive hypothesis to $e_i,\dots ,e_m$ . By transitivity, $A$ and $B$ are connected in $L(v)$ .

Now, suppose that $v \not \in e_i$ for every $1 \lt i \lt m$ . For $1 \leq i \leq m-2$ , let $C_i := e_i \cap e_{i+1} \cap e_{i+2}$ . Then $|C_i| \geq \ell -2$ and we may assume, by passing to a subset if necessary, that $|C_i| = \ell -2$ . Note that $v \notin C_i$ for every $1 \leq i \leq m-2$ . Set also $C_0 = A$ and $C_{m-1} = B$ . It suffices to show that for every $1 \leq i \leq m-1$ , $C_{i-1},C_{i}$ are connected in $L(v)$ . Indeed, this would imply, by transitivity, that $A = C_0$ and $B = C_{m-1}$ are connected in $L(v)$ . Observe that $A$ and $C_1$ are connected in $L(v)$ because $e_1 \! \setminus \! \{v\}$ is an edge of $L(v)$ containing $A, C_1$ . Similarly, $B$ and $C_{m-2}$ are connected in $L(v)$ because $v \in e_m$ and $B,C_{m-2} \subseteq e_m \! \setminus \! \{v\}$ . Now let $2 \leq i \leq m-2$ . By Lemma 2.8, there is a tight walk $f, g$ with $v \in f$ and $e_i \cap e_{i + 1} \subseteq g$ . By definition, $C_i,C_{i-1} \subseteq e_i \cap e_{i+1}$ , and hence $C_i,C_{i-1} \subseteq g$ . Now, by Lemma 2.2, $C_i,C_{i-1}$ are connected in $L(v)$ , as required.

Proof. Let $v \in V(G)$ . Fix any $A,B \subseteq V(G) \setminus \{v\}$ of size $\ell -2$ . As $G$ is tightly connected, $A \cup \{v\}$ and $B \cup \{v\}$ are connected in $G$ . Hence, by Lemma 2.9, $A$ and $B$ are connected in $L(v)$ . This shows that $L(v)$ is tightly connected.

Proof. Let $v \in V(G)$ , and suppose by contradiction that $a,b,c,d$ is an induced path in $L(v)$ . In order for $v,a,b,c$ not to span two edges, we must have $\{a,b,c\} \in E(G)$ . Similarly, $\{b,c,d\} \in E(G)$ and $\{a,b,d\},\{a,c,d\} \notin E(G)$ . But now $a,b,c,d$ span two edges, a contradiction.

Proof of Theorem 1.4. We use induction on $\ell$ . When $\ell = 3$ , the link of each vertex is a cograph by Lemma 2.11, and every cograph $H$ satisfies that $H$ or $\overline {H}$ is not connected. In particular, by the contrapositive of Theorem2.10, this implies that $G$ or $\overline {G}$ is not tightly connected. Now, suppose $\ell \gt 3$ . Fix any $v \in V(G)$ . By Lemma 2.1, $L_G(v)$ is disperse. Hence, by the induction hypothesis, $L_G(v)$ or $\overline {L_G(v)} = L_{\overline {G}}(v)$ is not tightly connected. Thus, $G$ or $\overline {G}$ is not tightly connected by the contrapositive of Theorem2.10, as desired.

Proof. We prove this claim by induction on the uniformity $\ell$ . First, suppose $\ell = 3$ and write $A_1 = \{v,x\}$ , $A_2 = \{v,y\}$ . By Lemma 2.9, $x$ and $y$ are in the same component of $L(v)$ . Since $L(v)$ is a cograph by Lemma 2.11, $L(v)$ is induced $P_4$ -free, and so, the shortest path from $x$ to $y$ in $L(v)$ has length at most 2. Thus, $G$ contains a tight walk from $\{v,x\}$ to $\{v,y\}$ of length at most 2, as required. Now, suppose $\ell \geq 4$ . Let $v \in A_1 \cap A_2$ be arbitrary. By Lemma 2.9, there is a tight walk from $A_1 \! \setminus \! \{v\}$ to $A_2 \! \setminus \! \{v\}$ in $L(v)$ . Hence, by the inductive hypothesis, there exists such a walk of length at most 2. Denote this walk by $e_1, e_2$ (possibly $e_1=e_2$ ). Then, $e_1 \cup \{v\}, e_2 \cup \{v\}$ is a tight walk of length at most 2 from $A_1$ to $A_2$ in $G$ , as desired.

Proof. By assumption, $G$ has a tight walk between $A_1$ and $A_2$ . By Lemma 2.12, there exists such a tight walk $W$ of length at most 2. If $W$ has length 1 then it consists of the single edge $e = A_1 \cup A_2 = B$ , and thus, any $(\ell - 1)$ -subset of $B$ clearly belongs to $\mathcal{C}$ . So suppose that $W$ has length 2 and write $W = (e_1,e_2)$ , where $A_1 \subseteq e_1$ , $A_2 \subseteq e_2$ , and $e_1 \neq e_2$ . Since $G$ is disperse, $G$ misses at most one edge on the $(\ell +1)$ -set $C := e_1 \cup e_2$ . This implies that for every $(\ell -1)$ -set $A \subseteq C$ , there is an edge $e \in E(G)$ with $A \subseteq e \subseteq C$ (indeed, otherwise $C$ misses at least two edges). As $|e \cap e_1| \geq \ell -1$ , we have a tight walk $(e_1,e)$ between $A_1$ and $A$ . Hence, $A$ belongs to $\mathcal{C}$ , as required.

Proof. We will prove this claim by induction on $m$ . The claim is trivial when $m = 1$ , as $e_1$ is a tight walk between any two $(\ell - 1)$ -subsets of $e_1$ . Now assume $m \geq 2$ . By the induction hypothesis, there is a tight component $\mathcal{C}$ containing all $(\ell - 1)$ -subsets of $W := \bigcup _{i = 1}^{m - 1}e_i$ . If $e_m \subseteq \bigcup _{i=1}^{m-1} e_i$ then there is nothing to prove, so suppose otherwise. Let $x$ be the unique vertex in $e_m \setminus e_{m-1}$ , and write $e_m = \{v_1, v_2, \dots , v_{\ell - 1}, x\}$ . It suffices to show that for every $w_1,\dots ,w_{\ell -2} \in W$ , the $(\ell - 1)$ -set $\{w_1, w_2, \ldots , w_{\ell - 2}, x\}$ belongs to $\mathcal{C}$ . We now prove by induction on $j$ that for every $0 \leq j \leq \ell -2$ , $A_j := \{w_1, \ldots , w_j, v_{j + 1}, \ldots , v_{\ell - 2}, x\} \in \mathcal{C}$ . For the base case $j=0$ , note that $A_0 = \{v_1,\dots ,v_{\ell -2},x\}$ is in the same tight component as $\{v_1,\dots ,v_{\ell -1}\} \in \mathcal{C}$ , because both of these $(\ell -1)$ -sets are contained in $e_m$ . For the induction step, let $1 \leq j \leq \ell -2$ . By the inductive hypothesis, $A_{j-1} = \{w_1, \ldots , w_{j - 1}, v_j, v_{j + 1}, \ldots , v_{\ell - 2}, x\} \in \mathcal{C}$ . Also, $\{w_1, \ldots , w_j, v_j, v_{j + 1}, \ldots , v_{\ell - 2}\} \in \mathcal{C}$ because all vertices of this $(\ell -1)$ -set belong to $W$ . Hence, by Lemma 2.13 for the set $B := \{w_1, \ldots , w_j, v_j, v_{j + 1}, \ldots , v_{\ell - 2}, x\}$ , every $(\ell -1)$ -subset of $B$ , and in particular $A_j$ , also belongs to $\mathcal{C}$ . This completes the induction step. Taking $j = \ell -2$ , we get that $A_{\ell -2} = \{w_1,\dots ,w_{\ell -2},x\} \in \mathcal{C}$ , as required.

Proof of Theorem 1.5. Let $\mathcal{C}$ be a tight component of $G$ . Let $U$ be the set of all $v \in V(G)$ such that $v \in A$ for some $A \in \mathcal{C}$ . We want to show that $\mathcal{C} = \binom {U}{\ell -1}$ . So suppose for the sake of contradiction that there exists some $A \subseteq U$ of size $\ell -1$ such that $A \not \in \mathcal{C}$ . Let $A'$ be a largest subset of $A$ such that $A' \subseteq B$ for some $B \in \mathcal{C}$ . As $A \notin \mathcal{C}$ , we have $|A'| \lt |A|$ . So fix $x \in A \! \setminus \! A'$ . As $x \in U$ , there is some $C \in \mathcal{C}$ with $x \in C$ (by the definition of $U$ ). As $B,C \in \mathcal{C}$ , there is a tight walk $e_1,\dots ,e_m$ in $G$ between $B$ and $C$ . By Lemma 2.14, each $(\ell -1)$ -tuple of vertices in $\bigcup _{i=1}^m e_i$ also belongs to $\mathcal{C}$ . As $A' \cup \{x\} \subseteq B \cup C \subseteq \bigcup _{i=1}^m e_i$ , there is an $(\ell -1)$ -tuple $D \in \mathcal{C}$ with $A' \cup \{x\} \subseteq D$ . However, this contradicts the maximality of $A'$ , completing the proof.

Proof. If $d \lt 1$ then $e(G) = \frac {nd}{\ell } \lt \frac {n}{\ell }$ , and deleting one vertex per edge gives an independent set of size at least $\frac {\ell -1}{\ell }n$ . Suppose now that $d \geq 1$ . Sample a random subset $U \subseteq V(G)$ with probability $p := d^{-1/(\ell -1)} \leq 1$ . Deleting one vertex per edge in $U$ gives an independent set of size at least $|U| - e(U)$ . By linearity of expectation, we have $\mathbb{E}[|U| - e(U)] = pn - p^{\ell } \frac {dn}{\ell } = pn(1 - \frac {p^{\ell -1}d}{\ell }) = \frac {\ell -1}{\ell }pn = \frac {\ell -1}{\ell }nd^{-1/(\ell -1)}$ .

Proof. We prove this by induction on $n$ . The case $n=1$ is trivial, so suppose $n \geq 2$ . By definition, we can write $V(G) = V(H_1) \cup V(H_2)$ , where $H_1$ and $H_2$ are vertex-disjoint cohypergraphs, and $G$ has either all edges or no edges which intersect both $V(H_1)$ and $V(H_2)$ . By the inductive hypothesis, $\alpha (H_i) \cdot \omega (H_i) \geq |V(H_i)|$ for $i=1,2$ . Let us assume that $G$ has all edges which intersect both $V(H_1)$ and $V(H_2)$ ; the other case is symmetrical (by switching to $\overline {G}$ ). Then $\omega (G) = \omega (H_1) + \omega (H_2)$ and $\alpha (G) = \max (\alpha (H_1), \alpha (H_2))$ , giving

\begin{align*} \alpha (G) \cdot \omega (G) &= \max (\alpha (H_1), \alpha (H_2)) \cdot (\omega (H_1) + \omega (H_2)) \\ &\geq \alpha (H_1) \cdot \omega (H_1) + \alpha (H_2) \cdot \omega (H_2) \\ &\geq |V(H_1)| + |V(H_2)| \\ &= |V(G)| = n, \end{align*}

as desired.

Proof. First, we will bound the number of edges of $G$ . Note that each edge of $G$ is contained in $V(\mathcal{C})$ for some tight component $\mathcal{C}$ . Thus, by summing over all tight components, we obtain that

\begin{align*} e(G) &\leq \sum _{\mathcal{C}}\binom {|V(\mathcal{C})|}{\ell } \\ &\leq \frac {m}{\ell }\sum _{\mathcal{C}}\binom {|V(\mathcal{C})|}{\ell - 1} \\ &= \frac {m}{\ell }\binom {n}{\ell - 1}, \end{align*}

where the equality uses the fact that every $(\ell -1)$ -subset of $V(G)$ is contained in $V(\mathcal{C})$ for exactly one tight component $\mathcal{C}$ , as the tight components partition $\binom {V(G)}{\ell -1}$ . Letting $d$ denote the average degree of $G$ , we get

\begin{equation*} d = \frac {\ell \cdot e(G)}{n} \leq \frac {m \binom {n}{\ell - 1}}{n} \leq m n^{\ell -2}. \end{equation*}

Thus, by Lemma 3.1,

\begin{align*} \alpha (G) &\geq \frac {\ell - 1}{\ell } \cdot \frac {n}{m^{\frac {1}{\ell -1}}n^{\frac {\ell -2}{\ell -1}}} = \frac {\ell - 1}{\ell } \cdot \left ( \frac {n}{m} \right )^{\frac {1}{\ell -1}}, \end{align*}

as desired.

Proof. We prove the claim by induction on $i$ . The base case $i=1$ holds by assumption as $e_G(X^{\ell - 1}, Y^1) = 0$ . Now let $2 \leq i \leq \ell -1$ and suppose that the result holds for $i-1$ . Without loss of generality, assume that $v_1,v_{\ell -1} \in Y$ (note that $|A \cap Y| = i \geq 2$ ). Suppose, for the sake of contradiction, that there are at least $2^{i - 1}$ edges in $E_G(X^{\ell -i},Y^i)$ which contain $A$ . As $|A \cap Y| = i$ , each such edge consists of $A$ and a vertex from $X$ . So let $X_A = \{x \in X \setminus A \, : \, \{x\} \cup A \in E(G)\}$ denote the set of vertices in $X$ such that $\{x\} \cup A \in E_G(X^{\ell -i},Y^i)$ . Then $|X_A| \geq 2^{i - 1}$ . Observe that for any $x_1, x_2 \in X_A$ , the set of vertices $\{x_1, x_2\} \cup A$ contains at least two edges (namely, the edges $\{x_j\} \cup A$ for $j=1,2$ ), and thus, avoids at most one edge (as $G$ is disperse). Colour the pair $x_1x_2 \in \binom {X_A}{2}$ blue if $\{x_1, x_2, v_2, \dots , v_{\ell -1}\} \in E(G)$ , and red otherwise. So, if $x_1x_2$ is coloured red, then $\{x_1, x_2, v_1, \dots , v_{\ell - 2}\} \in E(G)$ . Fix any $v \in X_A$ . By pigeonhole, at least $\lceil \frac {|X_A|-1}{2} \rceil \geq 2^{i-2}$ of the edges touching $v$ have the same colour. Without loss of generality, suppose this colour is red. That is, $\{v, x, v_1, \dots , v_{\ell - 2}\} \in E(G)$ for at least $2^{i-2}$ vertices $x$ . Note that the set $A' := \{v_1, \dots , v_{\ell - 2}, v\}$ satisfies $|A' \cap Y| = i-1$ , because $v_{\ell -1} \in Y$ and $v \in X$ . Hence, each edge $\{v, x, v_1, \dots , v_{\ell - 2}\}$ as above belongs to $E_G(X^{\ell - i + 1}, Y^{i - 1})$ . Therefore, the number of edges in $E_G(X^{\ell - i + 1}, Y^{i - 1})$ containing $A'$ is at least $2^{i - 2}$ , a contradiction to the induction hypothesis.

Proof. Observe that at each step of the algorithm, the hypergraph $H$ (Line 3) has no edges which cross $(X \cap U, Y \cap U)$ (because all such edges are added to $\mathcal{B}$ ). This implies that $G$ contains all or none of the edges crossing $(X \cap U, Y \cap U)$ (by the choice of $H$ ). Moreover, $|U \cap W| \leq \ell -1$ for every $W \in \mathcal{P}$ (here $\mathcal{P}$ is the partition at the end of the algorithm), because $U$ contains no $\ell$ -tuple from $\mathcal{B}$ , whereas $\binom {W}{\ell } \subseteq \mathcal{B}$ . It follows that $G[U]$ is a cohypergraph obtained by starting with the empty hypergraphs $G[U \cap W]$ , $W \in \mathcal{P}$ , and repeatedly joining two of the parts (by backtracking the algorithm).

Proof. We will bound $|\mathcal{B}_1|$ as follows. For each pair of sets $(X,Y)$ obtained in Line 7 of the algorithm, and each edge of $H$ $e$ which crosses $(X,Y)$ , we assign to $e$ an $(\ell -1)$ -subset $A \subseteq e$ with $A \cap Y = e \cap Y$ (if there are several such $A$ , namely if $|e \cap Y| \leq \ell -2$ , then choose one such $A$ arbitrarily). In other words, for any given $(\ell -1$ )-set $A$ , we will bound the number of edges $e$ of $H$ crossing $(X,Y)$ , which contain $A$ and satisfy $A \cap Y = e \cap Y$ . Summing over all $A \in \binom {V(G)}{\ell -1}$ and all choices for $(X,Y)$ will give us the desired bound on $|\mathcal{B}_1|$ .

So, let $A \subseteq V(G)$ be an arbitrary set of size $\ell -1$ . Let $(X_1, Y_1), \dots , (X_t, Y_t)$ be all pairs of sets $(X,Y)$ obtained in the course of the algorithm for which $A$ crosses $(X,Y)$ , ordered according to when the algorithm processed them. Since any two sets handled by the algorithm are disjoint or nested, and since all $Y_i$ intersect $A$ , we must have $Y_1 \supseteq Y_2 \supseteq \dots \supseteq Y_t$ . Also, since $|X_i| \geq n^{1-\epsilon }$ for every $i$ (see Line 7 of the algorithm), we have $|Y_i| \leq |Y_{i - 1}| - n^{1 - \epsilon }$ for all $2 \leq i \leq t$ , and thus, $t \leq n^{\epsilon }$ . Now, fix any $1 \leq i \leq t$ . As $e_H(X_i^{\ell -1},Y_i^1) = 0$ , Lemma 3.4 implies that the number of edges $e$ of $H$ crossing $(X_i,Y_i)$ and satisfying $A \cap Y_i = e \cap Y_i$ is less than $\sum _{i=1}^{\ell -1} 2^{i - 1} = 2^{\ell - 1} - 1$ . Summing over all choices of $A \in \binom {V(G)}{\ell -1}$ and $i=1,\dots ,t$ , we get that

\begin{align*} |\mathcal{B}_1| &\leq \binom {n}{\ell -1} \cdot n^{\epsilon } \cdot 2^{\ell } \\ &\leq 2^{\ell } n^{\ell - 1 + \epsilon }, \end{align*}

as desired.

Proof. By the definition of the algorithm, we have $|W| \leq n^{1-\gamma }$ for every $W \in \mathcal{P}$ (where $\mathcal{P}$ denotes the partition at the end of the algorithm). Hence,

\begin{align*} |\mathcal{B}_2| &\leq \sum _{W \in \mathcal{P}}\binom {|W|}{\ell } \\ &\leq (n^{1 - \gamma })^{\ell - 1} \cdot \sum _{W \in \mathcal{P}}|W| \\ &= n^{\ell - (\ell - 1)\gamma }, \end{align*}

as desired.