The Graph Ramsey Number $R(F_\ell,K_6)$

TL;DR

This paper proves a conjecture about the exact value of a specific graph Ramsey number involving a join of a single vertex and disjoint edges, confirming the formula for the case when n=6.

Contribution

It confirms the conjecture for the case n=6, establishing the exact Ramsey number for the pair (F_ell, K_6).

Findings

Proved the conjecture R(F_ell, K_6) = 2\,ell\, (6-1) + 1 for all \,ell \, ≥ 6.

Established the exact value of the Ramsey number for the specific graph pair.

Extended understanding of Ramsey numbers involving joins of graphs.

Abstract

For a given pair of two graphs $(F,H)$, let $R(F,H)$ be the smallest positive integer $r$ such that for any graph $G$ of order $r$, either $G$ contains $F$ as a subgraph or the complement of $G$ contains $H$ as a subgraph. Baskoro, Broersma and Surahmat (2005) conjectured that \[ R(F_\ell,K_n)=2\ell(n-1)+1 \] for $\ell\ge n\ge3$, where $F_\ell$ is the join of $K_1$ and $\ell K_2$. In this paper, we prove that this conjecture is true for the case $n=6$.

Equations

$$R(F_{\ell },K_n)=2\ell (n-1)+1$$

$$N(v)=$$

$$d(v)=|N(v)|,\delta (G)=\underset{v\in V}{\min }d(v),\Delta (G)=\underset{v\in V}{\max }d(v).$$

$$|N(v_0)\cap V(H)|=\max \{|N(v)\cap V(H)|\mid v\in V(G)\backslash V(H)\}.$$

$$\frac{11-t}{5-t}\le \frac{2\ell -1-t}{|U_2|}\le \underset{u\in U_2}{\max }|N(u)\cap V(H)|\le |N(v_0)\cap V(H)|=t,$$

$$U=\{u_1,u_2,u_3,u_4,u_5\}\subset V(G)$$

$$X_i:=\left\{v\in V(G)\right\}|N(v)\cap U|=i$$

$$V(G)=U\bigsqcup X_1\bigsqcup X_2\bigsqcup X_3\bigsqcup X_4\bigsqcup X_5.$$

$$\sum _{i=1}^5|X_i|=10\ell -4$$

$$\sum _{i=1}^{5}i|X_i|=\sum _{i=1}^{5}d(u_i)\le 10\ell +15.$$

$$|X_1|\ge \sum _{i=1}^5(2-i)|X_i|+3|X_5|\ge 10\ell -23+3|X_5|,$$

$$|X_2|\ge \sum _{i=1}^5(3-i)|X_i|-2|X_1|\ge 20\ell -27-2|X_1|.$$

$$X_1=X_{11}\bigsqcup X_{12}\bigsqcup X_{13}\bigsqcup X_{14}\bigsqcup X_{15}.$$

$$|X_{1i}|\le 2\ell -3,|X_1|\le 10\ell -15.$$

$$|X_{11}|\ge |X_{12}|\ge |X_{13}|\ge |X_{14}|\ge |X_{15}|$$

$$|X_{15}|=|X_1|-\sum _{i=1}^4|X_{1i}|\ge 2\ell -11\ge 1.$$

$$|N(u_i)\backslash X_{1i}|=d(u_i)-|X_{1i}|\ge 3.$$

$$Q_i\subset N(u_i)\backslash X_{1i}$$

$$6\le \lceil (4\ell -7)/3\rceil \le |X_{1i}\backslash N(d)|\le |X_{1i}|\le 2\ell -3$$

$$|X_{1j}\backslash N(d)|\ge (4\ell -7-|X_{1i}|)/2\ge \ell -2\ge 4.$$

$$\{a_1,a_2,\dots ,a_6\}\subset X_{1i}\backslash N(d),\{b_1,b_2,b_3,b_4\}\subset X_{1j}\backslash N(d)\text{and}c\in X_{1k}\backslash N(d).$$

$$J:=\Set{i\in I}{\text{a subset $Q_{i}\subset N(u_{i})\backslash X_{1i}$ satisfying (\ref{Q_cond}) exists}}.$$

$$\sum _{i\in J}|Q_i\cap X_2|\le |X_2|.$$

$$\sum _{i\in J}|Q_i\cap X_3|\le 2|X_3|.$$

$$2\ell -4\le |X_{1i}|\le 2\ell -3,2\ell -5\le |X_{1j}|\le 2\ell -3,2\ell -8\le |X_{1k}|\le 2\ell -3.$$

$$\{a_1,a_2,\dots ,a_8\}\subset X_{1i},\{b_1,b_2,\dots ,b_7\}\subset X_{1j}\text{and}c\in X_{1k}.$$

$$10\ell -4=\sum _{i=1}^4|X_{1i}|+(|X_{15}|+|X_2|+|X_3|+|X_5|)+|X_4|\le 10\ell -9+|X_4|,$$

$$2|X_4|\le \sum _{i=2}^5(i-2)|X_i|\le 8,$$

$$U=\{u_1,u_2,\dots ,u_5\}\subset V(G)$$

$$\sum _{i=1}^{5}d(u_i)=\sum _{i=1}^{5}i|X_i|\ge 2\sum _{i=1}^5|X_i|-|X_1|\ge 10\ell +11$$

Full text

[TABLE]

for $\ell\geq n\geq 3$, where $F_{\ell}$ is the join of $K_{1}$ and $\ell K_{2}$. In this paper, we prove that this conjecture is true for the case $n=6$.

1. Introduction

Throughout this paper, all graphs are finite and simple. For a given pair of two graphs $(F,H)$, let $R(F,H)$ be the smallest positive integer $r$ such that for any graph $G$ of order $r$, $G$ contains $F$ as a subgraph or the complement of $G$ contains $H$ as a subgraph. In general, it is quite difficult to calculate the exact values of $R(F,H)$. However, for sparse graphs $F$ and $H$, there are many results on the exact values of $R(F,H)$. In this paper, we consider fan graph $F_{\ell}$, which is the join of $K_{1}$ and $\ell K_{2}$. This fan graph is one example of such sparse graphs.

Recently, although $K_{n}$ itself is the densest graph, some authors succeeded in determining the value $R(F_{\ell},K_{n})$ for large $\ell$ and small $n$. As for this problem, Baskoro, Broersma and Surahmat [1] conjectured:

Conjecture** (The Baskoro-Broersma-Surahmat conjecture [1]).**

**

For any $\ell\geq n\geq 3$, we have $R(F_{\ell},K_{n})=2\ell(n-1)+1$.

In the last 20 years, this conjecture was proved to be true for the case $n=3,4,5$:

Theorem A** **(Gupta, Gupta and Sudan [5],

Li and Rousseau [6, Proposition 1]).

**

For any $\ell\geq 3$, we have $R(F_{\ell},K_{3})=4\ell+1$.

Theorem B** (Baskoro, Broersma and Surahmat [1]).**

**

For any $\ell\geq 4$, we have $R(F_{\ell},K_{4})=6\ell+1$.

Theorem C** (Chen and Zhang [2]).**

**

For any $\ell\geq 5$, we have $R(F_{\ell},K_{5})=8\ell+1$.

In this paper, we prove the Baskoro-Broersma-Surahmat conjecture for $n=6$.

Theorem 1**.**

**

For any $\ell\geq 6$, we have $R(F_{\ell},K_{6})=10\ell+1$.

Our main strategy is based on the approach of Chen and Zhang [2]. In particular, we shall heavily use their structural setting. However, we need some additional insight on the Chen-Zhang structure than their paper [2].

We briefly summarize our notation and terminology. Let $G=(V,E)$ be a graph. For any subset $S\subset V$, we use $G[S]$ to denote the subgraph induced by $S$. For any vertex $v\in V$, we denote the neighborhood of $v$ by $N(v)$, i.e.

[TABLE]

and we let $N[v]=N(v)\cup\{v\}$. We let

[TABLE]

Moreover, we denote by $\omega(G)$ the clique number of $G$, i.e. the order of the largest clique in $G$, and denote by $\alpha(G)$ the independence number of $G$, i.e. the order of the largest independent set in $G$. We refer to the book [4] for other graph theoretical notation and terminology not described in this paper.

2. The lower bound

The lower bound of $R(F_{\ell},K_{6})$ is given by the following theorem, which is just a special case of the Chvátal-Harary lemma [3, Lemma 4].

Theorem 2**.**

For any $\ell\geq 6$, we have $R(F_{\ell},K_{6})\geq 10\ell+1$.

Proof.

It is sufficient to give a graph $G$ of order $10\ell$ such that $G$ does not contain $F_{\ell}$ and $\overline{G}$ does not contain $K_{6}$. For example, the graph $G=5K_{2\ell}$ satisfies this condition. Hence the theorem follows. ∎

Hence we prove the upper bound $R(F_{\ell},K_{6})\leq 10\ell+1$ in the following sections.

Theorem 3**.**

For any $\ell\geq 6$, we have $R(F_{\ell},K_{6})\leq 10\ell+1$.

We prove this upper bound by contradiction. In the remaining part of this paper, we assume $\ell\geq 6$ and that there exists a graph $G$ of order $10\ell+1$ such that $F_{\ell}\not\subset G$ and $K_{6}\not\subset\overline{G}$. By Theorem C, we may assume that $\alpha(G)=5$.

3. Preliminary lemmas

In this section, we prove several lemmas on basic properties of the graph $G$. We start with the following simple observation, which is related to the number of independent edges in $G$.

Lemma 1**.**

There is no vertex $v\in G$ for which $G[N(v)]$ contains $\ell K_{2}$.

Proof.

If $\ell K_{2}\subset G[N(v)]$, then $F_{\ell}\subset G[N[v]]$. This is a contradiction. ∎

The next lemma is a special case of Stahl’s lemma [7].

Lemma 2**.**

For any $\ell\geq 1$, we have $R(\ell K_{2},K_{6})=2\ell+4$.

Proof.

See [7, pp. 586–587]. ∎

As a consequence of Lemma 2 and $\alpha(G)=5$, any subgraph $H\subset G$ of order $n\geq 6$ contains $\lfloor n/2\rfloor-2$ or more independent edges. The following is an immediate consequence of Theorem C and Lemma 2.

Lemma 3**.**

We have $2\ell\leq\delta(G)\leq\Delta(G)\leq 2\ell+3$.

Proof.

First we show that $\delta(G)\geq 2\ell$. Assume, to the contrary, that there exists a vertex $v$ with $d(v)\leq 2\ell-1$. Let $S=V(G)\backslash N[v]$. Then $|S|\geq 8\ell+1$. Also, since $\alpha(G)=5$, it follows that $\alpha(G[S])\leq 4$. Then $G[S]$ contains $F_{\ell}$ by Theorem C, a contradiction. To show that $\Delta(G)\leq 2\ell+3$, suppose that $d(v)\geq 2\ell+4$ for some vertex $v$. Since $\alpha(G)=5$, Lemma 2 guarantees the existence of $\ell K_{2}$ in $G[N(v)]$, producing $F_{\ell}$ in $G$. This cannot occur and so $\Delta(G)\leq 2\ell+3$, as claimed. ∎

The next lemma estimates the clique number $\omega(G)$.

Lemma 4**.**

We have $\omega(G)\leq 2\ell-2$.

Proof.

Assume, to the contrary, that $G$ contains a clique $H$ of order $2\ell-1$. Select a vertex $v_{0}\in V(G)\backslash V(H)$ such that

[TABLE]

The graph $G-H-v_{0}$ is of order $8\ell+1$ so that $\alpha(G-H-v_{0})=5$ by Theorem C. Let $U$ be a $5$-set of independent vertices in the graph $G_{0}=G-H-v_{0}$.

Since $U\cup\{v\}$ cannot be independent for each $v\in V(G)\backslash U$, there are at least $2\ell-1$ edges between $H$ and $U$. Hence, there exists a vertex $u_{0}\in U$ with $|N(u_{0})\cap V(H)|\geq(2\ell-1)/5>2$. Consequently, $|N(v_{0})\cap V(H)|\geq|N(u_{0})\cap V(H)|\geq 3$.

Next we show that, for each vertex $v\in V(G_{0})$, if $w\in N(v_{0})\cap N(v)\cap V(H)$, then $N(v)\cap V(H)=\{w\}$. If this is not the case, say there exists a vertex $w^{\prime}\in N(v)\cap V(H)$ with $w\neq w^{\prime}$, then $(\ell-1)K_{2}$ in $G[(V(H)\backslash\{w,w^{\prime}\})\cup\{v_{0}\}]$ and the edge $vw^{\prime}$ form $\ell K_{2}$ in $G[N(w)]$, which is impossible.

Now, let $S_{1}=N(v_{0})\cap V(H)=\{w_{1},w_{2},\ldots,w_{t}\}$, where $t=|N(v_{0})\cap V(H)|$. Then by the above observations, we can find a $t$-set $U_{1}=\{u_{1},u_{2},\ldots,u_{t}\}\subset U$ such that $N(u_{i})\cap V(H)=\{w_{i}\}$ for $1\leq i\leq t$. Also, let $S_{2}=V(H)\setminus S_{1}$ and $U_{2}=U\backslash U_{1}$. Recall that $|N(u_{0})\cap V(H)|\geq 3$, so $u_{0}\in U_{2}$. Hence, $t=3,4$.

Note that there is no edge between $U_{1}$ and $S_{2}$ and also note that $U\cup\{v\}$ cannot be independent for each $v\in S_{2}$. Thus there are at least $|S_{2}|=2\ell-1-t$ edges between $U_{2}$ and $S_{2}$. However then,

[TABLE]

which cannot occur. This completes the proof. ∎

4. Structural observation

In this section, we give some observation on the structure of the graph $G$ following the argument of Chen and Zhang [2]. Let

[TABLE]

be a 5-set of independent vertices and let

[TABLE]

for $1\leq i\leq 5$. Since $\alpha(G)\leq 5$, we have a partition

[TABLE]

4.1. On the sets $X_{i}$

Obviously, we have

[TABLE]

and, since $d(u_{i})\leq\Delta(G)\leq 2\ell+3$ by Lemma 3, we have

[TABLE]

We will use (1) and (2) throughout this paper. For example, by (1) and (2) we have

[TABLE]

which implies $|X_{1}|\geq 10\ell-23$. Also, we have

[TABLE]

Let $I=\{1,2,3,4,5\}$. For each pair $i,j\in I$, we define the set $X_{ij}=X_{i}\cap N(u_{j})$. Then we have a partition

[TABLE]

We next find that each $X_{1i}$ induces a clique in $G$.

Lemma 5**.**

For each $i\in I$, the graph $G[X_{1i}]$ is complete. Consequently, we have

[TABLE]

Proof.

Let $v,v^{\prime}\in X_{1i}$. Since the $6$-set $(U\backslash\{u_{i}\})\cup\{v,v^{\prime}\}$ cannot be independent, it follows that $v$ and $v^{\prime}$ are adjacent. Therefore, $G[X_{1i}\cup\{u_{i}\}]$ is a complete graph of order $|X_{1i}|+1\leq\omega(G)\leq 2\ell-2$. In addition, $|X_{1}|=\sum_{i=1}^{5}|X_{1i}|\leq 10\ell-15$. ∎

Assume, without loss of generality, that

[TABLE]

in the rest of our discussion. Thus (3) implies that

[TABLE]

In particular, each $X_{1i}$ is nonempty. The next lemma on the sets $X_{1i}$ is immediate yet useful. We will use this lemma mainly with the choice $t=2$.

Lemma 6**.**

Let $i,j,k\in I$ be distinct indices and $v\in X_{1i}$. Suppose that $|X_{1i}|\geq 2\ell-2t$ for some integer with $2\leq t\leq 6$. Then we have $|N(v)\cap(X_{1j}\cup X_{1k})|\leq 2t$ and $|N(v)\cap X_{1j}|\leq 2t-1$. In addition, if $|N(v)\cap X_{1j}|\geq 2$ also holds, then we have $|N(v)\cap X_{1k}|\leq 2t-3$.

Proof.

Note that $(\ell-t)K_{2}\subset G[(X_{1i}\backslash\{v\})\cup\{u_{i}\}]$. Thus, we must avoid $tK_{2}$ in $G[N(v)\cap(X_{1j}\cup X_{1k})]$ and so the result is immediate as each of $X_{1j}$ and $X_{1k}$ induces a clique. ∎

4.2. On the sets $N(u_{i})\backslash X_{1i}$

Besides the same kind of information as Chen and Zhang obtained, we need some new information on their structural setting. We start with some observations on the set $N(u_{i})\backslash X_{1i}$. By Lemma 3 and Lemma 5

[TABLE]

In general, suppose that $A$ and $B$ are disjoint nonempty sets of vertices in a graph. Then it is well-known that, if $|N(v)\cap B|\geq|A|$ for every $v\in A$, then there are at least $|A|$ independent edges between $A$ and $B$. The following is then immediate.

Lemma 7**.**

Let $i\in I$ and $t=2\ell-|X_{1i}|$. Every $t$-set $S\subset N(u_{i})\backslash X_{1i}$ contains a vertex $v$ for which $|N(v)\cap X_{1i}|<t$.

Proof.

If $S=\{v_{1},v_{2},\ldots,v_{t}\}\subset N(u_{i})\backslash X_{1i}$ and $|N(v_{j})\cap X_{1i}|\geq t$ for $1\leq j\leq t$, then $X_{1i}$ contains $t$ distinct vertices $w_{1},w_{2},\ldots,w_{t}$ such that $v_{j}w_{j}\in E(G)$ for $1\leq j\leq t$. However then, $(\ell-t)K_{2}$ in $G[X_{1i}\backslash\{w_{1},w_{2},\ldots,w_{t}\}]$ with the $t$ edges $v_{j}w_{j}$ $(1\leq j\leq t)$ forms $\ell K_{2}$ in $G[N(u_{i})]$, a contradiction. ∎

Let $i\in I$. Under certain conditions, there exists a $4$-set $Q_{i}\subset V(G)$ satisfying

[TABLE]

For example, such $Q_{i}$ exists if the degree of $u_{i}$ equals $2\ell+3$, which we verify next.

Lemma 8**.**

*Let $i\in I$. If $d(u_{i})=2\ell+3$, then there exists a $4$-set $Q_{i}$ satisfying *(6).

Proof.

Note that $|N(u_{i})\backslash X_{1i}|=d(u_{i})-|X_{1i}|\geq 6$ since $d(u_{i})=2\ell+3$.

Suppose first that $|X_{1i}|$ is odd, say $|X_{1i}|=2t-1$ for some positive integer $t$. Then $G[N(u_{i})\backslash X_{1i}]$ contains $(\ell-t)K_{2}$ by Lemma 2 with four remaining vertices $v_{1},v_{2},v_{3},v_{4}$. Let $Q_{i}=\{v_{1},v_{2},v_{3},v_{4}\}$. Then for any $w\in X_{1i}$, $G[X_{1i}\backslash\{w\}]$ contains $(t-1)K_{2}$ as $G[X_{1i}]$ is complete. Thus $Q_{i}\cup\{w\}$ must be independent in order to avoid $\ell K_{2}$ in $G[N(u_{i})]$.

Suppose next that $|X_{1i}|=2t$ for some positive integer $t$. Then $G[N(u_{i})\backslash X_{1i}]$ contains $(\ell-t-1)K_{2}$ again by Lemma 2 with five remaining vertices $v_{1},v_{2},\ldots,v_{5}$. Since $G[X_{1i}]$ obviously contains $tK_{2}$, the set $Q=\{v_{1},v_{2},\ldots,v_{5}\}$ must be independent to avoid $\ell K_{2}$ in $G[N(u_{i})]$. For any $w\in X_{1i}$, $|N(w)\cap Q|\geq 1$ since $Q\cup\{w\}$ cannot be independent. Also, there cannot be two or more independent edges between $X_{1i}$ and $Q$ to avoid $\ell K_{2}$ in $G[N(u_{i})]$. Thus we can find a vertex in $Q$, say $v_{5}$, such that every $w\in X_{1i}$ is adjacent to $v_{5}$. Let $Q_{i}=\{v_{1},v_{2},v_{3},v_{4}\}$. Recall that $|X_{1i}|=2t\geq 2$. Then $Q_{i}\cup\{w\}$ must be independent for every $w\in X_{1i}$ to avoid two independent edges between $X_{1i}$ and $Q$. ∎

4.3. On the sets $X_{2}$ and $X_{3}$

We next study the sets $X_{2}$ and $X_{3}$ in more detail. We start with the following lemma, which can be seen as an analogue of Lemma 5.

Lemma 9**.**

Let $i,j,k\in I$ be distinct indices.

1. (a)

If $a\in X_{1i}$, $b\in X_{1j}$, $c\in X_{2i}\cap X_{2j}$, then the set $\{a,b,c\}$ is not independent. 2. (b)

If $a\in X_{1i}$, $b\in X_{1j}$, $c\in X_{1k}$, $d\in X_{3i}\cap X_{3j}\cap X_{3k}$, then the set $\{a,b,c,d\}$ is not independent.

Proof.

First, (a) is immediate since the $6$-set $(U\backslash\{u_{i},u_{j}\})\cup\{a,b,c\}$ cannot be independent. Similarly, (b) holds by considering the $6$-set $(U\backslash\{u_{i},u_{j},u_{k}\})\cup\{a,b,c,d\}$. ∎

The next lemma is an analogue of Claim 1 of Chen and Zhang [2].

Lemma 10**.**

Let $i,j,k\in I$ be distinct indices.

1. (a)

If $|X_{1i}|=|X_{1j}|=2\ell-3$, then $X_{2i}\cap X_{2j}=\emptyset$. 2. (b)

If $|X_{1i}|=|X_{1j}|=|X_{1k}|=2\ell-3$, then $X_{3i}\cap X_{3j}\cap X_{3k}=\emptyset$.

Proof.

Let us begin with (a) by assuming, to the contrary, that $c\in X_{2i}\cap X_{2j}$. Since $G[N(c)]$ does not contain $\ell K_{2}$ while $G[X_{1i}\cup\{u_{i}\}]\cong G[X_{1j}\cup\{u_{j}\}]\cong K_{2\ell-2}$, it follows that $|N(c)\cap(X_{1i}\cup X_{1j})|\leq 2\ell-2$. Hence, $|(X_{1i}\cup X_{1j})\backslash N(c)|\geq 2\ell-4$. Without loss of generality, assume that $|X_{1i}\backslash N(c)|\geq(2\ell-4)/2=\ell-2\geq 4$. Note that $X_{1j}\backslash N(c)\neq\emptyset$ as $\omega(G)\leq 2\ell-2$. Let $b\in X_{1j}\backslash N(c)$. By Lemma 9 (a) then, $b$ is adjacent to every vertex in $X_{1i}\backslash N(c)$, i.e. $|N(b)\cap X_{1i}|\geq 4$, contradicting Lemma 6. This proves the assertion (a).

For (b), suppose that $d\in X_{3i}\cap X_{3j}\cap X_{3k}$. Then by a similar argument done in (a), one sees that $|(X_{1i}\cup X_{1j}\cup X_{1k})\backslash N(d)|\geq 4\ell-7$ in order to avoid $\ell K_{2}$ in $G[N(d)]$. Without loss of generality, suppose that $|X_{1i}\backslash N(d)|\geq|X_{1j}\backslash N(d)|\geq|X_{1k}\backslash N(d)|$. Then

[TABLE]

and

[TABLE]

Also, $X_{1k}\backslash N(d)\neq\emptyset$ in order to avoid $K_{2\ell-1}$ in $G$. Let

[TABLE]

By Lemma 6, none of $|N(c)\cap X_{1i}|,|N(c)\cap X_{1j}|,|N(b_{1})\cap X_{1i}|$ exceeds $3$. By this fact with Lemma 9 (b), we may assume that $b_{1}c\notin E(G)$, and $N(c)\cap X_{1i}=\{a_{1},a_{2},a_{3}\}$. Then again by Lemma 6, at most one of $b_{2},b_{3},b_{4}$ is adjacent to $c$ and so suppose that $b_{2}c\notin E(G)$. Thus, by Lemma 9, $N(b_{1})\cap X_{1i}=N(b_{2})\cap X_{1i}=\{a_{4},a_{5},a_{6}\}$. This gives us $(\ell-2)K_{2}$ in $G[(X_{1j}\backslash\{b_{1},b_{2}\})\cup\{u_{j}\}]$ with two edges $a_{4}a_{5}$ and $a_{6}b_{2}$ in $G[N(b_{1})]$, which is impossible. ∎

We let

[TABLE]

The next lemma describes how the sets $Q_{i}$ intersect each other.

Lemma 11**.**

**

1. (a)

Let $i,j\in J$ be indices with $i<j$. If $|X_{1i}|+|X_{1j}|\geq 2\ell-1$, then we have $Q_{i}\cap Q_{j}\cap X_{2}=\emptyset$. Consequently, if $|X_{14}|+|X_{15}|\geq 2\ell-1$, then

[TABLE] 2. (b)

Let $i,j,k\in J$ be indices with $i<j<k$. If $|X_{1i}|+|X_{1j}|+|X_{1k}|\geq 6\ell-14$, then $Q_{i}\cap Q_{j}\cap Q_{k}\cap X_{3}=\emptyset$. Consequently, if $|X_{13}|+|X_{14}|+|X_{15}|\geq 6\ell-14$, then

[TABLE]

Proof.

For (a), suppose that $c\in Q_{i}\cap Q_{j}\cap X_{2}$. Then no vertex in $X_{1i}\cup X_{1j}$ is adjacent to $c$ by (6), so $G[X_{1i}\cup X_{1j}]$ is a clique by Lemma 9 (a). However then, $\omega(G)\geq|X_{1i}|+|X_{1j}|\geq 2\ell-1$. As a result, $Q_{i}\cap Q_{j}\cap X_{2}=\emptyset$.

For (b), suppose that $d\in Q_{i}\cap Q_{j}\cap Q_{k}\cap X_{3}$. Then no vertex in $X_{1i}\cup X_{1j}\cup X_{1k}$ is adjacent to $d$ by (6). By the assumption $|X_{1i}|+|X_{1j}|+|X_{1k}|\geq 6\ell-14$, we have

[TABLE]

We consider two cases separately according to whether $|X_{1k}|\leq 2\ell-7$ or not.

We first consider the case $|X_{1k}|\leq 2\ell-7$. In this case, we have $|X_{1j}|\geq 2\ell-4$ by the assumption. Therefore we have $|X_{1i}|,|X_{1j}|\geq 8$. Take a vertex $c\in X_{1k}$. By Lemma 6 with $t=4$, we find that $|N(c)\cap X_{1i}|,|N(c)\cap X_{1j}|\leq 7$. Thus we can take vertices $a\in X_{1i}\setminus N(c)$ and $b\in X_{1j}\setminus N(c)$. By Lemma 6 with $t=2$, we find that $|N(a)\cap X_{1j}|,|N(b)\cap X_{1i}|\leq 3$, i.e. $|X_{1j}\setminus N(a)|,|X_{1i}\setminus N(b)|\geq 5$. By Lemma 9 (b), this implies $|N(c)\cap(X_{1i}\cup X_{1j})|\geq 10$, contradicting Lemma 6 with $t=4$.

We next consider the case $|X_{1k}|\geq 2\ell-6$. In this case, we have $|X_{1i}|\geq 8$, $|X_{1j}|\geq 7$ and $|X_{1k}|\geq 6$. Thus we can take vertices

[TABLE]

By Lemma 6 with $t=3$, we find that $|N(c)\cap X_{1i}|,|N(c)\cap X_{1j}|,|N(b_{1})\cap X_{1i}|\leq 5$. Thus we may assume that $a_{1}c,b_{1}c,a_{2}b_{1},a_{3}b_{1}\not\in E(G)$. By Lemma 9 (b), we find that $a_{1}\in N(b_{1})\cap X_{1i}$ and $a_{2},a_{3}\in N(c)\cap X_{1i}$. Also, by Lemma 6 with $t=2$, we see that $|N(a_{1})\cap X_{1j}|\leq 3$. Thus we may assume that $a_{1}b_{2},a_{1}b_{3},a_{1}b_{4},a_{1}b_{5}\not\in E(G)$. However, again by Lemma 9 (b), we find that $b_{2},b_{3},b_{4},b_{5}\in N(c)\cap X_{1j}$, contradicting Lemma 6 with $t=3$. ∎

4.4. On the set $X_{1}$

Based on the above observations, we next give a closer look at $X_{1}$. Recall that $10\ell-23\leq|X_{1}|\leq 10\ell-15$. The next lemma gives an improved upper bound for $|X_{1}|$.

Lemma 12**.**

We have $|X_{14}|\leq 2\ell-4$. Consequently, $|X_{1}|\leq 10\ell-17$.

Proof.

If the statement is false, then $|X_{1i}|=2\ell-3$ for $1\leq i\leq 4$. Thus, $X_{2},X_{3}\subset N(u_{5})$ by Lemma 10. Then $X_{15}\cup X_{2}\cup X_{3}\cup X_{5}\subset N(u_{5})$ and so $|X_{15}|+|X_{2}|+|X_{3}|+|X_{5}|\leq d(u_{5})\leq 2\ell+3$. This then implies that

[TABLE]

so that $|X_{4}|\geq 5$. On the other hand, using (1) and (2),

[TABLE]

so that $|X_{4}|\leq 4$. Since this is impossible, it follows that $|X_{14}|\leq 2\ell-4$ and so $|X_{1}|\leq 3(2\ell-3)+2(2\ell-4)=10\ell-17$. This completes the proof. ∎

Lemma 13**.**

The graph $G$ contains a $5$-set of independent vertices

[TABLE]

for which either (A) $|X_{1}|\in\{10\ell-18,10\ell-17\}$ or (B) $X_{5}\neq\emptyset$, i.e. each of the five vertices in $U$ is adjacent to a common vertex.

Proof.

Let $U=\{u_{1},u_{2},\ldots,u_{5}\}$ be a $5$-set of independent vertices in $G$ and suppose that $U$ does not satisfy (A). By Lemma 12, we have $|X_{1}|\leq 10\ell-19$. Then by (1) and (2), we have

[TABLE]

so that $d(u_{i})=2\ell+3$ for some $i\in I$. By Lemma 8, therefore, we may assume the existence of a $4$-set $Q_{i}$ that satisfies (6). Select a vertex $v\in X_{1i}$ and consider the $5$-set $U^{\prime}=Q_{i}\cup\{v\}$, which must be independent. Note also that each of the five vertices in $U^{\prime}$ is adjacent to $u_{i}$. Thus, $U^{\prime}$ satisfies (B). ∎

For the rest of this paper, $U$ denotes a fixed 5-set of independent vertices in $G$ satisfying either (A) or (B) in Lemma 13.

4.5. Additional lemmas when $|X_{13}|=2\ell-3$

We prepare some more lemmas on $X_{2}$ under the condition $|X_{13}|=2\ell-3$.

Lemma 14**.**

Suppose that $|X_{13}|=2\ell-3$. Then $|X_{2}|=|X_{24}|+|X_{25}|-|X_{24}\cap X_{25}|$. Furthermore, $|X_{24}\cap X_{25}|\leq 1$.

Proof.

The first assertion is immediate by Lemma 10 (a). By Lemma 10 (a) (b), we see that $X_{2},X_{3}\subset N(u_{4})\cup N(u_{5})$. Also, obviously $X_{4},X_{5}\subset N(u_{4})\cup N(u_{5})$. Thus

[TABLE]

so that $|X_{24}\cap X_{25}|\leq 1$. This proves the lemma. ∎

Lemma 15**.**

Suppose that $|X_{13}|=2\ell-3$ and $v\in X_{2i}$ for some $i\in\{4,5\}$.

1. (a)

If $v\notin X_{24}\cap X_{25}$, then $|N(v)\cap X_{1i}|\geq|X_{1i}|-3$. 2. (b)

If $v\notin X_{24}\cap X_{25}$ and $|X_{1i}|=2\ell-4$, then $G[X_{1i}\cup\{v\}]\cong K_{2\ell-3}$.

Proof.

Suppose that $v\notin X_{24}\cap X_{25}$, that is, $v\in X_{2j}$ for some $j\in\{1,2,3\}$. Since $\omega(G)\leq 2\ell-2$, there is $w\in X_{1j}\backslash N(v)$. By Lemma 9 (a), observe that $w$ is adjacent to each vertex in $X_{1i}\backslash N(v)$. Thus, $|X_{1i}\backslash N(v)|\leq|N(w)\cap X_{1i}|\leq 3$ by Lemma 6. This proves (a).

We next prove the assertion (b). Assume, to the contrary, that $v\notin X_{24}\cap X_{25}$, $|X_{1i}|=2\ell-4$ and $G[X_{1i}\cup\{v\}]$ is not complete. Suppose $v\in X_{2j}$ with $j\in\{1,2,3\}$. By the same argument above, we see that $|X_{1i}\backslash N(v)|,|X_{1j}\backslash N(v)|\leq 3$ since we can take some vertex $w^{\prime}\in X_{1i}\setminus N(v)$ by the assumption. However then, $|N(v)\cap(X_{1i}\cup X_{1j})|\geq|X_{1i}|+|X_{1j}|-6=4\ell-13\geq 2\ell-1$, producing $\ell K_{2}$ in $G[N(v)\cap(X_{1i}\cup X_{1j}\cup\{u_{i},u_{j}\})]$. Since this does not occur, (b) also holds. ∎

Lemma 16**.**

Suppose that $|X_{13}|=2\ell-3$. If $|X_{1i}|\geq 2\ell-5$ for some $i\in\{4,5\}$, then $|X_{1i}|+|X_{2i}|\leq 2\ell$.

Proof.

Note that $|X_{1i}|\in\{2\ell-5,2\ell-4\}$ by Lemma 12.

If $|X_{1i}|=2\ell-4$ and $|X_{2i}|\geq 5$, then $X_{2i}$ contains four vertices $v_{1},v_{2},v_{3},v_{4}$ such that $|N(v_{j})\cap X_{1i}|\geq|X_{1i}|-3\geq 5$ for $1\leq j\leq 4$ by Lemma 15 (a) since $|X_{24}\cap X_{25}|\leq 1$. This however contradicts Lemma 7. Thus, $|X_{2i}|\leq 4$, as desired.

Similarly, if $|X_{1i}|=2\ell-5$ and $|X_{2i}|\geq 6$, say $\{v_{1},v_{2},\ldots,v_{6}\}\subset X_{2i}$, then we may assume that $v_{1}v_{2}\in E(G)$ as $\alpha(G)=5$ and $|N(v_{j})\cap X_{1i}|\geq|X_{1i}|-3\geq 4$ for $3\leq j\leq 5$. Thus, there are three distinct vertices $w_{3},w_{4},w_{5}\in X_{1i}$ such that $v_{j}w_{j}\in E(G)$ for $3\leq j\leq 5$. These three edges with the edge $v_{1}v_{2}$ and $(\ell-4)K_{2}$ in $G[X_{1i}\backslash\{w_{3},w_{4},w_{5}\}]$ result in $\ell K_{2}$ in $G[N(u_{i})]$, again a contradiction. It follows that $|X_{2i}|\leq 5$. ∎

5. Completion of the proof

We are now prepared to prove Theorem 3. Recall that we are now considering a fixed 5-set $U$ of independent vertices in $G$ satisfying one of the conditions

(A) $|X_{1}|\in\{10\ell-18,10\ell-17\}$ or (B) $X_{5}\neq\emptyset$.

In the following subsections, we consider these two cases separately.

5.1. Case A: $|X_{1}|\in\{10\ell-18,10\ell-17\}$

In this subsection, we consider Case A. In this case, observe that $|X_{11}|=|X_{12}|=2\ell-3$ and $|X_{14}|=2\ell-4$ since otherwise we have $|X_{1}|\leq 10\ell-19$ by Lemma 12. Thus we also have $|X_{13}|+|X_{15}|\in\{4\ell-8,4\ell-7\}$. Recalling (4), we have

[TABLE]

We now consider two subcases, according to the value of $|X_{13}|$.

5.1.1.

Subcase A1: $|X_{13}|=2\ell-3$.

In this case, $|X_{15}|\in\{2\ell-5,2\ell-4\}$. Again by (4) with Lemmas 14 and 16,

[TABLE]

Thus,

[TABLE]

First, if $X_{24}\cap X_{25}\neq\emptyset$, then it is immediate by (7) and (8) that $|X_{1i}|=2\ell-|X_{2i}|=2\ell-4\geq 8$ for $i=4,5$. Let $X_{24}\cap X_{25}=\{v\}$. By Lemma 15 (b), each of the three vertices in $X_{2i}\backslash\{v\}$ is adjacent to every vertex in $X_{1i}$ for $i=4,5$. Since neither $G[N(u_{4})]$ nor $G[N(u_{5})]$ contains $\ell K_{2}$, we find that $N(v)\cap(X_{14}\cup X_{15})=\emptyset$. However then, $G[X_{14}\cup X_{15}]\cong K_{4\ell-8}$ by Lemma 9 (a), which cannot occur since $4\ell-8>2\ell-2\geq\omega(G)$.

Thus, we may assume that $X_{24}\cap X_{25}=\emptyset$. By (7) and (8) again, $|X_{1i}|=2\ell-|X_{2i}|=2\ell-4$ for some $i\in\{4,5\}$. Then Lemma 15 (b) implies that each of the four vertices in $X_{2i}$ is adjacent to every vertex in $X_{1i}$. Thus, $|N(v)\cap X_{1i}|=|X_{1i}|=2\ell-4\geq 8$ for every $v\in X_{2i}$. However, this contradicts Lemma 7.

5.1.2.

Subcase A2: $|X_{13}|=2\ell-4$.

In this case, $|X_{1i}|=2\ell-4$ for $3\leq i\leq 5$ and $|X_{1}|=10\ell-18$.

We first verify that, if $v\in X_{2}$, then $v\in X_{2j}$ and $|N(v)\cap X_{1j}|\geq 5$ for some $j\in\{3,4,5\}$. To see this, suppose that $v\in X_{2i}\cap X_{2j}$ with $1\leq i<j\leq 5$. If $i\leq 2$, then $3\leq j\leq 5$ as $X_{21}\cap X_{22}=\emptyset$ by Lemma 10. Since $\omega(G)\leq 2\ell-2$ and $|X_{1i}|=2\ell-3$, there is a vertex $w\in X_{1i}$ such that $vw\notin E(G)$. Then $w$ is adjacent to every vertex in $X_{1j}\backslash N(v)$ by Lemma 9 (a), implying that $|X_{1j}\backslash N(v)|\leq 3$ by Lemma 6. If $3\leq i<j\leq 5$, on the other hand, then a similar reasoning shows that $|X_{1i}\backslash N(v)|\leq 3$ or $|X_{1j}\backslash N(v)|\leq 3$. Consequently, $|X_{1j}\backslash N(v)|\leq 3$ for some $j\in\{3,4,5\}$ in each case, i.e. $|N(v)\cap X_{1j}|\geq|X_{1j}|-3\geq 5$.

We next verify that $|X_{2}|=9$. If this is not the case, then $|X_{2}|\geq 10$. By the pigeonhole principle, there exists an integer $j\in\{3,4,5\}$ and a $4$-set $S\subset X_{2j}$ such that $|N(v)\cap X_{1j}|\geq 5$ for each $v\in S$, contradicting Lemma 7.

Recalling (4), we have

[TABLE]

and so $|X_{4}|=|X_{5}|=0$, which further tells us that $|X_{3}|=5$. Also, $\sum_{i=1}i|X_{i}|=10\ell+15$, that is, $d(u_{i})=2\ell+3$ for $1\leq i\leq 5$. By Lemma 8, we can find five $4$-sets $Q_{1},Q_{2},\ldots,Q_{5}$ satisfying (6). By Lemma 11 (a) (b) then,

[TABLE]

which is clearly impossible.

As a result, Subcases A1 and A2 are both impossible, i.e. Case A never occurs.

5.2. Case B: $X_{5}\neq\emptyset$

In this subsection, we next consider Case B. In this case, we see that $|X_{1}|\geq 10\ell-23+3|X_{5}|\geq 10\ell-20$ by (3). Since Case A never occurs, we may assume that $|X_{1}|\in\{10\ell-20,10\ell-19\}$. We again consider two subcases.

5.2.1.

Subcase B1: $|X_{1}|=10\ell-20$.

By (3), we find that $|X_{3}|=|X_{4}|=0$ and $|X_{5}|=1$, which in turn implies that $|X_{2}|=15$. Also, $\sum_{i=1}^{5}i|X_{i}|=10\ell+15$ and so $d(u_{i})=2\ell+3$ for $1\leq i\leq 5$, which guarantees the existence of five $4$-sets $Q_{1},Q_{2},\ldots,Q_{5}$ satisfying (6). As done in (9), we can write

[TABLE]

and so $\sum_{i=1}^{5}|Q_{i}\cap X_{2}|=15$ and $\sum_{i=1}^{5}|Q_{i}\cap X_{5}|=5$. Hence, if we let $X_{5}=\{v_{0}\}$, then each $Q_{i}$ consists of three vertices in $X_{2}$ and $v_{0}$ so that $\bigcup_{i=1}^{5}Q_{i}=X_{2}\cup X_{5}$. However then, no vertex in $X_{1}\cup X_{2}$ is adjacent to $v_{0}$ as $v_{0}\in Q_{i}$, implying that $N(v_{0})=U$. Thus, $d(v_{0})=5<2\ell=\delta(G)$, a contradiction.

5.2.2.

Subcase B2: $|X_{1}|=10\ell-19$.

First, observe that

[TABLE]

by (1) and (2). Thus, we find that $|X_{4}|=0$, $|X_{5}|=1$ and either

(i) $|X_{2}|=13$ and $|X_{3}|=1$ or (ii) $|X_{2}|=14$ and $|X_{3}|=0$.

We consider these two cases separately. Let $X_{5}=\{v_{0}\}$.

If (i) occurs, then again $\sum_{i=1}^{5}i|X_{i}|=10\ell+15$ and so we have $4$-sets $Q_{1},Q_{2},\ldots,Q_{5}$ satisfying (6). Since $|X_{1}|=10\ell-19$, we can check that $|X_{13}|+|X_{14}|+|X_{15}|\geq 6\ell-13$. Thus by Lemma 11 (a) (b),

[TABLE]

and we arrive at the same contradiction as in Subcase B.1.

Finally, suppose that (ii) occurs. Then $\sum_{i=1}^{5}i|X_{i}|=10\ell+14$. Thus, four of the five vertices in $U$ have degree $2\ell+3$ and the remaining one has degree $2\ell+2$. Let $i_{0}\in I$ for which $d(u_{i_{0}})=2\ell+2$. We know that there is a $4$-set $Q_{i}$ satisfying (6) for each $i\in I\backslash\{i_{0}\}$.

In the set $N(u_{i_{0}})\backslash X_{1i_{0}}$, which contains $2\ell+2-|X_{1{i_{0}}}|\geq 5$ vertices, we show that there exists a $3$-set $Q\subset N(u_{i_{0}})\backslash X_{1i_{0}}$ such that $|N(v)\cap X_{1{i_{0}}}|\leq 2$ for each $v\in Q$. Recall Lemma 2. First, if $|X_{1{i_{0}}}|=2t-1$ for some positive integer $t$, then $G[N(u_{i_{0}})\backslash X_{1i_{0}}]$ contains $(\ell-t-1)K_{2}$ and five vertices $v_{1},v_{2},\ldots,v_{5}$. Since $G[N(u_{i_{0}})]$ cannot contain $\ell K_{2}$, there are at most two independent edges between $X_{1{i_{0}}}$ and $\{v_{1},v_{2},\ldots,v_{5}\}$. Thus, we may assume that $|N(v_{j})\cap X_{1{i_{0}}}|\leq 2$ for $1\leq j\leq 3$. Similarly, if $|X_{1{i_{0}}}|=2t$ for some positive integer $t$, then $G[N(u_{i_{0}})\backslash X_{1i_{0}}]$ contains $(\ell-t-1)K_{2}$ and four vertices $v_{1},v_{2},v_{3},v_{4}$. Then there cannot be two or more independent edges between $X_{1{i_{0}}}$ and $\{v_{1},v_{2},v_{3},v_{4}\}$. Hence, we may assume that $|N(v_{j})\cap X_{1{i_{0}}}|\leq 1$ for $1\leq j\leq 3$. As a result, we find a $3$-set $Q=\{v_{1},v_{2},v_{3}\}\subset N(u_{i_{0}})\backslash X_{1i_{0}}$ such that $|N(v)\cap X_{1{i_{0}}}|\leq 2$ for each $v\in Q$.

Write $Q^{\prime}_{i_{0}}=Q$ and $Q^{\prime}_{i}=Q_{i}$ for each $i\in I\backslash\{i_{0}\}$. We next verify that $\sum_{i=1}^{5}|Q^{\prime}_{i}\cap X_{2}|\leq|X_{2}|$ by proving that $Q^{\prime}_{i}\cap Q^{\prime}_{j}\cap X_{2}=\emptyset$ for distinct integers $i,j\in I$. First, since $|X_{1}|=10\ell-19$, it follows that $|X_{14}|+|X_{15}|=|X_{1}|-\sum_{i=1}^{3}|X_{1i}|\geq 4\ell-10\geq 2\ell+2$. Hence, by Lemma 11 (a), it suffices to verify the result for $i\in I\backslash{i_{0}}$ and $j=i_{0}$. If $v\in Q^{\prime}_{i}\cap Q^{\prime}_{j}\cap X_{2}$, then no vertex in $X_{1i}$ is adjacent to $v$. Also, we have just shown that $v$ is adjacent to at most two vertices in $X_{1j}$. Then by Lemma 9 (a), there are at least $|X_{1j}|-2$ vertices in $X_{1j}$ that are adjacent to every vertex in $X_{1i}$. Hence, $\omega(G)\geq|X_{1i}|+|X_{1j}|-2\geq|X_{14}|+|X_{15}|-2\geq 2\ell$, which cannot occur.

Therefore,

[TABLE]

so $v_{0}$ belongs to every $Q^{\prime}_{i}$. Thus, $N[v_{0}]\subset U\cup Q^{\prime}_{i_{0}}\cup X_{1{i_{0}}}$. In particular, $|N(v_{0})\cap X_{1{i_{0}}}|\leq 2$ and so $d(v_{0})\leq|U|+|Q^{\prime}_{i_{0}}\backslash\{v_{0}\}|+2=9<2\ell$. This is again impossible.

We conclude that Case B never occurs, either.

This completes the proof of Theorem 3 and Theorem 1.

Acknowledgements

The authors would like to thank Prof. Futaba Fujie for her comments and suggestions, which greatly improved the original manuscript.