## 1 Introduction

Unless stated otherwise, we use small letters such as to denote non-negative integers or elements of sets, and capital letters such as to denote sets or graphs. The set of positive integers is denoted by . For , denotes the set (that is, ). We take to be the empty set . Arbitrary sets are taken to be finite. For a set , denotes the set of -element subsets of (that is, ). We may represent a -element set by .

For standard terminology in graph theory, we refer the reader to [11]. Most of the notation and terminology used here is defined in [1], which lays the foundation for the work presented here.

Every graph is taken to be *simple*, that is, is a pair such that and (the vertex set and the edge set of ) are sets that satisfy . We call an *-vertex graph* if . We call an *-edge graph* if . For a vertex of , denotes the set of neighbours of in , denotes the closed neighbourhood of , and denotes the degree of . For a subset of , denotes the closed neighbourhood of , denotes the subgraph of induced by (that is, ), and denotes the graph obtained by deleting the vertices in from (that is, ). Where no confusion arises, the subscript may be omitted from any notation that uses it; for example, may be abbreviated to . If is a subgraph of , then we say that * contains *. A *component of * is a maximal connected subgraph of (that is, no other connected subgraph of contains it). Clearly, the components of are pairwise vertex-disjoint (that is, no two have a common vertex), and their union is . If is a copy of , then we write .

For , the graphs and are denoted by and , respectively. For , denotes the graph (). A copy of is called an *-clique* or a *complete graph*. A copy of is called an *-path* or simply a *path*. A copy of is called an *-cycle* or simply a *cycle*. We call a -cycle a *triangle*. Note that is the triangle .

If is a set of graphs and is a copy of a graph in , then we call an *-graph*. If is a graph and such that contains no -graph, then is called an *-isolating set of *. The size of a smallest -isolating set of is denoted by and called the *-isolation number of *. We abbreviate to .

The study of isolating sets was initiated recently by Caro and Hansberg [3]. It generalizes the study of the classical domination problem [4, 5, 6, 7, 8, 9] naturally. Indeed, is a *dominating set of * (that is, ) if and only if is a -isolating set of , so the *domination number* is the -isolation number. One of the earliest domination results is the upper bound of Ore [10] on the domination number of any connected -vertex graph (see [6]). While deleting the closed neighbourhood of a dominating set produces the graph with no vertices, deleting the closed neighbourhood of a -isolating set produces a graph with no edges. A major contribution of Caro and Hansberg in [3] is a sharp upper bound for this case. They proved that if is a connected -vertex graph, then unless or . Fenech, Kaemawichanurat, and the present author [2] generalized these bounds by showing that for any , unless or and is a -cycle. This sharp bound settled a problem of Caro and Hansberg [3].

Let denote the set of cycles. The author [1] obtained a sharp upper bound on , and consequently settled another problem of Caro and Hansberg [3]. Before stating the result, we recall the explicit construction used for an extremal case.

###### Construction 1 ([1])

*For any and any -vertex graph , we construct a connected -vertex graph as follows. If , then let . If , then let , let (so ), let be copies of such that are pairwise vertex-disjoint, and let be the graph with and .*

###### Theorem 1 ([1])

If is a connected -vertex graph that is not a triangle, then

Moreover, equality holds if .

For , let denote the set of connected graphs that have at least edges. For a graph , we may abbreviate to . For , as every -isolating set of is an -isolating set of . In this paper, we introduce the problem of determining a sharp upper bound on similar to the bounds mentioned above. The above-mentioned result of Caro and Hansberg [3] solves the problem for . We solve the problem for and for .

###### Theorem 2 ([3])

If is a connected -vertex graph that is neither a -clique nor a -cycle, then

The bound in Theorem 2 is attained if .

For , let denote the star . Let denote the graph . Let denote the graph . Let .

If such that , then is called a *leaf of *.
Let denote the set of leaves of . Let denote .

If , then let . If , then let be the graph with and . Thus, if , then is the connected -vertex graph obtained from the union of by joining (a vertex of ) to the vertex of of degree for each , and joining to each .

In Section 2, we prove the following result.

###### Theorem 3

If is a connected -vertex graph, is not an -graph, and , then

Moreover, equality holds if .

It is easy to check that the inequality in Theorem 3 does not hold if is an -graph. Note that deleting the closed neighbourhood of an -isolating set of produces a graph such that no two edges have a common vertex (and hence each component is a copy of or of ).

In Section 3, we prove the next result.

###### Theorem 4

If is a connected -vertex graph that is neither a triangle nor a -cycle, then

Moreover, equality holds if .

## 2 Proof of Theorem 3

In this section, we prove Theorem 3.

We start with two lemmas from [1] that will be used repeatedly. We provide their proofs for completeness.

###### Lemma 1

If is a graph, is a set of graphs, , and , then

Proof. Let be an -isolating set of of size . Clearly, for each -graph that is a subgraph of and not a subgraph of . Thus, is an -isolating set of . The result follows.

For a graph , let denote the set of components of .

###### Lemma 2

If is a graph and is a set of graphs, then

Proof. For each , let be a smallest -isolating set of . Then, is an -isolating set of , so . Let be a smallest -isolating set of . For each , is an -isolating set of . We have . The result follows.

For a graph and a subgraph of , let , , and

We may abbreviate to .

###### Lemma 3

(a) If are pairwise vertex-disjoint subgraphs of such that , then

(b) If is a connected subgraph of with , then .

Proof. (a) Trivially, and

(b) Suppose . Since and is connected, for some . Since , . Thus, , and hence .

If , then is an -isolating set of , and is an -isolating set of . Consequently,

if , , and , then . | (1) |

The next lemma settles Theorem 3 for .

###### Lemma 4

If is a connected -vertex graph, , and is not an -graph, then .

Proof. Since is an integer, it suffices to prove that . If , then . Since is not an -graph, . Suppose . Let be a vertex of of largest degree. Since is connected, and for each . If , then is a path or a cycle, so by (1). Suppose . Let . Let . Then, .

Suppose . Then, each vertex in is a neighbour of . Since , for some . Thus, . Since is an -isolating set of , .

Suppose . If no component of has more than vertices, then is an -isolating set of , so . Suppose that a component of has at least vertices. Since and , we have , , , , and either or . Since is connected, for some and some . Since , for some such that . Since is connected and , for some . If , then, since is an -isolating set of , . Suppose . If we show that has an -isolating set of size , then . Let . Then, .

Suppose . Then, . We may assume that . If for some , then we take . Suppose for each . Then, . If or , then we take . Suppose and . Let be the vertex in . If , then we take . Suppose . If we assume that , then we obtain , a contradiction. Thus, . If we assume that , then we obtain , a contradiction. Thus, . Consequently, is the unique neighbour of the vertex in . If , then we take . If , then we take .

Now suppose . Then, and . Let be the vertex in . If for some , then we take . Suppose for each . Since , we have and . If is the vertex in and is or , then we take . Now suppose and . Let such that , let such that , and let be the vertex in . If we assume that and , then we obtain , a contradiction. Thus, or . We take .

###### Lemma 5

If is an -graph, then

Proof. Trivially, , , , , , and .

Proof of Theorem 3. Let . If , then . If , then is an -graph. Suppose . Since is a dominating set of , . If is an -isolating set of of size , then for each , as does not contain the copy of . Thus, . Let . Then, . Since , . Since ,

We have shown that the bound in the theorem is attained if .

We now prove the inequality in the theorem, using induction on . Since is an integer, it suffices to prove that . If , then the result is given by Lemma 4. Suppose . Let . Since is connected, . Let such that . If , then is a path or a cycle, so by (1). Suppose . Then, . If , then is an -isolating set of , so . Suppose . Let and . Then, and . Let be the set of components of . Let

We have . Since is connected and , has a neighbour that has a neighbour in , so . We have

(2) |

Now suppose . If with , then . If such that and is not an -graph, then by the induction hypothesis and Lemma 3(b). Thus, each member of is an -graph. Let , , , , , and . If , , and for some and some in the -element set , then , so . Similarly, if , is a copy of or of , is the leaf of , and for some , then , so . Therefore,

(3) |

Note that

if and , then . | (4) |

For any and any such that for some , we say that is *linked to * and that is *linked to *. Since is connected, each member of is linked to at least one member of . For each , let , , and . For each , let be an -isolating set of of size .

For any and any , let . By (3), for some . If , then, since , by (4). Thus, if , then contains a -cycle such that , and we take . If , then we take . If , then, since , . Clearly, is an -isolating set of , and

(5) |

Since , . Since and , .

*Case 1: for some .* For each , let such that is linked to . Let . Note that . Let

We have , for each , and for each , so is an -isolating set of . Let

Then,

(6) |

as and . We have

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