1 Introduction

Many real-world networks can be conveniently modeled by graphs and studied by graph theory methods. A network can be defined as a graph in which vertices and edges of the graph correspond to nodes and links between the nodes in the network. Graph connectivity is an important parameter of graphs and networks. Through the study of the graph connectivity, people have a further understanding of the structure and properties of graphs, which then results in solutions to many practical problems such as reliable communication network design. Independence number and factors in graphs have attracted a great deal of attention due to their applications in matching theory, network design, combinatorial design, etc. In this paper, we study the existence of degree constraint factors in terms of independence number and connectivity.

All graphs considered in this paper are simple. Let G be a graph with vertex set V(G), edge set E(G) and order \(n =|V(G)|\). Given a vertex \(v \in V(G)\), let \(N_G ( v )\) denote the set of vertices adjacent to v in G and \(d_G(v)=|N_G(v)|\). We write \(N_{G}[v] =N_{G}(v)\cup \{v\}\). For any subset \(X\subseteq V(G)\), let G[X] denote the vertex induced subgraph of G induced by X and the subgraph induced by vertex set \(V(G)-X\) is also denoted by \(G- X\). For vertex \(v \in V(G)\), the number of edges which link v and X is denoted by \(e_G ( v, X)\), and \(e_G (Y , X) =\sum _{v\in Y} e_G ( v, X)\). A graph F is a spanning subgraph of G if \(V (F) = V (G)\) and \(E(F) \subseteq E(G)\).

A subset \(S\subseteq V (G)\) is a cut-set of a connected graph G if \(G-S\) is disconnected. The cardinality of the smallest cut-set in G is called the connectivity of G and is denoted by \(\kappa (G)\). A subset S of V(G) is called an independent set of graph G if any two vertices of S are non-adjacent in G. A maximum independent set is an independent vertex set of a graph G containing the largest possible number of vertices; the cardinality of this set is called the independence number of graph G and denoted by \(\alpha (G)\). The join \(G = G_1 + G_2\) is the graph obtained from two vertex disjoint graphs \(G_1\) and \(G_2\), by joining each vertex in \(G_1\) to every vertex in \(G_2\).

Let g and f be two integer-valued functions defined on V(G) such that \(0 \le g ( x ) \le f ( x )\) for all \(x \in V ( G ) \). We will say that G has all (g, f)-factors if G has an h-factor for every \(h:V(G)\rightarrow {\mathbb {N}}\) such that \(g(v)\le h(v)\le f(v)\) for all \(v\in V(G)\) and \(\sum _{v\in V(G)}h(v)\equiv 0\pmod 2\). Let \(b>a\) be two positive integers. All (g, f) factors are called all [a, b]-factors if \(g ( x ) \equiv a\) and \(f ( x )\equiv b\) for all \(x \in V(G)\). A (g, f)-parity factor of G is a spanning subgraph F of G satisfying \(g(v)\equiv f(v)\pmod 2\) and \(g ( x ) \le d_F ( x ) \le f ( x )\) for all \(x \in V ( G )\). A (g, f)-parity factor is an f-factor if \(f ( v ) = g( v )\) for all \(v \in V(G)\). If \(f ( v ) = k\) for all \(v \in V(G)\), then an f-factor is a k-factor. A (g, f)-parity factor is called an (a, b)-parity factor if \(g ( x ) \equiv a\) and \(f ( x )\equiv b\). For any real function h on V(G) and any subset \(S\subseteq V(G)\), we denote \(\sum _{x\in S}h(x)\) by h(S).

The following theorem gives a criteria for a graph G to have a (g, f)-parity factor.

Theorem 1.1

(Lovász, [6]) A graph G has a (g, f)-parity factor if and only if for any disjoint subsets S, T of V(G),

$$\begin{aligned} \eta (S,T)=f(S)-g(T)+\sum _{x\in T}d_{G-S}(x)-\tau (S,T)\ge 0, \end{aligned}$$

where \(\tau (S,T)\) denotes the number of connected components C of \(G-(S\cup T)\) such that \(e_G(V(C),T)+g(V(C))\equiv 1\pmod 2\).

There are some sufficient conditions for a graph to have all [a, b]-factors [2, 3]. In particular, Niessen [8] gave a characterization for a graph to have all (g, f)-factors.

Theorem 1.2

(Niessen, [8]) A graph G has all (g, f)-factors if and only if

$$\begin{aligned} \delta (S,T)=g(S)-f(T)+\sum _{x\in T}d_{G-S}(x)-q(S,T)\ge {\left\{ \begin{array}{ll} -1,&{} \ f\ne g\\ 0,&{} \text {f=g} \end{array}\right. } \end{aligned}$$

for all disjoint sets \(S, T\subseteq V(G)\), where q(S, T) denotes the number of connected components C of \(G-(S\cup T)\) such that there exists a vertex \(v\in V(C)\) with \(g(v)<f(v)\) or \(e_G(V(C),T)+f(V(C))\equiv 1\pmod 2\).

Lu et al. [7] showed that it is NP-hard to determine whether a graph has all (g, f)-factors, which answers a problem proposed by Niessen [8].

A classical result of Chvátal and Erdős [1] says that every graph G whose vertex connectivity is at least as large as its independence number is Hamiltonian. Motivated by Chvátal and Erdős’s result, Nishimura [9] obtained some sufficient conditions for a graph to have a k-factor.

Theorem 1.3

(Nishimura, [9]) Let \(k\ge 1\) be an odd integer and G be a graph such that |V(G)| is even. If \(\kappa (G)\ge (k+1)^2/2\) and \((k+1)^2\alpha (G)\le 4k \kappa (G)\), then G has a k-factor.

Theorem 1.4

(Nishimura, [9]) Let \(k\ge 2\) be an even integer and G be a graph. If \(\kappa (G)\ge k(k+2)/2\) and \((k+2)\alpha (G)\le 4\kappa (G)\), then G has a k-factor.

For the existence of f-factors, a sufficient condition on the independence number and connectivity was given by Katerinis and Tsikopoulos [4]. Kouider and Lonc [5] gave a sufficient condition for the existence of [a, b]-factors in terms of the minimum degree, the connectivity and the independence number. Furthermore, Zhou [11] obtained a sufficient condition for the existence of [a, b]-factors based on independence number and connectivity.

In this paper, we generalize Nishimura’s results in two aspects. Firstly, we extend extended the results from k-factors to all [a, b]-factors and obtained sufficient conditions for a graph to have all [a, b]-factors in terms of the independence number and the connectivity. The result is stated in the following.

Theorem 1.5

Let G be a graph, and let \(b> a\ge 1 \) be two integers. If one of the following two conditions holds, then G contains all [a, b]-factors.

  1. (i)

    b is odd and

    $$\begin{aligned}\kappa (G)\ge \max \left\{ \frac{(b+1)^2}{2}, \frac{\alpha (G)(b+1)^2}{4a}\right\} , \end{aligned}$$
  2. (ii)

    b is even and

    $$\begin{aligned}\kappa (G)\ge \max \left\{ \frac{b(b+2)}{2},\frac{\alpha (G)b(b+2)}{4a} \right\} . \end{aligned}$$

Secondly, we have to extend Nishimura’s results to (a, b)-parity factors, and the result is listed below.

Theorem 1.6

Let b, a be two positive integers such that \(b>a\) and \(b\equiv a\pmod 2\). Let G be a graph such that b|V(G)| is even. If one of the following two conditions holds, then G contains an (a, b)-parity factor.

  1. (i)

    b is odd and

    $$\begin{aligned}\kappa (G)\ge \max \left\{ \frac{(a+1)^2}{2},\frac{\alpha (G)(a+1)^2}{4b} \right\} , \end{aligned}$$
  2. (ii)

    b is even and

    $$\begin{aligned}\kappa (G)\ge \max \left\{ \frac{a(a+2)}{2},\frac{\alpha (G)a(a+2)}{4b} \right\} . \end{aligned}$$

Remark 1

The condition “\(\kappa (G)\ge \frac{\alpha (G)(b+1)^2}{4a}\)” in Theorem 1.5 (i) is best possible in some cases. Let \(K_p\) denote the complete graph with p vertices. Let a, b, s, k be four positive integers such that \( b \equiv 1\pmod 2\), \(b>a\), \(s< \frac{k(b+1)^2}{4a}-1/a\) and \(k\ge (b+1)^2/2\). Suppose that \(\frac{k(b+1)^2}{4a}-1/a\) is not an integer. Consider the join \(G=K_s+kK_{\frac{b+1}{2}}\), one can see that \(\alpha (G)=k\) and \(\kappa (G)=s<\frac{\alpha (G)(b+1)^2}{4a}-1/a\). Let \(S:=V(K_s)\) and \(T:=V(G)-S\). Then we have

$$\begin{aligned} \delta (S,T)&=a|S|-b|T|+\sum _{x\in T}d_{G-S}(x)<\frac{k(b+1)^2}{4}-1-\frac{bk(b+1)}{2}\\&\quad +\frac{k(b+1)}{2}\cdot \frac{(b-1)}{2}= -1. \end{aligned}$$

By Theorem 1.2, the property that G contains all (a, b)-factors does not hold. Let a, b, s, k be four positive integers such that \(b \equiv 0\pmod 2\), \(b>a\), \(s< \frac{kb(b+2)}{4a}-1/a\) and \(k\ge b(b+2)/2\). Suppose that \(\frac{kb(b+2)}{4a}-1/a\) is not an integer. Consider the join \(G=K_s+kK_{\frac{b+2}{2}}\). Along the same line of reasoning, we can see that the condition “\(\kappa (G)\ge \frac{\alpha (G)b(b+2)}{4a}\)” in Theorem 1.5 (ii) is also best possible in some cases.

Remark 2

The condition “\(\kappa (G)\ge \frac{\alpha (G)(a+1)^2}{4b}\)” in Theorem 1.6 (i) is best possible. Let a, b, s, k be four positive integers such that \(a \equiv b\equiv 1\pmod 2\), \(b>a\), \(s< \frac{k(a+1)^2}{4b}\) and \(s+k(a+1)/2 \equiv 0\pmod 2\). Consider the join \(G=K_s+kK_{\frac{a+1}{2}}\), one can see that \(\alpha (G)=k\) and \(\kappa (G)=s<\frac{\alpha (G)(a+1)^2}{4b}\). Let \(S:=V(K_s)\) and \(T:=V(G)-S\). Then we have

$$\begin{aligned} \eta (S,T)&=b|S|-a|T|+\sum _{x\in T}d_{G-S}(x) <\frac{k(a+1)^2}{4}-\frac{ak(a+1)}{2}\\&\quad +\frac{k(a+1)}{2}\cdot \frac{(a-1)}{2}=0. \end{aligned}$$

By Theorem 1.1, G contains no (a, b)-parity factors. Let a, b, s, k be four positive integers such that \(a \equiv b\equiv 0\pmod 2\), \(b>a\) and \(s< \frac{ka(a+2)}{4b}\). Consider the join \(G=K_s+kK_{\frac{a+2}{2}}\), along the same line of reasoning, we can see that the condition “\(\kappa (G)\ge \frac{\alpha (G)a(a+2)}{4b}\)” in Theorem 1.6 (ii) is also best possible.

2 Proofs of Theorem 1.5

Proof of Theorem 1.5

Firstly, suppose that condition (i) holds and the theorem does not hold. Since \(b>a\), by Theorem 1.2, there exist two disjoint vertex subsets S and T of V(G) such that

$$\begin{aligned} \delta (S,T)=as-bt+\sum _{x\in T}d_{G-S}(x)-q\le -2, \end{aligned}$$
(1)

where \(s = |S|\), \(t = |T |\) and \(q:=q(S,T)\) denotes the number of connected components of \(G-(S\cup T)\).

If \(S\cup T =\emptyset \), we have \(q \ge 2\) by (1), which implies that G consists of at least two connected components, which contradicts to the condition that \(\kappa (G)\ge 1\). So we assume that \(S\cup T\ne \emptyset .\) If \(q\ge 1\), let \(C_1,C_2,\ldots ,C_q\) denote the connected components of \(G-(S\cup T)\). We choose S and T satisfying (1) and, subject to this, \(S\cup T\) is maximal. For convenience, let \(W:=\bigcup _{1\le i\le q}V(C_i)\) and \(p:=\kappa (G-S)\).

Now we claim that \(T\ne \emptyset \). Suppose that \(T=\emptyset \). By (1) and the definition of W, we have \(\alpha (G)\ge q\ge as+2\). Note that \(\kappa (G)\le s\) and \((b+1)^2>4\). So we have

$$\begin{aligned} as+2\le q\le \alpha (G)\le \frac{4a\kappa (G)}{(b+1)^2}< as, \end{aligned}$$

a contradiction.

Define \(T_1:=G[T]\). We choose \(x_1\in V(T_1)\) such that \(d_{T_1}(x_1)=\min \{d_{T_1}(x)~|~x\in V(T_1)\}\) and let \(N_1:=N_{T_1}[x_1]\) and \(T_2:=G[V(T_1)-N_1]\). For \(j\ge 2\), we choose \(x_j\in V(T_j)\) such that \(d_{T_j}(x_j)=\min \{d_{T_j}(x)~|~x\in V(T_j)\}\) and let \(N_{j}:=N_{T_{j}}[x_j]\) and \(T_{j+1}: =G[V(T_{j})- N_j]\). We continue these procedures until we reach the situation where \(V(T_{j+1})=\emptyset \) for some j. Without loss of generality, suppose that \(N_k\ne \emptyset \) and \(N_{k+1}=\emptyset \), where \(k\ge 1\). Then we have a sequence of non-empty sets of vertices \(N_1,\ldots , N_k\). Let \(n_j:=|N_j|\) for \(j\in \{1,2,...,k\}\). From the discussion, we can infer that

$$\begin{aligned} |T|=|N_1\cup \ldots \cup N_k|=\sum _{j=1}^k |N_j|=\sum _{j=1}^k n_j, \end{aligned}$$
(2)

and since all vertices in \(N_j\) have degree at least \(n_j-1\) in \(T_j\), thus

$$\begin{aligned} \sum _{j=1}^k\sum _{x\in N_j}d_{T_j}(x)\ge \sum _{j=1}^k n_j(n_j-1). \end{aligned}$$
(3)

Note that an edge joining a vertex x in \(N_j\) and a vertex y in \(N_l\) \((1\le j < l\le k)\) is counted only once; that is to say, it is counted in the term \(d_{T_j}(x)\) but not in the term \(d_{T_l}(y)\). From this observation and (3), we have

$$\begin{aligned} \sum _{x\in T} d_{G-S}(x)\ge \sum _{j=1}^k n_j(n_j-1)+\sum _{1\le j<l\le k} e_{G-S}(N_j, N_l)+e_{G}(T,W). \end{aligned}$$
(4)

On the other hand, since \(G-S\) is p-connected, then for each \(N_j\) (\(1\le j\le k\)),

$$\begin{aligned} e_{G-S}(N_j,T-N_j)+e_{G}(N_j,W)\ge p. \end{aligned}$$

Summing up these inequalities for all j, we have

$$\begin{aligned} \sum _{j=1}^k e_{G-S}(N_j,T-N_j)+e_{G}(T,W)\ge kp, \end{aligned}$$

which implies that

$$\begin{aligned} 2\sum _{1\le j<l\le k} e_{G-S}(N_j,N_l)+e_{G}(T,W)\ge kp. \end{aligned}$$
(5)

Combining (4) and (5), we get

$$\begin{aligned} \sum _{x\in T} d_{G-S}(x)\ge \sum _{j=1}^k n_j(n_j-1)+\frac{kp+e_G(T,W)}{2}. \end{aligned}$$
(6)

Thus we have,

$$\begin{aligned} -2 \ge \delta (S, T )&= as-bt+\sum _{x\in T}d_{G-S}(x)-q\\&\ge as-bt+\sum _{j=1}^k n_j(n_j-1)+\frac{kp+e_G(T,W)}{2}-q\quad (\hbox {by}\ (6))\\&= as+\sum _{i=1}^k n_j(n_j-b-1)+\frac{kp+e_G(T,W)}{2}-q\quad (\hbox {by}\ (2))\\&\ge as-\frac{(b+1)^2k}{4}+\frac{kp+e_G(T,W)}{2}-q, \end{aligned}$$

i.e.,

$$\begin{aligned} \delta (S, T )\ge as-\frac{(b+1)^2k}{4}+\frac{kp+e_G(T,W)}{2}-q. \end{aligned}$$
(7)

Claim 1

For every vertex \(u\in W\), we have \(e_G(u, T)\le a\) with the equality holds only if u is an isolated vertex of G[W].

We show this claim by contradiction. Suppose that there exists a vertex \(u\in W\) such that one of following two conditions holds:

(a):

\(e_G(u, T)\ge a+1\);

(b):

\(e_G(u,T)=a\) and u is not an isolated vertex of G[W].

Let \(S'=S\cup \{u\}\). Note that \(\omega (G[W-u])\ge q-1\) with the equality holds if and only if u is an isolated vertex of G[W], where \(\omega (G[W-u])\) denotes the number of connected components of graph \(G[W-u]\). So we have

$$\begin{aligned} \delta (S',T)&=a|S'|-bt+\sum _{x\in T}d_{G-S'}(x)-q(S',T)\\&= as-bt+\sum _{x\in T}d_{G-S}(x)-\omega (G[W-u])+a-e_G(u, T)\\&\le as-bt+\sum _{x\in T}d_{G-S}(x)-q \quad \hbox {(by (a) and (b))}\\&=\delta (S,T)\le -2, \end{aligned}$$

contradicting to the maximality of \(S\cup T\). This completes the proof of Claim 1.

Claim 2

For every vertex \(u\in W\), we have \(d_{G-S}(u)\ge b\) with equality only if u is an isolated vertex of G[W].

Suppose that there exists a vertex \(u\in W\) such that one of following two conditions holds:

(c):

\(d_{G-S}(u)\le b-1\);

(d):

\(d_{G-S}(u)=b\) and u is not an isolated vertex of G[W].

Let \(T'=T\cup \{u\}\). Note that \(\omega (G[W-u])\ge q-1\) with the equality holds if and only if u is an isolated vertex of G[W]. So we have

$$\begin{aligned} \delta (S,T')&=as-b|T'|+\sum _{x\in T'}d_{G-S}(x)-q(S,T')\\&=as-bt+\sum _{x\in T}d_{G-S}(x)-\omega (G[W]-u)-b+d_{G-S}(u)\\&\le as-bt+\sum _{x\in T}d_{G-S}(x)-q \quad \hbox {(by (c) and (d))}\\&=\delta (S,T)\le -2, \end{aligned}$$

contradicting to the maximality of \(S\cup T\). This completes the proof of Claim 2.

Since \(b>a\), by Claims 1 and 2, G[W] does not contain isolated vertices. Hence \(d_{G-S}(u)\ge b+1\) and \(e_{G}(u, T)\le a-1\) hold for all \(u\in W\). So we have \(d_{G[W]}(u)\ge b+1 -(a-1)= b-a+2\). The degree of each vertex in G[W] is at least \(b-a+2\); thus, each connected component of G[W] contains at least \(b-a+3\) vertices, i.e., the vertex itself with all \(b-a+2\) neighbors. If \(V(C_i)\subseteq N_G(\{x_1,\ldots ,x_k\})\), then we have \(e_G(V(C_i),T)\ge b-a+3\) for \(i\in \{1,2,...,q\}\). Let \(r_1\) denote the number of connected components \(C_i\) such that \(V(C_i)\subseteq N_G(\{x_1,\ldots ,x_k\})\) and let \(r_2\) denote the number of connected components \(C_i\) such that \(V(C_i)\nsubseteq N_G(\{x_1,\ldots ,x_k\})\). Recall that \(\kappa (G-S)=p\), then we have

$$\begin{aligned} e_G(T,W)\ge (b-a+3)r_1+pr_2. \end{aligned}$$
(8)

Note that \( q=r_1+r_2, \)

$$\begin{aligned} \alpha (G)\ge k+r_2 \end{aligned}$$
(9)

and

$$\begin{aligned} \kappa (G)\le s+p. \end{aligned}$$
(10)

Thus, by (7), we have

$$\begin{aligned} -2\ge \delta (S,T)&\ge as-\frac{(b+1)^2k}{4}+\frac{kp+e_G(T,W)}{2}-q\\&\ge as - \frac{(b + 1)^2k}{4} +\frac{kp+(b-a+3)r_1+r_2p}{2}- (r_1+r_2)\quad (\hbox {by}\ (8)) \\&= as - \frac{(b + 1)^2k}{4} +\frac{ (k+r_2)p}{2}+\frac{(b-a+1)r_1}{2}-r_2\\&\ge as - \frac{(b + 1)^2k}{4} +\frac{ (k+r_2)p}{2}-r_2 \\&> as - \frac{(b + 1)^2(k+r_2)}{4}+\frac{(k+r_2)p}{2}\quad (\hbox { since}\ (b+1)^2>4)\\&=as - (k+r_2)\left( \frac{(b + 1)^2}{4}-\frac{p}{2}\right) \\&\ge a(\kappa (G)-p)- \alpha (G)\left( \frac{(b + 1)^2}{4}-\frac{p}{2}\right) \\&\quad \left( \hbox {since}\ \alpha (G)\le \frac{4a\kappa (G)}{(b+1)^2}\ \hbox {and}\ \hbox {by}\ (9)\ \hbox {and}\ (10)\right) \\&\ge ap\left( \frac{2 \kappa (G)}{(b+1)^2}-1\right) \ge 0, \quad \left( \hbox { since}\ \kappa (G)\ge \frac{(b+1)^2}{2}\right) \end{aligned}$$

a contradiction. This completes the proof of Theorem 1.5 (i).

To show the theorem is true when condition (ii) holds, since \(n_j(n_j-b-1)\ge -\frac{b(b + 2)}{4}\) for even b, we may replace \(-\frac{(b+1)^2k}{4}\) by \(-\frac{b(b+2)k}{4}\) in (7), then following the same discussion as for Theorem 1.5 (i), we can obtain the proof of Theorem 1.5 (ii). \(\square \)

3 Proof of Theorem 1.6

Proof of Theorem 1.6

The proof of this theorem is similar to the proof of Theorem 1.5, so we omit some details in the proof just to avoid repeating ourselves. Firstly, suppose that condition (i) holds and G contains no (a, b)-parity factors. By Theorem 1.1, there exist two disjoint subsets S and T of V(G) such that

$$\begin{aligned} \eta (S,T)=bs-at+\sum _{x\in T}d_{G-S}(x)-\tau \le -2, \end{aligned}$$
(11)

where \(s: = |S|\), \(t: = |T |\) and \(\tau :=\tau (S,T)\) denotes the number of the connected components C of \(G-(S\cup T)\) where \(a|V(C)|+e_G(V(C),T)\equiv 1\pmod 2\) (these components are also called a-odd components). If \(\tau \ge 1\), let \(C_1,C_2,\ldots ,C_{\tau }\) denote these a-odd components of \(G-(S\cup T)\). Let \(W:=\bigcup _{1\le i\le \tau }V(C_i)\). With the same discussion of the proof of Theorem 1.5, we may assume that \(S\cup T\ne \emptyset \). We choose S and T satisfying (11) and \(S\cup T\) is maximal.

Now we claim that \(T\ne \emptyset \). Suppose that \(T=\emptyset \). By (11) and the definition of W, we have \(\alpha (G)\ge \tau \ge bs+2\). Note that \(\kappa (G)\le s\) and \((a+1)^2\ge 4\). So we have

$$\begin{aligned} bs+2\le \tau \le \alpha (G)\le \frac{4b\kappa (G)}{(a+1)^2}\le bs, \end{aligned}$$

a contradiction. Let \(p,x_1,\ldots ,x_k\), \(n_1,\ldots ,n_k\) be defined as in the proof of Theorem 1.5. With the same discussion as (6) in Theorem 1.5, we also have

$$\begin{aligned} \sum _{x\in T} d_{G-S}(x)\ge \sum _{j=1}^k n_j(n_j-1)+\frac{kp+e_G(T,W)}{2}. \end{aligned}$$

Next, we estimate \(e_G(T,W)\).

Claim 3

For every vertex \(u\in W\), \(d_{G-S}(u)\ge a+1\).

We show this claim by contradiction. Suppose that there exists a vertex \(u\in W\) such that \(d_{G-S}(u)\le a\). Let \(T'=T\cup \{u\}\), then we have

$$\begin{aligned} \eta (S,T')&=bs-a|T'|+\sum _{x\in T'}d_{G-S}(x)-\tau (S,T')\\&=bs-at-a+\sum _{x\in T}d_{G-S}(x)+d_{G-S}(u)-\tau (S,T')\\&\le bs-at+\sum _{x\in T}d_{G-S}(x)-(\tau -1)\le -1, \end{aligned}$$

which implies by parity

$$\begin{aligned} \eta (S,T')&=bs-a|T'|+\sum _{x\in T'}d_{G-S}(x)-\tau (S,T')\le -2, \end{aligned}$$

contradicting to the maximality of \(S\cup T\). This completes the proof of Claim 3.

Now we show if \(V(C_i)\subseteq N_G(\{x_1,...,x_k\})\), where \(i\in \{1,2,\ldots , \tau \}\), we have \(e_G(V(C_i),T)\ge 2\). If \(V(C_i)=\{u\}\), by Claim 3, we have \(e_G(V(C_i),T)=d_{G-S}(u)\ge a+1\ge 2\). If \(|V(C_i)|\ge 2\), then we have \(e_G(V(C_i),T)\ge |V(C_i)|\ge 2\).

Recall that \(r_1\) (\(r_2\)) denotes the number of connected components \(C_i\) such that \(V(C_i)\subseteq N_G(\{x_1,\ldots ,x_k\})\) (\(V(C_i)\nsubseteq N_G(\{x_1,\ldots ,x_k\})\), respectively). Then we have

$$\begin{aligned} e_G(T,W)\ge 2r_1+pr_2. \end{aligned}$$
(12)

Note that (9) and (10) also hold and \(\tau =r_1+r_2.\) Thus, by (11), we have

$$\begin{aligned} -2\ge \eta (S,T)&= bs-at+\sum _{x\in T}d_{G-S}(x)-\tau \\&\ge bs-at+\sum _{j=1}^k n_j(n_j-1)+\frac{kp+e_G(T,W))}{2}-\tau \quad \hbox {(by} (6)\hbox {)}\\&\ge bs+\sum _{j=1}^k n_j(n_j-a-1)+\frac{kp+2r_1+pr_2}{2}-r_1-r_2\quad \left( \hbox {by}\ \hbox {(}2\hbox {) and}\ \hbox {(}12\hbox {)}\right) \\&\ge bs-\frac{k(a+1)^2}{4}+\frac{p(k+r_2)}{2}-r_2\\&\ge bs-\frac{(k+r_2)(a+1)^2}{4}+\frac{p(k+r_2)}{2} \quad (\hbox { since}\ (a+1)^2\ge 4)\\&= bs-(k+r_2)\left( \frac{(a+1)^2}{4}-\frac{p}{2}\right) \\&\ge b(\kappa (G)-p)-\alpha (G)\left( \frac{(a+1)^2}{4}-\frac{p}{2}\right) \\&\quad \left( \hbox {by}\ (9)\ \hbox {and}\ (10),\ \hbox {and since}\ \alpha (G)\le \frac{4b\kappa (G)}{(a+1)^2}\right) \\&\ge bp\left( \frac{2\kappa (G)}{(a+1)^2}-1\right) \ge 0 \quad (\hbox { since}\ \kappa (G)\ge \frac{(a+1)^2}{2}), \end{aligned}$$

a contradiction. This contradiction completes the proof of Theorem 1.6 (i).

Secondly, suppose that condition (ii) holds. Replacing “\(n_i(n_i-a-1)\ge (a+1)^2/4\)” in the proof of Theorem 1.6 (i) by “\(n_i(n_i-a-1)\ge a(a+2)/4\)”, Theorem 1.6 (ii) can be proved with the same discussion as Theorem 1.6 (i). \(\square \)