Isolation of $k$-cliques

TL;DR

This paper introduces a bound on the minimum size of a vertex set whose removal eliminates all $k$-cliques from a connected graph, generalizing classical and recent results and solving an open problem.

Contribution

It establishes a sharp upper bound on the size of such vertex sets for any positive integer $k$, extending previous specific cases.

Findings

Bound is tight for general graphs

Includes classical and recent special cases

Solves an open problem in graph theory

Abstract

For any positive integer $k$ and any $n$-vertex graph $G$, let $\iota(G,k)$ denote the size of a smallest set $D$ of vertices of $G$ such that the graph obtained from $G$ by deleting the closed neighbourhood of $D$ contains no $k$-clique. Thus, $\iota(G,1)$ is the domination number of $G$. We prove that if $G$ is connected, then $\iota(G,k) \leq \frac{n}{k+1}$ unless $G$ is a $k$-clique or $k = 2$ and $G$ is a $5$-cycle. The bound is sharp. The case $k=1$ is a classical result of Ore, and the case $k=2$ is a recent result of Caro and Hansberg. Our result solves a problem of Caro and Hansberg.

Equations

$$B_{n,k}=\left(V(P_{b_{n,k}})\cup \bigcup _{i=1}^{a_{n,k}}V(F_i),E(P_{b_{n,k}})\cup \{iv:i\in [a_{n,k}],v\in V(F_i)\}\cup \bigcup _{i=1}^{a_{n,k}}E(F_i)\right).$$

$$\iota (n,k)=\max \{\iota (G,k):G\text{mbox}isaconnectedgraph,V(G)=[n],G\not\simeq K_k\}.$$

$$\iota (n,2)=\max \{\iota (G,2):G\text{mbox}isaconnectedgraph,V(G)=[n],G\not\simeq K_2,G\not\simeq C_5\}.$$

$$\iota (G,k)\le \frac{n}{k+1}.$$

$$\iota (n,k)=\iota (B_{n,k},k)=\lfloor \frac{n}{k+1}\rfloor .$$

$$\underset{n\to \infty }{\mathrm{limsup}}\frac{\iota (n,k)}{n}=\frac{1}{k+1}.$$

$$n\ge n^{\prime }+k+1$$

$$\iota (G,k)\le 1+\iota (G^{\prime },k)=1+\sum _{H\in \mathcal{H}}\iota (H,k)\le 1+\sum _{H\in \mathcal{H}}\frac{|V(H)|}{k+1}=\frac{k+1+n^{\prime }}{k+1}\le \frac{n}{k+1}.$$

$${\mathcal{H}}_x\subseteq \mathcal{H}\backslash {\mathcal{H}}^{\prime },$$

$$\iota (G,k)\le |D|=|D_v^{\ast }|+|D^{\prime }|+\sum _{H\in {\mathcal{H}}_x}|D_H|\le |D_v^{\ast }|+\frac{|X|}{k+1}+\sum _{H\in {\mathcal{H}}_x}\frac{|V(H)|}{k+1}.$$

$$\iota (G,k)\le |D^{\prime \prime }|+\sum _{H\in {\mathcal{H}}_x}|D_H|<\frac{|X\cup V(G_v^{\ast })|}{k+1}+\sum _{H\in {\mathcal{H}}_x}\frac{|V(H)|}{k+1}=\frac{n}{k+1}.$$

$$V(C_Y)\subseteq (V(G_v^{\ast })\backslash \{v\})\cup (V(H^{\prime })\backslash X^{\prime }).$$

$$Z\subseteq N[z].$$

$$|Z|=|V(G_v^{\ast })|+|V(C_Y)\backslash V(G_v^{\ast })|=k+|V(C_Y)\cap V(H^{\prime })|\ge k+1.$$

$$|D^{\prime }|=|D_X|+\sum _{H\in {\mathcal{H}}_2}|D_H|\le \frac{|X|}{k+1}+\sum _{H\in {\mathcal{H}}_2}\frac{|V(H)|}{k+1}=\frac{n-n^{\ast }}{k+1}.$$

$$|D|=1+|D^{\prime }|\le 1+\frac{n-n^{\ast }}{k+1}=1+\frac{n-5}{3}<\frac{n}{k+1}.$$

Full text

1 Introduction

Unless stated otherwise, we use small letters such as $x$ to denote non-negative integers or elements of a set, and capital letters such as $X$ to denote sets or graphs. The set of positive integers is denoted by $\mathbb{N}$. For $n\in\{0\}\cup\mathbb{N}$, the set $\{i\in\mathbb{N}\colon i\leq n\}$ is denoted by $[n]$. Note that $[0]$ is the empty set $\emptyset$. Arbitrary sets are assumed to be finite. For a set $X$, the set of $2$-element subsets of $X$ is denoted by ${X\choose 2}$ (that is, ${X\choose 2}=\{\{x,y\}\colon x,y\in X,x\neq y\}$).

If $Y$ is a subset of ${X\choose 2}$ and $G$ is the pair $(X,Y)$, then $G$ is called a graph, $X$ is called the vertex set of $G$ and is denoted by $V(G)$, and $Y$ is called the edge set of $G$ and is denoted by $E(G)$. A vertex of $G$ is an element of $V(G)$, and an edge of $G$ is an element of $E(G)$. We call $G$ an $n$-vertex graph if $|V(G)|=n$. We may represent an edge $\{v,w\}$ by $vw$. If $vw\in E(G)$, then we say that $w$ is a neighbour of $v$ in $G$ (and vice-versa). For $v\in V(G)$, $N_{G}(v)$ denotes the set of neighbours of $v$ in $G$, $N_{G}[v]$ denotes $N_{G}(v)\cup\{v\}$, and $d_{G}(v)$ denotes $|N_{G}(v)|$ and is called the degree of $v$ in $G$. For $S\subseteq V(G)$, $N_{G}[S]$ denotes $\bigcup_{v\in S}N_{G}[v]$ (the closed neighbourhood of $S$), $G[S]$ denotes the graph $(S,E(G)\cap{S\choose 2})$ (the subgraph of $G$ induced by $S$), and $G-S$ denotes $G[V(G)\backslash S]$ (the graph obtained from $G$ by deleting $S$). We may abbreviate $G-\{v\}$ to $G-v$. Where no confusion arises, the subscript $G$ is omitted from any of the notation above that uses it; for example, $N_{G}(v)$ is abbreviated to $N(v)$.

If $G$ and $H$ are graphs, $f:V(H)\rightarrow V(G)$ is a bijection, and $E(G)=\{f(v)f(w)\colon vw\in E(H)\}$, then we say that $G$ is a copy of $H$, and we write $G\simeq H$. Thus, a copy of $H$ is a graph obtained by relabeling the vertices of $H$.

For $n\geq 1$, the graphs $([n],{[n]\choose 2})$ and $([n],\{\{i,i+1\}\colon i\in[n-1]\})$ are denoted by $K_{n}$ and $P_{n}$, respectively. A copy of $K_{n}$ is called a complete graph or an $n$-clique. A copy of $P_{n}$ is called an $n$-path or simply a path.

If $G$ and $H$ are graphs such that $V(H)\subseteq V(G)$ and $E(H)\subseteq E(G)$, then $H$ is called a subgraph of $G$, and we say that $G$ contains $H$.

If $\mathcal{F}$ is a set of graphs and $F$ is a copy of a graph in $\mathcal{F}$, then we call $F$ an $\mathcal{F}$-graph. If $G$ is a graph and $D\subseteq V(G)$ such that $G-N[D]$ contains no $\mathcal{F}$-graph, then $D$ is called an $\mathcal{F}$-isolating set of $G$. Let $\iota(G,\mathcal{F})$ denote the size of a smallest $\mathcal{F}$-isolating set of $G$. The study of isolating sets was introduced recently by Caro and Hansberg [1]. It is an appealing and natural generalization of the classical domination problem [2, 3, 4, 5, 6, 7]. Indeed, $D$ is a $\{K_{1}\}$-isolating set of $G$ if and only if $D$ is a dominating set of $G$ (that is, $N[D]=V(G)$), so $\iota(G,\{K_{1}\})$ is the domination number of $G$ (the size of a smallest dominating set of $G$). In this paper, we obtain a sharp upper bound for $\iota(G,\{K_{k}\})$, and consequently we solve a problem of Caro and Hansberg [1].

We call a subset $D$ of $V(G)$ a $k$-clique isolating set of $G$ if $G-N[D]$ contains no $k$-clique. We denote the size of a smallest $k$-clique isolating set of $G$ by $\iota(G,k)$. Thus, $\iota(G,k)=\iota(G,\{K_{k}\})$.

If $G_{1},\dots,G_{t}$ are graphs such that $V(G_{i})\cap V(G_{j})=\emptyset$ for every $i,j\in[t]$ with $i\neq j$, then $G_{1},\dots,G_{t}$ are vertex-disjoint. A graph $G$ is connected if, for every $v,w\in V(G)$, $G$ contains a path $P$ with $v,w\in V(P)$. A connected subgraph $H$ of $G$ is a component of $G$ if, for each connected subgraph $K$ of $G$ with $K\neq H$, $H$ is not a subgraph of $K$. Clearly, two distinct components of $G$ are vertex-disjoint.

For $n,k\in\mathbb{N}$, let $a_{n,k}=\left\lfloor\frac{n}{k+1}\right\rfloor$ and $b_{n,k}=n-ka_{n,k}$. We have $a_{n,k}\leq b_{n,k}\leq a_{n,k}+k$. If $n\leq k$, then let $B_{n,k}=P_{n}$. If $n\geq k+1$, then let $F_{1},\dots,F_{a_{n,k}}$ be copies of $K_{k}$ such that $P_{b_{n,k}},F_{1},\dots,F_{a_{n,k}}$ are vertex-disjoint, and let $B_{n,k}$ be the connected $n$-vertex graph given by

[TABLE]

Thus, $B_{n,k}$ is the graph obtained by taking $P_{b_{n,k}},F_{1},\dots,F_{a_{n,k}}$ and joining $i$ (a vertex of $P_{b_{n,k}}$) to each vertex of $F_{i}$ for each $i\in[a_{n,k}]$.

For $n,k\in\mathbb{N}$ with $k\neq 2$, let

[TABLE]

For $n\in\mathbb{N}$, let

[TABLE]

In Section 2, we prove the following result.

Theorem 1.1

If $G$ is a connected $n$-vertex graph, then, unless $G$ is a $k$-clique or $k=2$ and $G$ is a $5$-cycle,

[TABLE]

Consequently, for any $k\geq 1$ and $n\geq 3$,

[TABLE]

A classical result of Ore [8] is that the domination number of a graph $G$ with $\min\{d(v)\colon v\in V(G)\}\geq 1$ is at most $\frac{n}{2}$ (see [4]). Since the domination number is $\iota(G,1)$, it follows by Lemma 2.2 in Section 2 that Ore’s result is equivalent to the bound in Theorem 1.1 for $k=1$. The case $k=2$ is also particularly interesting; while deleting the closed neighbourhood of a $\{K_{1}\}$-isolating set yields the graph with no vertices, deleting the closed neighbourhood of a $\{K_{2}\}$-isolating set yields a graph with no edges. In [1], Caro and Hansberg proved Theorem 1.1 for $k=2$, using a different argument. Consequently, they established that $\frac{1}{k+1}\leq\limsup_{n\rightarrow\infty}\frac{\iota(n,k)}{n}\leq\frac{1}{3}$. In the same paper, they asked for the value of $\limsup_{n\rightarrow\infty}\frac{\iota(n,k)}{n}$. The answer is given by Theorem 1.1.

Corollary 1.2

For any $k\geq 1$,

[TABLE]

Proof. By Theorem 1.1, for any $n\geq 3$, $\frac{1}{k+1}-\frac{k}{(k+1)n}=\frac{1}{n}\left(\frac{n-k}{k+1}\right)\leq\frac{\iota(n,k)}{n}\leq\frac{1}{k+1}$, and, if $n$ is a multiple of $k+1$, then $\frac{\iota(n,k)}{n}=\frac{1}{k+1}$. Thus, $\lim_{n\rightarrow\infty}\sup\left\{\frac{\iota(p,k)}{p}\colon p\geq n\right\}=\lim_{n\rightarrow\infty}\frac{1}{k+1}=\frac{1}{k+1}$.  $\Box$

2 Proof of Theorem 1.1

In this section, we prove Theorem 1.1. We start with two lemmas that will be used repeatedly.

If a graph $G$ contains a $k$-clique $H$, then we call $H$ a $k$-clique of $G$. We denote the set \{V(H)\colon H\mbox{ is a k-clique of }G\} by $\mathcal{C}_{k}(G)$.

Lemma 2.1

If $v$ is a vertex of a graph $G$, then $\iota(G,k)\leq 1+\iota(G-N_{G}[v],k)$.

Proof. Let $D$ be a $k$-clique isolating set of $G-N_{G}[v]$ of size $\iota(G-N_{G}[v],k)$. Clearly, $C\cap N_{G}[v]\neq\emptyset$ for each $C\in\mathcal{C}_{k}(G)\backslash\mathcal{C}_{k}(G-N_{G}[v])$. Thus, $D\cup\{v\}$ is a $k$-clique isolating set of $G$. The result follows.   $\Box$

Lemma 2.2

If $G_{1},\dots,G_{r}$ are the distinct components of a graph $G$, then $\iota(G,k)=\sum_{i=1}^{r}\iota(G_{i},k)$.

Proof. For each $i\in[r]$, let $D_{i}$ be a smallest $k$-clique isolating set of $G_{i}$. Then, $\bigcup_{i=1}^{r}D_{i}$ is a $k$-clique isolating set of $G$. Thus, $\iota(G,k)\leq\sum_{i=1}^{r}|D_{i}|=\sum_{i=1}^{r}\iota(G_{i},k)$. Let $D$ be a smallest $k$-clique isolating set of $G$. For each $i\in[r]$, $D\cap V(G_{i})$ is a $k$-clique isolating set of $G_{i}$, so $|D_{i}|\leq|D\cap V(G_{i})|$. We have $\sum_{i=1}^{r}\iota(G_{i},k)=\sum_{i=1}^{r}|D_{i}|\leq\sum_{i=1}^{r}|D\cap V(G_{i})|=|D|=\iota(G,k)$. The result follows.  $\Box$

Proof of Theorem 1.1. We use induction on $n$. If $G$ is a $k$-clique, then $\iota(G)=1=\frac{n+1}{k+1}$. If $k=2$ and $G$ is a $5$-cycle, then $\iota(G)=2=\frac{n+1}{k+1}$. Suppose that $G$ is not a $k$-clique and that, if $k=2$, then $G$ is not a $5$-cycle. Suppose $n\leq 2$. If $k\geq 3$, then $\iota(G)=0$. If $k=2$, then $G\simeq K_{1}$, so $\iota(G)=0$. If $k=1$, then $G\simeq K_{2}$, so $\iota(G)=1=\frac{n}{k+1}$. Now suppose $n\geq 3$. If $\mathcal{C}_{k}(G)=\emptyset$, then $\iota_{k}(G)=0$. Suppose $\mathcal{C}_{k}(G)\neq\emptyset$. Let $C\in\mathcal{C}_{k}(G)$. Since $G$ is connected and $G$ is not a $k$-clique, there exists some $v\in C$ such that $N[v]\backslash C\neq\emptyset$. Thus, $|N[v]|\geq k+1$ as $C\subset N[v]$. If $V(G)=N[v]$, then $\{v\}$ is a $k$-clique isolating set of $G$, so $\iota(G)=1\leq\frac{n}{k+1}$. Suppose $V(G)\neq N[v]$. Let $G^{\prime}=G-N[v]$ and $n^{\prime}=|V(G^{\prime})|$. Then,

[TABLE]

and $V(G^{\prime})\neq\emptyset$. Let $\mathcal{H}$ be the set of components of $G^{\prime}$. If $k\neq 2$, then let $\mathcal{H}^{\prime}=\{H\in\mathcal{H}\colon H\simeq K_{k}\}$. If $k=2$, then let $\mathcal{H}^{\prime}=\{H\in\mathcal{H}\colon H\simeq K_{k}\mbox{ or }H\simeq C_{5}\}$. By the induction hypothesis, $\iota(H,k)\leq\frac{|V(H)|}{k+1}$ for each $H\in\mathcal{H}\backslash\mathcal{H}^{\prime}$. If $\mathcal{H}^{\prime}=\emptyset$, then, by Lemmas 2.1 and 2.2,

[TABLE]

Suppose $\mathcal{H}^{\prime}\neq\emptyset$. For any $H\in\mathcal{H}$ and any $x\in N(v)$, we say that $H$ is linked to $x$ if $xy\in E(G)$ for some $y\in V(H)$. Since $G$ is connected, each member of $\mathcal{H}$ is linked to at least one member of $N(v)$. One of Case 1 and Case 2 below holds.

Case 1: For each $H\in\mathcal{H}^{\prime}$, $H$ is linked to at least two members of $N(v)$. Let $H^{\prime}\in\mathcal{H}^{\prime}$ and $x\in N(v)$ such that $H^{\prime}$ is linked to $x$. Let $\mathcal{H}_{x}$ be the set of members of $\mathcal{H}$ that are linked to $x$ only. Then,

[TABLE]

and hence, by the induction hypothesis, each member $H$ of $\mathcal{H}_{x}$ has a $k$-clique isolating set $D_{H}$ with $|D_{H}|\leq\frac{|V(H)|}{k+1}$.

Let $X=\{x\}\cup V(H^{\prime})$ and $G^{*}=G-X$. Then, $G^{*}$ has a component $G^{*}_{v}$ with $N[v]\backslash\{x\}\subseteq V(G^{*}_{v})$, and the other components of $G^{*}$ are the members of $\mathcal{H}_{x}$. Let $D^{*}_{v}$ be a $k$-clique isolating set of $G^{*}_{v}$ of size $\iota(G^{*}_{v},k)$. Since $H^{\prime}$ is linked to $x$, $xy\in E(G)$ for some $y\in V(H^{\prime})$. If $H^{\prime}$ is a $k$-clique, then let $D^{\prime}=\{y\}$. If $k=2$ and $H^{\prime}$ is a $5$-cycle, then let $y^{\prime}$ be one of the two vertices in $V(H^{\prime})\backslash N_{H^{\prime}}[y]$, and let $D^{\prime}=\{y,y^{\prime}\}$. We have $X\subseteq N[D^{\prime}]$ and $|D^{\prime}|=\frac{|X|}{k+1}$. Let $D=D^{\prime}\cup D^{*}_{v}\cup\bigcup_{H\in\mathcal{H}_{x}}D_{H}$. Since the components of $G^{*}$ are $G^{*}_{v}$ and the members of $\mathcal{H}_{x}$, we have $V(G)=X\cup V(G^{*}_{v})\cup\bigcup_{H\in\mathcal{H}_{x}}V(H)$, and, since $X\subseteq N[D^{\prime}]$, $D$ is a $k$-clique isolating set of $G$. Thus,

[TABLE]

Subcase 1.1: $G^{*}_{v}$ is neither a $k$-clique nor a $5$-cycle. Then, $|D^{*}_{v}|\leq\frac{|V(G^{*}_{v})|}{k+1}$ by the induction hypothesis. By (1), $\iota(G,k)\leq\frac{1}{k+1}\left(|V(G^{*}_{v})|+|X|+\sum_{H\in\mathcal{H}_{x}}|V(H)|\right)=\frac{n}{k+1}$.

Subcase 1.2: $G^{*}_{v}$ is a $k$-clique. Since $|N[v]|\geq k+1$ and $N[v]\backslash\{x\}\subseteq V(G^{*}_{v})$, we have $V(G^{*}_{v})=N[v]\backslash\{x\}$. If $H^{\prime}$ is a $k$-clique, then let $X^{\prime}=\{y\}$ and $D^{\prime\prime}=\{x\}$. If $k=2$ and $H^{\prime}$ is a $5$-cycle, then let $X^{\prime}$ be the set whose members are $y$, $y^{\prime}$, and the two neighbours of $y^{\prime}$ in $H^{\prime}$, and let $D^{\prime\prime}=\{x,y^{\prime}\}$. Let $Y=(X\cup V(G^{*}_{v}))\backslash(\{v,x\}\cup X^{\prime})$. Let $G_{Y}=G-(\{v,x\}\cup X^{\prime})$. Then, the components of $G_{Y}$ are the components of $G[Y]$ and the members of $\mathcal{H}_{x}$.

If $G[Y]$ has no $k$-clique, then, since $\{v,x\}\cup X^{\prime}\subseteq N[D^{\prime\prime}]$, $D^{\prime\prime}\cup\bigcup_{H\in\mathcal{H}_{x}}D_{H}$ is a $k$-clique isolating set of $G$, and hence

[TABLE]

This is the case if $k=1$ as we then have $Y=\emptyset$.

Suppose that $k\geq 2$ and $G[Y]$ has a $k$-clique $C_{Y}$. We have

[TABLE]

Thus, $|V(C_{Y})\cap V(G^{*}_{v})|=|V(C_{Y})\backslash(V(H^{\prime})\backslash X^{\prime})|\geq k-(k-1)=1$ and $|V(C_{Y})\cap V(H^{\prime})|=|V(C_{Y})\backslash(V(G^{*}_{v})\backslash\{v\})|\geq k-(k-1)=1$. Let $z\in V(C_{Y})\cap V(G^{*}_{v})$ and $Z=V(G^{*}_{v})\cup V(C_{Y})$. Since $z$ is a vertex of each of the $k$-cliques $G^{*}_{v}$ and $C_{Y}$,

[TABLE]

We have

[TABLE]

Let $G_{Z}=G-Z$. Then, $V(G_{Z})=\{x\}\cup(V(H^{\prime})\backslash V(C_{Y}))\cup\bigcup_{H\in\mathcal{H}_{x}}V(H)$. We have that the components of $G_{Z}-x$ are $G_{Z}[V(H^{\prime})\backslash V(C_{Y})]$ (which is a clique or a path, depending on whether $H^{\prime}$ a $k$-clique or a $5$-cycle) and the members of $\mathcal{H}_{x}$, $y\in V(H^{\prime})\backslash V(C_{Y})$ (by (2)), $y\in N_{G_{Z}}[x]$, and, by the definition of $\mathcal{H}_{x}$, $N_{G_{Z}}(x)\cap V(H)\neq\emptyset$ for each $H\in\mathcal{H}_{x}$. Thus, $G_{Z}$ is connected, and, if $\mathcal{H}_{x}\neq\emptyset$, then $G_{Z}$ is neither a clique nor a $5$-cycle.

Suppose $\mathcal{H}_{x}\neq\emptyset$. By the induction hypothesis, $\iota(G_{Z},k)\leq\frac{|V(G_{Z})|}{k+1}$. Let $D_{G_{Z}}$ be a $k$-clique isolating set of $G_{Z}$ of size $\iota(G_{Z},k)$. By (3), $\{z\}\cup D_{G_{Z}}$ is a $k$-clique isolating set of $G$. Thus, $\iota(G,k)\leq 1+\iota_{k}(G_{Z})\leq 1+\frac{|V(G_{Z})|}{k+1}$, and hence, by (4), $\iota(G,k)\leq\frac{|Z|}{k+1}+\frac{|V(G_{Z})|}{k+1}=\frac{n}{k+1}$.

Now suppose $\mathcal{H}_{x}=\emptyset$. Then, $G^{*}=G^{*}_{v}$, so $V(G)=V(G^{*}_{v})\cup\{x\}\cup V(H^{\prime})$. Recall that either $H^{\prime}$ is a $k$-clique or $k=2$ and $H^{\prime}$ is a $5$-cycle.

Suppose that $H^{\prime}$ is a $k$-clique. Then, $n=2k+1$. By (3), $|V(G-N[z])|\leq|V(G-Z)|=n-|Z|=2k+1-|Z|$. Suppose $|Z|\geq k+2$. Then, $|V(G-N[z])|\leq k-1$, and hence $\{z\}$ is a $k$-clique isolating set of $G$. Thus, $\iota(G,k)=1<\frac{n}{k+1}$. Now suppose $|Z|\leq k+1$. Then, by (4), $|Z|=k+1$ and $|V(C_{Y})\cap V(H^{\prime})|=1$. Let $z^{\prime}$ be the element of $V(C_{Y})\cap V(H^{\prime})$, and let $Z^{\prime}=V(C_{Y})\cup V(H^{\prime})$. Since $z^{\prime}$ is a vertex of each of the $k$-cliques $C_{Y}$ and $H^{\prime}$, $Z^{\prime}\subseteq N[z^{\prime}]$. We have $|Z^{\prime}|=|V(C_{Y})|+|V(H^{\prime})|-|V(C_{Y})\cap V(H^{\prime})|=2k-1$ and $|V(G-N[z^{\prime}])|\leq|V(G-Z^{\prime})|=n-|Z^{\prime}|=(2k+1)-(2k-1)=2$. If $k\geq 3$, then $\{z^{\prime}\}$ is a $k$-clique isolating set of $G$, and hence $\iota(G)=1<\frac{n}{k+1}$. Suppose $k=2$. Then, $H^{\prime}$, $G^{*}_{v}$, and $C_{Y}$ are the $2$-cliques with vertex sets $\{y,z^{\prime}\}$, $\{v,z\}$, and $\{z,z^{\prime}\}$, respectively. Thus, $V(G)=\{v,z,z^{\prime},y,x\}$, and $G$ contains the $5$-cycle with edge set $\{xv,vz,zz^{\prime},z^{\prime}y,yx\}$. Since $G$ is not a $5$-cycle, $d(w)\geq 3$ for some $w\in V(G)$. Since $|V(G-N[w])|=5-|N[w]|\leq 1$, $\{w\}$ is a $k$-clique isolating set of $G$, and hence $\iota(G,k)=1<\frac{5}{3}=\frac{n}{k+1}$.

Now suppose that $k=2$ and $H^{\prime}$ is a $5$-cycle. Then, $V(G^{*}_{v})=\{v,z\}$ and $E(H^{\prime})=\{yy_{1},y_{1}y_{2},y_{2}y_{3},y_{3}y_{4},y_{4}y\}$ for some $y_{1},y_{2},y_{3},y_{4}\in V(G)$. Recall that $|V(C_{Y})\cap V(H^{\prime})|\geq 1$. Let $z^{\prime}\in V(C_{Y})\cap V(H^{\prime})$. Since $z$ and $z^{\prime}$ are vertices of $C_{Y}$, $zz^{\prime}\in E(G)$. We have $V(G)=\{v,z,x,y,y_{1},y_{2},y_{3},y_{4}\}$, $N(v)=\{x,z\}$, $z^{\prime}\in\{y_{1},y_{2},y_{3},y_{4}\}$ (as $y\notin V(C_{Y})$ by (2)), and $\{vx,vz,xy,zz^{\prime}\}\cup E(H^{\prime})\subseteq E(G)$. If $z^{\prime}$ is $y_{1}$ or $y_{4}$, then $V(G-N[\{y,z^{\prime}\}])$ is $\{v,y_{3}\}$ or $\{v,y_{2}\}$. If $z^{\prime}$ is $y_{2}$ or $y_{3}$, then $V(G-N[\{y,z^{\prime}\}])=\{v\}$. Thus, $\{y,z^{\prime}\}$ is a $k$-clique reducing set of $G$, and hence $\iota(G,k)=2<\frac{8}{3}=\frac{n}{k+1}$.

Subcase 1.3: $G^{*}_{v}$ is a $5$-cycle. If $k\neq 2$, then the result follows as in Subcase 1.1. Suppose $k=2$. We have $E(G^{*}_{v})=\{vv_{1},v_{1}v_{2},v_{2}v_{3},v_{3}v_{4},v_{4}v\}$ for some $v_{1},v_{2},v_{3},v_{4}\in V(G)$. Let $Y=\{v_{2},v_{3},v_{4}\}$. Recall that the components of $G^{*}$ are $G^{*}_{v}$ and the members of $\mathcal{H}_{x}$. Thus, $G-Y$ is connected and $V(G-Y)=\{v,v_{1},x\}\cup V(H^{\prime})\cup\bigcup_{H\in\mathcal{H}_{x}}V(H)$.

Suppose that $G-Y$ is not a $5$-cycle. By the induction hypothesis, $G-Y$ has a $k$-clique isolating set $D$ with $|D|\leq\frac{|V(G-Y)|}{k+1}=\frac{n-3}{3}=\frac{n}{3}-1$. Since $Y\subseteq N[v_{3}]$, $\{v_{3}\}\cup D$ is a $k$-clique isolating set of $G$, so $\iota(G,k)\leq\frac{n}{3}=\frac{n}{k+1}$.

Now suppose that $G-Y$ is a $5$-cycle. Then, $H^{\prime}$ is a $2$-clique and $V(G-Y)=\{v,v_{1},x,y,z\}$, where $\{z\}=V(H^{\prime})\backslash\{y\}$. Since $v_{1}v,vx,xy,yz\in E(G-Y)$ and $G-Y$ is a $5$-cycle, $E(G-Y)=\{v_{1}v,vx,xy,yz,zv_{1}\}$. We have $V(G-N[\{v,v_{1}\}])\subseteq\{v_{3},y\}$. If $v_{3}y\notin E(G)$, then $\{v,v_{1}\}$ is a $k$-clique isolating set of $G$. If $v_{3}y\in E(G)$, then $V(G-N[v,v_{3}])\subseteq\{z\}$, so $\{v,v_{3}\}$ is a $k$-clique isolating set of $G$. Therefore, $\iota(G,k)=2<\frac{8}{3}=\frac{n}{k+1}$.

Case 2: For some $x\in N(v)$ and some $H^{\prime}\in\mathcal{H}^{\prime}$, $H^{\prime}$ is linked to $x$ only. Let $\mathcal{H}_{1}=\{H\in\mathcal{H}^{\prime}\colon H$ is linked to $x$ only$\}$ and $\mathcal{H}_{2}=\{H\in\mathcal{H}\backslash\mathcal{H}^{\prime}\colon H$ is linked to $x$ only$\}$. Let $h_{1}=|\mathcal{H}_{1}|$ and $h_{2}=|\mathcal{H}_{2}|$. Since $H^{\prime}\in\mathcal{H}_{1}$, $h_{1}\geq 1$. For each $H\in\mathcal{H}_{1}$, $y_{H}\in N(x)$ for some $y_{H}\in V(H)$. Let $X=\{x\}\cup\bigcup_{H\in\mathcal{H}_{1}}V(H)$.

For each $k$-clique $H\in\mathcal{H}_{1}$, let $D_{H}=\{x\}$. If $k=2$, then, for each $5$-cycle $H\in\mathcal{H}_{1}$, let $y_{H}^{\prime}$ be one of the two vertices in $V(H)\backslash N_{H}[y_{H}]$, and let $D_{H}=\{x,y_{H}^{\prime}\}$. Let $D_{X}=\bigcup_{H\in\mathcal{H}_{1}}D_{H}$. Then, $D_{X}$ is a $k$-clique isolating set of $G[X]$. If $k\neq 2$, then $D_{X}=\{x\}$, so $|D_{X}|=1\leq\frac{1+k|\mathcal{H}_{1}|}{k+1}=\frac{|X|}{k+1}$. If $k=2$ and we let $h_{1}^{\prime}=|\{H\in\mathcal{H}_{1}\colon H\simeq C_{5}\}|$, then $|D_{X}|=1+h_{1}^{\prime}\leq\frac{1+5h_{1}^{\prime}+2(h_{1}-h_{1}^{\prime})}{3}=\frac{|X|}{k+1}$.

Let $G^{*}=G-X$. Then, $G^{*}$ has a component $G^{*}_{v}$ with $N[v]\backslash\{x\}\subseteq V(G^{*}_{v})$, and the other components of $G^{*}$ are the members of $\mathcal{H}_{2}$. By the induction hypothesis, $\iota(H,k)\leq\frac{|V(H)|}{k+1}$ for each $H\in\mathcal{H}_{2}$. For each $H\in\mathcal{H}_{2}$, let $D_{H}$ be a $k$-clique isolating set of $H$ of size $\iota(H,k)$.

If $G^{*}_{v}$ is a $k$-clique, then let $D^{*}_{v}=\{x\}$. If $k=2$ and $G^{*}_{v}$ is a $5$-cycle, then let $v^{\prime}$ be one of the two vertices in $V(G^{*}_{v})\backslash N_{G^{*}_{v}}[v]$, and let $D^{*}_{v}=\{x,v^{\prime}\}$. If neither $G^{*}_{v}$ is a $k$-clique nor $k=2$ and $G^{*}_{v}$ is a $5$-cycle, then, by the induction hypothesis, $G^{*}_{v}$ has a $k$-clique isolating set $D^{*}_{v}$ with $|D^{*}_{v}|\leq\frac{|V(G^{*}_{v})|}{k+1}$.

Let $D=D^{*}_{v}\cup D_{X}\cup\bigcup_{H\in\mathcal{H}_{2}}D_{H}$. By the definition of $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$, the components of $G-x$ are $G^{*}_{v}$ and the members of $\mathcal{H}_{1}\cup\mathcal{H}_{2}$. Thus, $D$ is a $k$-clique isolating set of $G$ since $x\in D$, $v\in V(G^{*}_{v})\cap N[x]$, and $D_{X}$ is a $k$-clique isolating set of $G[X]$. Let $D^{\prime}=D_{X}\cup\bigcup_{H\in\mathcal{H}_{2}}D_{H}$ and $n^{*}=|V(G^{*}_{v})|$. We have

[TABLE]

If $G^{*}_{v}$ is a $k$-clique, then $|D|=|D^{\prime}|<\frac{n}{k+1}$. If $k=2$ and $G^{*}_{v}$ is a $5$-cycle, then

[TABLE]

If neither $G^{*}_{v}$ is a $k$-clique nor $k=2$ and $G^{*}_{v}$ is a $5$-cycle, then $|D|=|D^{*}_{v}|+|D^{\prime}|\leq\frac{n^{*}}{k+1}+\frac{n-n^{*}}{k+1}=\frac{n}{k+1}$.  $\Box$