Defining cores

Our cores will be disjoint subsets of size $s-3$ . Let us try to give a rough intuition for the purpose of our cores. For every edge $e$ , we aim to find a set $C$ of size $s-3$ such that $G[e\cup C]\cong K_s^3$ . Further, we want that either $H[e\cup C]$ will form a clique without $e$ , and then $e$ can be immediately activated, or that after several steps of activation of other edges, $H[e\cup C]$ together with the previously activated edges will form a clique without $e$ . To allow for this activation process, we will not choose cores for the edges themselves (as this will be inefficient and require too many edges in $H$ ), but rather assign cores to sets of vertices $S$ of size at most two. Each such core allows us to either activate immediately edges containing $S$ , or to do so through some ‘chain’ of activation. As each core will contribute additional edges to $H$ , we aim to minimise the number of cores. We now turn to choosing the cores.

We begin by choosing $C_0$ , where the only requirement on the set $C_0$ is that $G[C_0]\cong K_{s-3}^3$ (whp such a set exists in $G$ ). We remark that we will use $C_0$ to activate edges $e\subseteq [n]\setminus C_0$ satisfying that $G[e\cup C_0]=K_s^3$ .

We now turn to define a core for every singleton $v\in [n]\setminus C_0$ (noting that cores for different vertices may coincide). For every $v\in [n]\setminus C_0$ , if $G[C_0\cup \{v\}]\cong K_{s-2}^3$ , we choose the core $C_v \, :\! = C_0$ . Otherwise, if $G[C_0\cup \{v\}]\ncong K_{s-2}^3$ , we choose a core $C_v$ , disjoint from $v$ , with the following property.

(1) \begin{align} \text{Every edge $f\subseteq C_v\cup \{v\}\cup C_0$ with $f\cap C_v\neq \varnothing $ is in $G$.} \end{align}

Indeed, whp this is possible for every such $v\in [n]\setminus C_0$ .

Let us mention two observations.

1. (O1) We have that $G[C_0\cup C_v]\cong K^3_{|C_0\cup C_v|}$ , and $|C_0\cup C_v|=s-3$ when $C_v=C_0$ , and $|C_0\cup C_v|=2(s-3)$ otherwise. Indeed, every $e\subseteq C_0$ is in $G$ since $G[C_0]\cong K^3_{s-3}$ , and every remaining edge $e\subseteq C_0\cup C_v$ with $e\cap C_v\neq \varnothing$ is in $G$ by Property (1).

2. (O2) If $u\in C_v$ for some $v$ , we have that $G[C_0\cup \{u\}]\cong K_{s-2}^3$ , and therefore $C_u=C_0$ .

We remark that in the final activation step of the argument, for every $v\in [n]$ , using the edges activated through $C_0$ , we will be able to activate edges of the form $e=\{v, x,y\}\subseteq [n]\setminus C_v$ where $G[e\cup C_v]\cong K_s^3$ . Crucially, note that for such edges $e$ (when $C_v\neq C_0$ ), we have $G[e\cup C_0]\ncong K_s^3$ , and thus choosing an additional core was indeed necessary.

Still, there will be edges $e=\{v_1,v_2,v_3\}$ , where for every $i\in [3]$ , $G[e\cup C_{v_i}]\ncong K_s^3$ . We thus turn to defining cores for pairs of vertices. We will define cores for all pairs of vertices $S=\{v_1,v_2\}$ which satisfy that $v_1\notin C_{v_2}$ and $v_2\notin C_{v_1}$ (the reason for defining cores for such vertices will become clearer in the last two steps of the argument: when counting the number of edges in $H$ and when considering the activation of edges). We will ensure that $C_S$ satisfies the following property.

(2) \begin{align} \text{Every edge $f \subseteq S \cup C_0 \cup C_{v_i} \cup C_S$ with $f \cap C_S \neq \varnothing $ is in $G$, for every $i \in \{1, 2\}$.} \end{align}

In all the cases when one of the sets $C_0,C_{v_1},C_{v_2}$ can be chosen for the role of $C_S$ so that Property (2) is satisfied, we will indeed set $C_S$ to be equal to such a core. However, in other cases, we will have to assign a new core for $S$ . Let us now describe in detail how we define the core $C_S$ .

If $G[C_0 \cup \{v_1, v_2\}] \cong K^3_{|C_0\cup \{v_1,v_2\}|}$ , then we set $C_{S} = C_0$ . Assume now that it is not the case. If for some $i\in \{1,2\}$ , every edge $f\subseteq C_{v_i}\cup S\cup C_0$ with $f\cap C_{v_i}\neq \varnothing$ is in $G$ , then we set $C_S=C_{v_i}$ (since verifying that (2) holds for such a choice of $C_S$ is somewhat technical, we omit the explanation here, and refer to the complete proof in Section 3). Otherwise, a new choice for the core $C_S$ should be made to later activate edges containing $S$ . Whp for any such $S$ we may find a core $C_S$ , disjoint from $S$ , which satisfies Property (2).

Let us mention a few observations. Below, we fix $S=\{v_1,v_2\}$ .

1. (O3) For every $i\in \{1,2\}$ and $c\in C_S$ , we have that $C_{\{v_i,c\}}=C_{v_i}$ . Indeed, we need to show that every edge $f\subseteq C_{v_i}\cup \{v_i,c\}\cup C_0$ with $f\cap C_{v_i}\neq \varnothing$ is in $G$ . Now, if $c\in f$ , then $f\cap C_S\neq \varnothing$ and thus, by Property (2), we have that $f$ is in $G$ . Otherwise, $f\subseteq C_{v_i}\cup \{v_i\}\cup C_0$ (and intersects with $C_{v_i}$ ), and thus, by Property (1), we have that $f$ is in $G$ .

2. (O4) For every $i\in \{1,2\}$ , we have that $G[C_0\cup C_{v_i}\cup C_{S}]\cong K^3_{|C_0\cup C_{v_i}\cup C_S|}$ . Indeed, by Observation (O1) we have that $G[C_0\cup C_{v_i}]\cong K^3_{|C_0\cup C_{v_i}|}$ , and by Property (2), every edge $f\subseteq C_0\cup C_{v_i}\cup C_S$ with $f\cap C_S\neq \varnothing$ is in $G$ .

3. (O5) If $X\subseteq [n]\setminus C_0$ satisfies that $G[C_0\cup X]\cong K^3_{|C_0\cup X|}$ , then $C_S=C_0$ for every $S\subseteq X$ of size at most two. In particular, if $Y\subseteq C_{v_i}\cup C_S$ with $|Y|\le 2$ , by (O4) we have that $G[C_0\cup Y]\cong K^3_{|C_0\cup Y|}$ and therefore $C_{Y}=C_0$ .

Constructing $H$

Let $H\subseteq G$ be a subhypergraph on $V(G)=[n]$ , consisting of the following edges:

1. (1) Every edge $e\subseteq C_0$ is in $H$ .

2. (2) For every $v\in [n]\setminus C_0$ , we add to $H$ all edges of the form $\{v\}\cup C'$ for every $C'\subseteq C_v$ of size two. These edges are in $G$ by Property (1).

3. (3) For every $v\in [n]\setminus C_0$ , we add to $H$ all edges of the form $\{v,c_0,c\}$ for every $c_0\in C_0$ and $c\in C_v$ . These edges are in $G$ by Property (1).

4. (4) For every $S=\{v_1, v_2\}\subseteq [n]$ for which $C_S$ is defined, we add to $H$ all edges of the form $S\cup \{c\}$ for every $c\in C_S$ . These edges are in $G$ by Property (2).

While Steps (1) and (2) are rather transparent, we refer to Figure 1 for an illustration of edges added to $H$ at Steps (3) and (4).

Figure 1. Illustration of a chain of cores, $C_0$ , $C_{v_2}$ , and $C_{\{v_1,v_2\}}$ , together with edges that were added to $H$ . The edge $\{v_1, v_2, c_1\}$ is added to $H$ by (4). The edge $\{v_2, c_1, c_2\}$ is added to $H$ by (4), noting that $C_{\{v_2, c_1\}}=C_{v_2}$ by (O3). The edge $\{c_1, c_2, c_3\}$ is added to $H$ by (4) since $C_{\{c_1, c_2\}}=C_0$ by (O5). Finally, the edge $\{v_2, c_2, c_3\}$ is added to $H$ by (3).

Number of edges in $H$

The number of edges we have added to $H$ is at most

\begin{equation*}\binom {s-3}{3}+\binom {n-s+3}{1}\binom {s-3}{2}+\binom {n-s+3}{2}\binom {s-3}{1}=\binom {n}{3}-\binom {n-s+3}{3}.\end{equation*}

Indeed, the first term comes from the edges induced by $C_0$ , which we have added to $H$ at Step (1).

The second term comes from the edges which were added at Step (2). For every $v\in [n]\setminus C_0$ , we have chosen $C_v$ (possibly $C_v=C_0$ ), $|C_v|=s-3$ , and added all edges of the form $\{v\}\cup C'$ where $C'\subseteq C_v$ with $|C'|=2$ . Thus, we have $\binom {n-s+3}{1}$ choices for $v \in [n] \setminus C_0$ and $\binom {s-3}{2}$ choices for $C' \subseteq C_v$ of size two.

Finally, the third term comes from the edges which were added at Step (3) and at Step (4). To see this, consider a pair of vertices $S = \{u, v\} \subseteq [n] \setminus C_0$ . Recall that we define $C_S$ only if $v\notin C_{u}$ and $u\notin C_{v}$ . Thus, if $C_S$ is not defined, then WLOG $v \in C_{u}$ . In this case, since $v \notin C_0$ , we must have $C_u \cap C_0 = \varnothing$ (recall that the cores are either identical or vertex-disjoint). In Step (3), we add all the edges $\{v, u, c_0\}$ for each of the $\binom {s-3}{1}$ possibilities for $c_0 \in C_0$ (note here, that when considering all pairs $\{u,v\}$ in this manner, we in fact cover all edges added at Step (3)). Otherwise, $C_S$ is defined, and at Step (4) we add all edges $\{u, v, c\}$ for each of the $\binom {s-3}{1}$ possibilities for $c \in C_S$ .

Activating the edges in $E(G)\setminus E(H)$

First, note that we can activate all edges $e\subseteq [n]\setminus C_0$ , such that $G[C_0\cup e]\cong K_s^3$ . Indeed, by Observation (O5), for every non-empty $S \subsetneq e$ we have $C_S = C_0$ . Thus, by Steps (1), (2), and (4), all edges induced by $C_0\cup \{e\}$ , except $e$ , are in $H$ . We can thus activate $e$ , which closes a copy of $K_s^3$ with $C_0$ . See Figure 2 for illustration.

Figure 2. The edge $e$ closes a copy with $C_0$ and can thus be activated. The three types of edges that are in $H$ appear in shaded colours. The edges induced by $C_0$ appear in blue, and were added to $H$ at Step (1). The edges that contain one vertex from $e$ and two vertices from $C_0$ appear in red, and were added to $H$ at Step (2). The edges that contain two vertices from $e$ and one vertex from $C_0$ appear in green, and were added to $H$ at Step (4).

We can now activate the remaining edges $e=\{v_1, v_2, v_3\}\subseteq [n]\setminus C_0$ . To that end, whp we can choose an auxiliary clique $\tilde {C} \, :\! = \tilde {C}(e)$ disjoint with $e\cup \bigcup _{i,j\in [3]}(C_{v_i}\cup C_{\{v_i,v_j\}})$ (in fact, the union is over $i,j\in [3]$ for which such cores are defined), such that $G[\tilde {C}\cup e]\cong K^3_s$ and for every $i\neq j \in \{1,2,3\}$ , every edge $f\subseteq \tilde {C} \cup \{v_i, v_j\}\cup C_{\{v_i, v_j\}}\cup C_{v_i}\cup C_0$ with $f\cap \tilde {C}\neq \varnothing$ is in $G$ . We will utilise the following observation.

1. (O6) For every $c\in \tilde {C}$ and every $i\in [3]$ , we have that $c\notin C_{v_i}$ and thus $C_{\{v_i,c\}}$ is defined. Further, $C_{\{v_i,c\}}=C_{v_i}$ . To show that, we need to argue that every edge $f\subseteq C_{v_i}\cup \{v_i,c\}\cup C_0$ which intersects with $C_{v_i}$ is in $G$ . Indeed, if $c\in f$ , then this follows from our choice of $\tilde {C}$ , and otherwise, this follows from (1).

Our goal is to show that all edges induced by $e\cup \tilde {C}$ , except for $e$ , are either in $H$ or can be activated. We will then have that $e$ closes a copy of $K^3_s$ with $\tilde {C}$ , and can thus be activated as well. We do so in several activation steps, where some activation steps may depend on previous activation steps (see also Figure 3 for illustration). In what follows, we always assume $i,j\in [3], i\neq j$ .

1. (A1) Edges $f$ induced by $\tilde {C}\cup C_{\{v_i, v_j\}}\cup C_{v_i}$ .

   We claim that such $f$ closes a copy of $K^3_s$ with $C_0$ and can thus be activated. By the choice of $\tilde {C}$ and Observation (O4),

   \begin{equation*}G\left [C_0\cup \left (\tilde {C}\cup C_{\{v_i, v_j\}}\cup C_{v_i}\right )\right ]\cong K^3_{\left |C_0\cup \left (\tilde {C}\cup C_{\{v_i, v_j\}}\cup C_{v_i}\right )\right |}.\end{equation*}
   Now, consider an edge $f'=\{c_0,x,y\}\subseteq C_0\cup \left (\tilde {C}\cup C_{\{v_i, v_j\}}\cup C_{v_i}\right )$ with $c_0\in C_0$ and $f'\nsubseteq C_0$ . We have that, in particular, $G[\{x,y\}\cup C_0]\cong K_{|\{x,y\}\cup C_0|}^3$ , and thus $C_{\{x,y\}}=C_0$ by (O5). Hence, we added $f'$ to $H$ at Step (4) (where $\{x,y\}$ play the role of $S$ ). Therefore, we can now activate any edge $f\subseteq \tilde {C}\cup C_{\{v_i, v_j\}}\cup C_{v_i}$ since it closes a copy of $K^3_s$ with $C_0$ (together with the edges in $H$ ).

2. (A2) Edges of the form $f=\{v, c_1, c_2\}$ where $v\in \{v_i,v_j\}, c_1,c_2\in C_S\cup \tilde {C}$ .

   We claim that $f$ closes a copy of $K^3_s$ with $C_v$ and can thus be activated. Set $S=\{v_i,v_j\}$ . We have already activated edges induced by $C_S\cup C_{v}\cup \tilde {C}$ during (A1). In Step (2), we added to $H$ all edges of the form $\{v, x_1,x_2\}$ where $x_1,x_2\in C_v$ . As for edges $\{v, c, x\}$ where $c\in \{c_1,c_2\}$ and $x\in C_v$ , note that by (O3) and (O6), $C_{\{v,c\}}=C_v$ . Thus, $\{v,c,x\}$ is in $H$ by Step (4). Thus, $f$ closes a copy of $K_s^3$ with $C_v$ (together with edges of $H$ and edges activated during (A1)) and may thus be activated. See the left-hand side of Figure 3 for illustration.

3. (A3) Edges of the form $f=\{v_i,v_j,c\}$ where $c\in \tilde {C}$ . Set $S=\{v_i,v_j\}$ . We claim that $f$ closes a copy of $K^3_s$ with $C_S$ and can thus be activated. Edges of the form $\{v_i,v_j,x\}$ where $x\in C_S$ have been added to $H$ at Step (4). Edges induced by $\{c\}\cup C_S$ were activated in (A1). Edges of the form $\{v, x_1,x_2\}$ where $v\in S$ and $x_1,x_2\in C_S$ were activated in (A2). Furthermore, edges of the form $\{v, c, x\}$ where $v\in S$ and $x\in C_S$ were activated in (A2). Thus, $f$ closes a copy of $K_s^3$ with $C_S$ (together with edges of $H$ and edges activated during (A1) and (A2)), and may thus be activated. See the right-hand side of Figure 3 for illustration.

Figure 3. Illustration of a chain of activation, with the complexity of the construction evident already when $r=3$ . Towards activating an edge $\{v_1, v_2, v_3\}$ with $\tilde {C}$ , we need to activate edges of the form $\{v_1, v_2, c\}$ where $c\in \tilde {C}$ . To that end, we first activate all edges that form a clique with $C_0$ , and in particular, all edges induced by $C_{v_1}\cup C_{\{v_1,v_2\}}\cup \tilde C$ . Then, as the left side illustrates, we can activate the edges of the form $\{v_1, c_1, c_2\}$ as they form a clique with $C_{v_1}$ . Indeed, the edge $\{c_1, c_2, c_3\}$ forms a clique with $C_0$ , and the edge $\{v_1, c_2, c_3\}$ is in $H$ since $C_{\{v_1, c_2\}}=C_{v_1}$ (and was added at Step (4)). We can then, as the right side illustrates, turn our attention to edges of the form $\{v_1, v_2, c\}$ , which will close a clique with $C_{\{v_1, v_2\}}$ . Here, the edge $\{c, c_1, c_2\}$ closes a clique with $C_0$ and thus has already been activated, and the edge $\{v_1, v_2, c_2\}$ is in $H$ by Step (4).

Thus, $e$ closes a copy of $K^3_s$ together with $\tilde {C}$ (with the edges of $H$ and edges activated during (A1)–(A3)).

Finally, we are left with edges $e=\{v_1,v_2,v_3\}$ which intersect with $C_0$ . The argument for these edges is similar to before, and we omit the details here, referring to the complete proof present in Section 3.

3.1 Laying the groundwork

Before we delve into the fine details, we note that what follows will be a natural extension of the construction in Section 2 to general $r$ (together with their formalisation). We will once again define a core $C_0$ , and inductively define cores for sets $S\subseteq V(G)\setminus C_0$ with size $1\le |S| \le r-1$ . The construction of the cores will naturally have a chain property, that is, the core of $S$ may (and will) depend on the cores of $S'\subsetneq S$ . Once again, there will be several sets $S$ for which we will not define a core, and instead draw relevant edges with vertices from $C_0$ .

We begin by partitioning $V(G)$ into $\lfloor \frac {n}{\ell }\rfloor$ sets of size $\ell$ , denoted by $Q_1, \ldots , Q_{\lfloor \frac {n}{\ell }\rfloor }$ . Let us further write $\mathcal{Q} \, :\! =\{Q_1, \ldots , Q_{\lfloor \frac {n}{\ell }\rfloor }\}$ . Throughout the proof, we will use the following probabilistic lemma.

Proof. Fix $S$ and fix $i$ such that $Q_i\cap S=\varnothing$ . The probability that every edge $e\subseteq Q_i\cup S$ with $e\cap Q_i\neq \varnothing$ is in $G$ is at least $p^{\binom {\ell +k}{r}}$ . There are at least $\lfloor \frac {n}{\ell }\rfloor -k$ different $i$ such that $Q_i\cap S=\varnothing$ . Therefore, the probability there doesn’t exist such a $Q_i$ is at most

\begin{align*} \left (1-p^{\binom {\ell +k}{r}}\right )^{\lfloor \frac {n}{\ell }\rfloor -k}={\text{exp}}\left (-\Theta (n)\right ), \end{align*}

where we used our assumptions that $k, r, \ell$ , and $p$ are constants. There are $\binom {n}{k}$ ways to choose $S$ , and thus by the union bound, the probability of violating the statement of the lemma is at most

\begin{align*} \binom {n}{k}{\text{exp}}\left (-\Theta (n)\right )=o(1), \end{align*}

as required.

Defining the cores

Let $i_0$ be the first index in $\left [\lfloor \frac {n}{\ell }\rfloor \right ]$ such that $G[Q_{i_0}]\cong K^r_{\ell }$ , and let us set $C_0 \, :\! = Q_{i_0}$ (note that by Lemma 3.2 whp such an $i_0$ exists). For every $j\gt 0$ , let $i_j$ be the $j$ -th index in $\left [\lfloor \frac {n}{\ell }\rfloor \right ]$ for which we have that $G[C_0\cup Q_{i_j}]\cong K^r_{2\ell }$ (that is, the $j$ -th occurrence of some $Q\in \mathcal{Q}$ for which it holds). We then set $C_j \, :\! = Q_{i_j}$ (once again, note that since $G[C_0]\cong K^r_{\ell }$ , by Lemma 3.2 whp such a $C_j$ exists). We call these sets cores, and we enumerate them $C_0, C_1,\ldots , C_m$ . Note that for every $i\neq j\in [m]$ , $C_i\neq C_j$ , as they are two different $Q\neq Q'\in \mathcal{Q}$ . We continue assuming these cores have been defined deterministically.

Assigning cores to sets

Set $C_{\varnothing }=C_0$ . For every vertex $v \in V \setminus C_0$ , let $i(v)$ be the first index such that the following holds.

1. 1. Every edge $e \subseteq \{v\} \cup C_0 \cup C_{i(v)}$ with $e \cap C_{i(v)} \neq \varnothing$ is in $G$ .

2. 2. $\{v\} \cap C_{i(v)} = \varnothing$ .

We then set $C_{\{v\}} = C_{i(v)}$ . Note that by the properties of the cores and by Lemma 3.2, whp such a $C_{\{v\}}$ exists for every $v\in V(G)\setminus C_0$ . Furthermore, observe that the properties of $C_{\{v\}}$ imply that $G[\{v\}\cup C_{\{v\}}]\cong K^r_{\ell +1}$ .

Now, we define cores for suitable subsets outside of $C_0$ by induction on their size. For every $j\in [2, r-1]$ and for every $S \subseteq V(G) \setminus C_0$ of size $j$ , if

$$\textrm {there is no } S'\subsetneq S, \text{ such that } C_{S'} \text{ is defined and } S\setminus S'\subseteq C_{S'}, \quad S\,\textrm{is core-definable}$$

we define $C_S$ in the following way. Let $i(S)$ be the first index in $[m]$ such that the following holds for every integer $t\in [j-1]$ and a sequence $\varnothing =S_0 \subsetneq \ldots \subsetneq S_t \subsetneq S$ for which $C_{S_0}, \ldots , C_{S_t}$ are defined.

1. (P1) Every edge $e \subseteq S \cup C_{i(S)} \cup C_{S_0} \cup \ldots \cup C_{S_t}$ with $e \cap C_{i(S)} \neq \varnothing$ is in $G$ .

2. (P2) $G\left [C_{i(S)} \cup C_{S_0} \cup \ldots \cup C_{S_t} \right ]\cong K^r_{|C_{i(S)} \cup C_{S_0} \cup \ldots \cup C_{S_t}|}$ .

3. (P3) $S \cap C_{i(S)} = \varnothing$ .

We then set $C_S = C_{i(S)}$ . Note that we allow $S_0=\varnothing$ in order to include edges intersecting with $C_{\varnothing }=C_0$ .

Note that the condition in Property (P1) applies only to edges intersecting with $C_{i(S)}$ . Furthermore, by induction, we have that $G\left [S_0 \cup C_{S_0} \cup \ldots \cup C_{S_t} \right ]\cong K^r_{|S_0 \cup C_{S_0} \cup \ldots \cup C_{S_t}|}$ , and thus the condition in Property (P2) also concerns only edges intersecting with $C_{i(S)}$ . Thus, by these properties and by Lemma 3.2, whp for every $S\subseteq V(G)\setminus C_0$ of size $j$ such that there is no $S'\subsetneq S$ with $C_{S'}$ defined and $S\setminus S'\subseteq C_{S'}$ , we can find such a $C_S$ . In what follows, we assume this holds deterministically.

Note that we have defined a core $C_{S}$ for every set $S\subseteq V(G)\setminus C_0$ of size at most $r-1$ which is core-definable. Furthermore, if $C_{S}$ is not defined, then there exists $S'\subseteq S$ for which $C_{S'}$ is defined and $S\setminus S'\subseteq C_{S'}$ . Finally, note that

(3) \begin{align} i(S') \le i(S) \quad \text{for every $S' \subseteq S$ such that $C_{S'}$ and $C_S$ are defined.} \end{align}

Indeed, the Properties (P1) through (P3) are closed under inclusion, and thus any index satisfying the properties for $S$ must already satisfy the properties for $S'$ .

Constructing $H$

Let $H \subseteq G$ be a subhypergraph consisting of the following edges:

1. (E1) Every edge $e \subseteq C_0$ is in $H$ .

2. (E2) For every $\varnothing \neq S \subseteq V(G) \setminus C_0$ for which $C_S$ is defined, we add to $H$ all edges of the form $S \cup C'$ for every $C' \subseteq C_S$ of size $r - |S|$ .

3. (E3) For every $\varnothing \neq S \subseteq V(G) \setminus C_0$ for which $C_S$ is defined, we add to $H$ all edges of the form $S \cup C' \cup C'_0$ for every $\varnothing \neq C' \subseteq C_S$ and $\varnothing \neq C'_0 \subseteq C_0$ satisfying $|C' \cup C'_0| = r - |S|$ .

Note that edges of type (E1) are in $G$ by the definition of the cores, and edges of type (E2) and type (E3) are in $G$ by Property (P1) (and the first property when defining cores for singletons).

Number of edges in $H$

Let us bound from above the number of edges added in each step of the construction.

First, at Step (E1), we add $\binom {\ell }{r}$ edges to $H$ induced by $C_0$ . We now turn to Steps (E2) and (E3). Let us set

\begin{align*} A &= \left \{S \colon \varnothing \neq S \subseteq V(G) \setminus C_0 \wedge C_S \textit { is defined}\right \} \quad \text{and} \\ B &= \left \{S \cup C' \colon \varnothing \neq S \subseteq V(G) \setminus C_0 \wedge C_S \textit { is defined} \wedge C' \subseteq C_S\right \}. \end{align*}

We have the following.

• • In Step (E2), we add the edges $S \cup C'$ for every $S \in A$ and $C' \subseteq C_S$ of size $r - |S|$ .

• • In Step (E3), we add the edges $X \cup C'_0$ for every $X \in B$ and $C'_0 \subseteq C_0$ of size $r - |X|$ .

Thus, the number of edges considered at Step (E3) is at most

\begin{equation*} \sum _{S \in A} \binom {\ell }{r-|S|} = \sum _{\substack {S\subseteq V(G)\setminus C_0\\ C_S\text{ is defined}}}\binom {\ell }{r-|S|}. \end{equation*}

Note that, for every $S \subseteq V(G) \setminus C_0$ such that $C_S$ is defined and for every $\varnothing \neq C' \subseteq C_S$ , we have that $C_{S \cup C'}$ is not defined. Indeed, in that case, we have that $S \subseteq S \cup C'$ , $C_{S}$ is defined, and $(S \cup C') \setminus S \subseteq C_S$ . Thus, by definition, $S \cup C'$ is not core-definable. Hence, we have that

\begin{equation*} B \subseteq \{X\subseteq V(G)\setminus C_0 \colon C_X\text{ is not defined}\}. \end{equation*}

Moreover, the number of added edges at Step (E3) is at most

\begin{equation*} \sum _{\substack {X\subseteq V(G)\setminus C_0\\ C_X\text{ is not defined}}}\binom {\ell }{r-|X|}. \end{equation*}

Therefore, the number of edges in $H$ is at most

\begin{align*} \binom {\ell }{r}+\sum _{i=1}^{r-1}\left (\sum _{\substack {S\subseteq V(G)\setminus C_0, \quad |S|=i\\ C_S\text{ is defined}}}\binom {\ell }{r-i}+\sum _{\substack {X\subseteq V(G)\setminus C_0, \quad |X|=i\\ C_X\text{ is not defined}}}\binom {\ell }{r-i}\right )\\ =\binom {\ell }{r}+\sum _{i=1}^{r-1}\binom {n-\ell }{i}\binom {\ell }{r-i}=\binom {n}{r}-\binom {n-\ell }{r}. \end{align*}

3.2 Activating the remaining edges

The argument for activating the remaining edges will be a natural extension of the argument given for the case $r=3$ in Section 2, together with its formalisation.

Assigning an auxiliary core to every edge

Let us fix an edge $e\in E(G)$ . Note that we only assign cores to sets of size at most $r-1$ , that is, there is no $C_e$ defined. By Lemma 3.2 applied with

we can choose an auxiliary core $\tilde {C} \, :\! = \tilde {C}(e)$ from $Q_1,\ldots , Q_{\lfloor \frac {n}{\ell }\rfloor }$ , such that the following holds. For every $S\subsetneq e$ , every $t\in [0,|S|-1]$ , and every $S_0\subsetneq \cdots \subsetneq S_t\subsetneq S$ such that $C_S$ is defined and $C_{S_i}$ is defined for every $i\in [0,t]$ :

1. ( $\overline {P}1$ ) Every edge $f \subseteq S \cup C_S \cup C_{S_0} \cup \cdots \cup C_{S_t}\cup \tilde {C}$ with $f \cap \tilde {C} \neq \varnothing$ is in $G$ .

2. ( $\overline {P}2$ ) $G\left [S_0\cup C_{S} \cup \tilde {C}\cup C_{S_0} \cup \cdots \cup C_{S_t} \right ]\cong K^r_{|S_0\cup C_{S} \cup \tilde {C}\cup C_{S_0} \cup \cdots \cup C_{S_t}|}$ .

3. ( $\overline {P}3$ ) $(S \cup C_{S}\cup C_{S_0}\cup \cdots \cup C_{S_t})\cap \tilde {C} = \varnothing$ .

Lemma 3.2 guarantees the (typical) existence of edges intersecting with $\tilde {C}$ . Indeed, the above properties concern only edges intersecting with $\tilde {C}$ . To see that in Property ( $\overline {P}2$ ), note that $G\left [S_0\cup C_{S} \cup C_{S_0} \cup \cdots \cup C_{S_t} \right ]\cong K^r_{|S_0\cup C_{S} \cup \tilde {C}\cup C_{S_0} \cup \cdots \cup C_{S_t}|}$ by Property (P2) and thus Property ( $\overline {P}2$ ) indeed concern only edges intersecting with $\tilde {C}$ . Lemma 3.2 further guarantees the non-intersection of $\tilde {C}$ with $S$ , which yields Property ( $\overline {P}3$ ).

Core extension property

Recall that, by (3), given $S\subseteq S\cup U\subseteq V(G)$ such that $C_S$ and $C_{S\cup U}$ are defined, then $i(S)\le i(S\cup U)$ . If $U$ satisfies $i(S\cup U)=i(S)$ , then $C_{S\cup U}=C_S$ . In the following claim, we show that for a given $S$ for which $C_S$ is defined, we have that $U$ is satisfies $i(S\cup U)=i(S)$ whenever $U\subseteq C_{S_0}\cup \cdots \cup C_{S_t}$ , for some $S\subsetneq S_0\subsetneq \ldots \subsetneq S_t$ . This ‘extension’ property of the cores will be important for us throughout the activation process, and in fact, we required Properties (P1) and (P2) when choosing our cores so that this extension property will hold.

Claim 3.3. Let $e_0$ be an edge in $G$ . Let $S \subseteq V(G) \setminus C_0$ be such that $C_S$ is defined. For every set $U$ and $t\in [r-1]$ such that $|S\cup U|\le r-1$ and $U \subseteq C_{S_0} \cup \cdots \cup C_{S_t}\cup \tilde {C}(e_0)$ for some $S \subsetneq S_0 \subsetneq \ldots \subsetneq S_t\subsetneq e$ for which $C_{S_0}, \ldots , C_{S_t}$ are defined and different than $C_{S}$ , we have that

\begin{equation*} C_{S \cup U} = C_{S}. \end{equation*}

Activating the edges

With this at hand, we are ready to prove that we can activate all the edges of $G$ outside of $H$ . The argument here will be a natural extension, and formalisation, of the argument for $r=3$ detailed in Section 2. We consider separately edges that intersect with $C_0$ and edges that do not.

4.1 Outline of the upper bound of Theorem1(b)

We will utilise two lemmas. The first one generalises a result of Krivelevich [Reference Krivelevich16] to the case of hypergraphs:

Lemma 4.1. Let $r \ge 3$ and $t \ge r$ be integers. There exist positive constants $c_0$ and $c_1$ such that if

\begin{equation*} \rho \ge c_1 n^{-\frac {t + 1 - r}{\binom {t + 1}{r} - 1}}\quad \text{and}\quad k \geq c_0 n^{\frac {\binom {t}{r}(t + 1 - r)}{\left (\binom {t+1}{r} - 1\right )(t-1)}} \log ^{\frac {1}{t-1}} n, \end{equation*}

then whp $G^{r}(n,\rho )$ contains a subhypergraph $H$ on $[n]$ such that

• • $H$ is $K^r_{t+1}$ -free,

• • every induced $k$ -subhypergraph of $H$ contains a copy of $K^r_t$ .

The proof of Lemma 4.1 follows, in general, ideas present in [Reference Krivelevich16], utilising Janson’s inequality.

We also require the following lemma, which shows that for $\rho$ chosen as in that in Lemma 4.1, typically every two vertices $u$ and $v$ are not both contained in ‘many’ copies of $K_t^r$ in $G^r(n,\rho )$ . This lemma is a new ingredient in the proof – this was not needed in the graph setting of [Reference Korándi and Sudakov15], but is necessary in the hypergraph setting.

Proof. Fix two vertices $u$ and $v$ . Denote by $X$ the number of copies of $K_t^r$ in $G \sim G^{r}(n,\rho )$ which contain $u$ and $v$ . By the union bound over all the choices of $u$ and $v$ , it suffices to prove that

\begin{equation*} \mathbb{P}\left (X \gt 2\binom {n}{t-2} p^{\binom {t}{r}}\right ) \ll n^{-2}. \end{equation*}

Set $L = K \log n$ where $K$ is a sufficiently large constant. For every subset $T \subset V(G)$ of size $t$ , denote by $Y_T$ the indicator random variable of the event $G[T] \cong K_t^r$ . Below, we consider only those $t$ -subsets of $V(G)$ that contain $u$ and $v$ (in particular, sum up over such sets in the computation of the $L$ -th moment of $X$ below). We have

(4) \begin{align} \mathbb{E}[X^L] &= \sum _{\substack {T_1, \ldots , T_{L} \subset V(G) \\ |T_1| = \cdots = |T_{L}| = t}} \mathbb{E}[Y_{T_1} \cdot \cdots \cdot Y_{T_{L}}]\notag \\ &= \sum _{\substack {T_1, \ldots , T_{L-1} \subset V(G) \\ |T_1| = \cdots = |T_{L-1}| = t}} \mathbb{E}[Y_{T_1} \cdot \cdots \cdot Y_{T_{L-1}}] \sum _{T_L \subset V(G), |T_L| = t} \mathbb{E}[Y_{T_L} \mid Y_{T_1} = \cdots = Y_{T_{L-1}} = 1]\notag \\ &= \sum _{\substack {T_1, \ldots , T_{L-1} \subset V(G) \\ |T_1| = \ldots = |T_{L-1}| = t}} \mathbb{E}[Y_{T_1} \cdot \ldots \cdot Y_{T_{L-1}}] \cdot \mathbb{E}[X \mid Y_{T_1} = \ldots = Y_{T_{L-1}} = 1]. \end{align}

Let $H$ be a hypergraph with $e(H) \le L \cdot t$ . By FKG inequality, $\mathbb{E}[X \mid H \subset G]\geq \mathbb{E}X$ . On the other hand,

\begin{align*} \mathbb{E}[X \mid H \subset G] \le \binom {n}{t-2} \rho ^{\binom {t}{r}} + \sum _{j = r}^{t} e(H)^{\binom {j}{r}} n^{t - j} \rho ^{\binom {t}{r} - \binom {j}{r}}. \end{align*}

The first summand in the right-hand side above is an upper bound to the expectation of the number of copies of $K_t^r$ which does not share any edges with $H$ while the second summand is an upper bound to the expectation of the number of copies of $K_t^r$ which does share at least one edge with $H$ . The index $j$ indicates how many vertices in the copy of $K_t^r$ belongs to $H$ . So we have at most $\binom {j}{r}$ edges which are already in $H$ . The term $e(H)^{\binom {j}{r}}$ is an upper bound to the number of choices of the common $j$ vertices, $n^{t-j}$ is an upper bound to the number of choices of the remaining $t - j$ vertices and $p^{\binom {t}{r} - \binom {j}{r}}$ is an upper bound to the probability that this copy appears in $G$ conditioned on $H \subseteq G$ . We will show that the second summand on the right-hand side of the inequality is asymptotically smaller than

(5) \begin{equation} \mathbb{E}[X]=(1+o(1))\binom {n}{t-2}\rho ^{\binom {t}{r}}, \end{equation}

implying $\mathbb{E}[X\mid H\subset G]\leq (1+o(1))\mathbb{E}[X]$ . Note that, for every $j = r, r+1, \ldots , t$ , we have

\begin{align*} \frac {n^{t - j} \rho ^{\binom {t}{r} - \binom {j}{r}}}{n^{t-2} \rho ^{\binom {t}{r}}} = n^{2 - j} \rho ^{-\binom {j}{r}} = \Theta \left (n^{2 - j + \frac {\binom {j}{r}(t+1-r)}{\binom {t+1}{r}-1}}\right ). \end{align*}

Thus, it is sufficient to show that the last expression tends to 0 as $n\to \infty$ , which is true if

\begin{equation*} \frac {\binom {j}{r}(t+1-r)}{\binom {t+1}{r}-1} \lt j - 2. \end{equation*}

To show the latter, we will use the following claim.

Claim 4.3. $\frac {\binom {t}{r} (t + 1 - r)}{ \left(\binom {t+1}{r} - 1 \right)(t-1)} \lt 1 - \frac {r - 1}{t}$ .

Proof. We have

\begin{align*} \frac {\binom {t}{r} (t + 1 - r)}{ \left(\binom {t+1}{r} - 1\right )(t-1)} &\lt \frac {\binom {t}{r}}{\binom {t+1}{r} - 1} = \frac {\binom {t}{r}}{\binom {t}{r} + \binom {t}{r - 1} - 1} \\ &= \frac {\binom {t}{r}}{\binom {t}{r} + \binom {t}{r} \cdot \frac {r}{t-r+1} - 1} = \frac {1}{1 + \frac {r}{t-r+1} - \frac {1}{\binom {t}{r}}}\\ &\le 1 - \frac {r}{t-r+1} + \frac {1}{\binom {t}{r}} \leq 1 - \frac {r}{t - r + 1} + \frac {1}{t - r + 1} \\ &=1-\frac {r-1}{t-r+1}\lt 1 - \frac {r - 1}{t}. \end{align*}

By Claim 4.3,

\begin{align*} \frac {\binom {j}{r}(t+1-r)}{\binom {t+1}{r}-1} &\lt \frac {\binom {j}{r}}{\binom {t}{r}} \left (1 - \frac {r - 1}{t}\right )(t-1) \le \frac {\binom {j}{r}}{\binom {t}{r}} (t-2)= \frac {j(j-1)\ldots (j-r+1)}{t(t-1)\ldots (t-r+1)}(t-2) \\&\stackrel {r\geq 3} \leq j-2, \end{align*}

as needed.

Thus, we conclude that $\mathbb{E}[X\mid H\subset G]=(1+o(1))\mathbb{E}[X]$ uniformly over all $H$ with $e(H)\leq L\cdot t$ . By induction, from (4), we have

\begin{equation*} \mathbb{E}[X^L] = \biggl ((1 + o(1))\mathbb{E}[X]\biggr )^L. \end{equation*}

From (5), by Markov’s inequality, we get

\begin{align*} \mathbb{P}\left (X \gt 2\binom {n}{t-2} \rho ^{\binom {t}{r}}\right ) \le \frac {\mathbb{E}[X^L]}{((2+o(1))\mathbb{E}[X])^L} = \left (\frac {(1 + o(1))\mathbb{E}[X]}{2\mathbb{E}[X]}\right )^L = (1/2+o(1))^L \ll n^{-2}, \end{align*}

where the last asymptotical inequality is true if $K$ is sufficiently large.

With these two lemmas at hand, we can now describe our construction. We say that an edge $e\in E(G)\setminus E(H)$ can be completed if it closes a copy of $K_s^r$ in $H\cup \{e\}$ . Recall that we aim to construct a $K_s^r$ -free subhypergraph $H\subseteq G$ with $(1+o(1))\binom {n}{r-1}\log _{\frac {1}{1-p^{r-1}}}n$ edges, such that every edge $e\in E(G)\setminus E(H)$ can be completed. Throughout the proof, we make use of the following observation, already appearing in [Reference Korándi and Sudakov15].

We set $\alpha =(1-p^{r-1})^{-1}$ and $\beta =(1-p^{\binom {s}{r}-\binom {s-r}{r}-1})^{-1}$ . Throughout the rest of this section, unless explicitly stated otherwise, the base of the logarithms is $\alpha$ . We then set aside three disjoint subsets of $[n]$ , $A_1$ , $A_2$ , and $A_3$ , of sizes $a_1 \, :\! = \frac {1}{p}\log n\left (1+\frac {3}{\log \log n}\right )$ , $a_2 \, :\! = sr\log _{\beta }n$ , and $a_3=\frac {a_2}{\log ^{1/r}a_2}$ , respectively. We let $B=[n]\setminus (A_1\cup A_2\cup A_3)$ , shorthand $t=s-r$ , and recall that $G\sim G^r(n,p)$ .

We now define $H$ to be a subhypergraph of $G$ with the following edges:

• • all edges of $G$ intersecting both $A_1$ and $B$ with at most $t$ vertices from $A_1$ ; and,

• • if $t\ge r$ , we take $H[A_1]\subsetneq G[A_1]$ to be $K_{t+1}^r$ -free, such that there exists a copy of $K_t^r$ in every induced subhypergraph of $H[A_1]$ of size at least $c_0a_1^{\frac {\binom {t}{r}(t+1-r)}{\left (\binom {t+1}{r}-1\right )(t-1)}}\log ^{1/(t-1)}a_1$ . Moreover, every two vertices in $A_1$ are contained in at most $2\binom {a_1}{t-2}\left (c_1a_1\right )^{-\frac {t+1-r}{\binom {t+1}{r}-1}}$ copies of $K_t^r$ in $H[A_1]$ . Otherwise, if $t\lt r$ , we take $H[A_1]$ to be the empty hypergraph.

Note that, whp, by Lemmas 4.1 and 4.2 the subhypergraph described in the second bullet exists. Furthermore, the number of edges we have in $H$ now is indeed $(1+o(1))\binom {n}{r-1}\log _{\frac {1}{1-p^{r-1}}}n$ . Moreover, since $H[A_1]$ is $K_{t+1}^r$ -free, if $t+1\ge r$ we see that $H$ is $K_s^r$ -free. If $t+1\le r-1$ , we note that we do not have edges intersecting both $A_1$ and $B$ with at least $t+1$ vertices in $A_1$ , and thus $H$ is $K_s^r$ -free.

Given an $(r-1)$ -set of vertices $S\subseteq [n]$ , the neighbourhood of $S$ in $X\subseteq [n]$ is given by $N_X(S) \, :\! = \left \{v\in X\colon \{v\}\cup S\in E(G)\right \}$ . Given an $(r-1)$ -set of vertices $S\subseteq B$ , we say that $S$ is good if $|N_{A_1}(S)|\ge pa_1-\frac {\log n}{\log \log n}$ , and otherwise we say that $S$ is bad. Finally, we say that an edge $e\in G[B]$ is good if there exists at least one good $(r-1)$ -subset $S\subseteq e$ , and otherwise say that the edge $e$ is bad.

We first utilise the subhypergraph in $H[A_1]$ in order to activate almost all the good edges $e\in E(G[B])$ . Observe that good edges $e$ have some $S\subseteq e$ , with $|S|=r-1$ and such that $S$ has a large neighbourhood in $A_1$ . Fix an $r$ -subset $e\subseteq B$ . Note that given $S\subseteq e$ with $|S|=r-1$ and its neighbourhood in $A_1$ , we have that

\begin{align*} N_{S,e} \, :\! = \left |\left \{v\in N_{A_1}(S)\colon \{v\}\cup S'\in E(G)\text{ for all }S'\subseteq e, |S'|=r-1\right \}\right |\sim Bin(|N_{A_1}(S)|,p^{r-1}). \end{align*}

By the choice of $H[A_1]$ , we can then expose the edges of the form $S'\cup \{v\}$ in $G$ , where $S'\subseteq e$ and $v\in N_{A_1}(S)$ , and then using Chernoff’s bound find in $N_{S,e}$ many copies of $K_t^r$ . Now, note that given two copies of $K_t^r$ in $H[N_{S,e}]$ the events that each of them closes a copy of $K_s^r$ with $e$ are dependent only if these copies intersect in two vertices. To bound these dependencies, we utilise the fact that in $H[A_1]$ every two vertices do not have many copies of $K_t^r$ sharing them. This, together with Janson’s inequality, allows us to show that whp, for all but $o(n^{r-1}\log n)$ good edges in $e\in E(G[B])$ , there is at least one copy of $K_t^r$ that closes a copy of $K_s^r$ with $e$ . Note that by Observation 4.4, this in fact completes our treatment for all the good edges in $G[B]$ .

Before turning to bad edges in $G[B]$ , let us discuss the difference in the treatment of good edges above from the setting of graphs in [Reference Korándi and Sudakov15]. Indeed, in the case of graphs, a key part of the argument for the upper bound is to consider vertices $v$ with a large neighbourhood in $A_1$ . Then, fixing $v$ and considering edges $uv\in E(G[B]))$ , it suffices to find a copy of $K_{s-2}^2$ in $N_v \cap N_u$ – whose size is distributed according to $Bin(|N_v|,p)$ . On the other hand, when considering hypergraphs, we consider instead an $(r-1)$ -set $S\subseteq e$ with a substantial neighbourhood in $A_1$ , $N_{S,e}$ . Then, when considering edges $\{u\}\cup S\in E(G[B])$ , it no longer suffices to find a copy of $K_{s-r}^r$ in $\tilde {N}_{S, u} = \bigcap _{S' \subset S \cup \{u\}, |S'| = r - 1} N_{S',e}$ . Indeed, one needs to further consider edges with $1\le \ell \lt r-1$ vertices from $S \cup \{u\}$ , and $r-\ell$ vertices from $\tilde {N}_{S, u}$ . As noted above, this further creates possible dependencies which do not appear in the case of graphs, and, in particular, this is why Lemma 4.2 is important to us in the hypergraph setting.

We now turn to bad edges in $G[B]$ . As there is a non-negligible portion of them, we require the set $A_2$ to complete them. By Lemma 4.1 there exists a subhypergraph $H_2\subseteq G[A_2]$ which is $K_{t+1}^r$ -free and every large enough subset in it induces a copy of $K_t^r$ , when $t\ge r$ . Moreover, there are at least $r\log _{\beta }n$ vertex-disjoint copies of $K_t^r$ in $H_2[A_2]$ . We can thus add to $H$ the edges of $H_2[A_2]$ , edges intersecting both $A_2$ and $B$ in $G$ with at least two and at most $t$ vertices from $A_2$ , and edges of the form $S\cup \{v\}$ where $S\subseteq B$ is bad and $v\in A_2$ . The first two are of order $o(n^{r-1}\log n)$ , and using Chernoff’s bound, one can show that there are at most $\frac {n^{r-1}}{\log n}$ bad $(r-1)$ -subsets $S\subseteq B$ , and thus the last set of edges is also of order $o(n^{r-1}\log n)$ . Similarly to before, $H$ is still $K_s^r$ -free. Using Markov’s inequality, whp we can now close all bad edges in $G[B]$ (recall that a bad edge has that all its $(r-1)$ -subsets are bad).

Note that we have not added any edge to $H$ intersecting both $B$ and $A_2$ , whose intersection with $B$ is a good $(r-1)$ -subset. As there are $\Omega (n^{r-1})$ good $(r-1)$ -subsets in $B$ , and $\Omega (\log n)$ vertices in $A_2$ , we cannot ignore these edges, and this is the reason for setting aside the set $A_3$ . Once again, we find in $G[A_3]$ a subhypergraph $H_3$ which is $K_{t+1}^r$ -free, and every large enough subset in it induces a copy of $K_t^r$ , when $t\ge r$ .

The choice of edges which we add to $H$ now is slightly more delicate. We add to $H$ the edges of $H_3$ , all edges intersecting both $B$ and $A_3$ with at least two and at most $t$ vertices from $A_3$ , edges of the form $S\cup \{v\}$ where $S\subseteq B$ is good and $v\in A_3$ , and edges intersecting both $B\cup A_2$ and $A_3$ with one vertex from $A_2$ and at least one and at most $t$ vertices from $A_3$ . This careful choice of edges allows us to show that the hypergraph $H$ is still $K_s^r$ -free. As the argument here is slightly more delicate, let us state it explicitly. Since there are no edges intersecting both $A_1$ and $A_2\cup A_3$ , it suffices to consider $X\subseteq B\cup A_{2}\cup A_{3}$ of size $s$ , and suppose towards contradiction that $X$ induces a clique in $H$ . Since $H[B]$ is an empty hypergraph, we must have that $|X\cap B|\le r-1$ . Since we only added edges intersecting $A_3$ , we may assume that $X$ contains vertices from $A_3$ . If $|X\cap A_3|\ge t+1$ , we are done by similar arguments to before. Furthermore, if there are at least two vertices in $A_2\cap X$ , note that there are no edges in $H$ containing two vertices from $A_2$ and at least one vertex from $A_3$ , leading to a contradiction. Thus, we are left with the case where $|X\cap A_{2}|=1, |X\cap A_{3}|=t, |X\cap B|=r-1$ . Now, if $X\cap B$ is bad, there are no edges of the form $(X\cap B)\cup \{v\}$ , $\{v\}\in A_3$ , in $H$ , and if it is good, then there are no edges of the form $(X\cap B)\cup \{v\}$ , $\{v\}\in A_2$ , in $H$ – either way, we get that $X$ cannot induce a clique. Let us note that this last part of the argument is also quite different from the argument in the setting of graphs. Indeed, in [Reference Korándi and Sudakov15], to ensure that $H$ remains $K_s^2$ -free, one needs to consider a partition of $A_2$ into independent sets, and partition $A_3$ accordingly. Here, adding edges intersecting both $B\cup A_2$ and $A_3$ which intersect $A_2$ in at most one vertex suffices to show that $H$ remains $K_s^r$ -free.

Furthermore, due to the size of $A_3$ , the number of edges we added at this stage is of order $o(n^{r-1}\log n)$ . Using similar arguments to before, we can show that we can activate all but $o(n^{r-1}\log n)$ of the edges of the form $S\cup \{v\}$ where $S\subseteq B$ is good and $v\in A_2$ . Again, by Observation 4.4, this completes our treatment for all edges of the form $S\cup \{v\}$ where $S\subseteq B$ is good and $v\in A_2$ .

Finally, note that we have not completed all the edges: we did not complete edges of the form $S\cup \{v\}$ where $S\subseteq B$ is bad and $v\in A_3$ as well as some of the edges induced by $A_1\cup A_2\cup A_3$ . Furthermore, if $t\lt r$ , we have not completed edges instersecting both $B$ and $A_1\cup A_2\cup A_3$ with more than $t$ vertices in either $A_1$ , $A_2$ , or $A_3$ . However, there are at most $o(n^{r-1}\log n)$ edges of these types, and we are thus done by Observation 4.4.