Abstract
In this paper, a graph is assigned to any probability measure on the -algebra of Borel sets of a topological space. Using this construction, it is proved that given any number (finite or infinite) there exists a nonregular graph such that its clique, chromatic, and dominating number equals .
1. Introduction
The distance between two vertices in a graph is the number of edges in a shortest path connecting them. Diameter of a graph is the longest path between two vertices of the graph.
A clique is a subset of vertices of an undirected graph, such that its induced subgraph is complete; that is, every two distinct vertices in the clique are adjacent. Clique number of a graph is the number of vertices of maximum clique in the graph and denoted by . The chromatic polynomial counts the number of ways a graph can be colored using no more than a given number of colors and denoted by .
A set of vertices in a graph is a dominating set if every vertex is an element of or adjacent to an element of . The domination number of a graph is the minimum cardinality of a dominating set of .
An independent set or stable set is a set of vertices in a graph, with no two of which being adjacent. A maximum independent set is an independent set of largest possible size for a given graph . This size is called the independence number of and denoted .
Lots of graphs are constructed from algebraic objects such as semigroups ([1]), groups ([2, 3]), and rings ([4, 5]). Recently some graphs have been constructed from Hilbert and topological spaces ([6, 7]) which have nice graph theoretical properties. This motivates us to construct a bigger category of graphs which has more properties. In this paper, we introduce a class of graphs generated by measures. Using this construction, we conclude that, given any number (finite or infinite), there exists a nonregular graph such that its clique, chromatic, and dominating number coincides and is equal to .
Given a measure space , a set is called an atom for , if has a positive measure and for every , either or is zero. It is easily obtained by definition that if is an atom for and , then is also an atom for . is an atomic measure iff every measurable set of positive measure contains an atom. We may say that is nonatomic if there are no atoms for . Therefore, in a nonatomic measure , every nonatomic measurable set of positive measure can split into two disjoint measurable sets, where both of them have positive measure. One can easily see that the zero measure is the only measure which is atomic as well as nonatomic measure.
Let be a measurable space and and be two measures on it. We say that is absolutely continuous with respect to , denoted by iff, for every measurable set , whenever .
We say that is singular with respect to and denoted by if, given any , there exists some , such that and . One can easily see that in this case It is easily seen that the following lemma holds.
Lemma 1. Two atomic measures are mutually singular if and only if the corresponding sets of atoms do not have a common element.
Johnson in [8] proved the following useful results about atomic and nonatomic measures.
Result 1. If is an atom for and , then either or is an atom for .
Result 2. For a given measure on a algebra , there are unique measures and , such that , , is atomic, and is nonatomic.
2. Definitions and Examples
Let be a topological space and be the -algebra of Borel subsets of . Let also be a probability measure on . Set In other words, is the collection of all absolutely continuous Borel measures with respect to . Define the following relation on : in other words, if and only if for any measurable set we have Clearly, is an equivalence relation on . Denoting the equivalence class of by , let For simplicity, we write for .
We define a graph with the set of vertices , and two vertices and are adjacent if and only if and are mutually singular. Clearly, the adjacency defined above is well-defined. The collection of all edges of is denoted by . Therefore, we may assign a graph to any probability Borel measure on .
Example 2. Let . If is the dirac measure concentrated on , then and ; so consists of a vertex with no edges.
Example 3. Let , , and . Let . Then where and are adjacent.
Example 4. Let be such that and be three distinct elements of . If then where Also, each is adjacent to for . is adjacent to if . No other vertices are adjacent.
The following example is indeed the generalization of the previous examples and presents the graph corresponding to a finitely atomic measure.
Example 5. Let be a finitely atomic measure with finite number of atoms where ’s are distinct points in and is such that . Then whereBy Lemma 1, two vertices and are adjacent if and only if for all and .
3. The Graphs of Finitely Atomic Measures
In the rest of the paper, by an atomic measure, we mean a finitely atomic measure with finite number of atoms. In this section, we study the graph corresponding to atomic measures and state some of its properties. The case of atomic measures is of special interest since the corresponding graph is finite and may be visualized and enumerated.
The following theorem satisfies for atomic measures.
Theorem 6. For an atomic measure one has the following.(1), where and ’s are disjoint.(2) for any .
Proof. Part (1) holds, since the support of each element of should be contained in .
To prove part (2), let and then . As mentioned in Example 5, a vertex is adjacent to if and only if . Hence should be of the form .
The number of such vertices iswhich completes the proof.
Theorem 7. For an atomic measure , one has the following: (1).(2).
Proof. () Since is a clique set for , it follows that .
Since all elements of are pairwise adjacent then . To prove the equality, we present a coloring of with exactly colors. For , let be a color assigned to the vertex . Now, for , assign the color , where Clearly, there are exactly colors used for the previous coloring. Also, no adjacent vertices have the same colors, since the two vertices and are adjacent if and only if ; therefore so . It completes the proof of .
Finally, since is a dominating set for then .
Now suppose , is a dominating set for . Set the set of atoms in which their coefficients in are nonzero. is a poset with the inclusion relation. By changing the indices we can write , where any is a minimal set for every and is not minimal set for other indices. By the axiom of choice, for every , one can choose an atom , where (module ) and for every one can choose .
Now define . It is easy to verify that is neither adjacent nor equal to any element of , which is contradiction to the assumption that is a dominating set.
() Since each vertex of is adjacent to at least an element of the set and all vertices in are adjacent, then for all . On the other hand, set and . Clearly, the path is of minimum length; therefore ; consequently
The following theorem determines the automorphism group of , where is an atomic measure.
Theorem 8. For any atomic measure one haswhere is the group of permutations on .
Proof. Let . Define the mapping by Clearly is a bijection; so . Now, define by It is easily seen that is an injective group homomorphism. To complete the proof, it is enough to show that is surjective.
To do this, let . For any given , in light of Theorem 6, since the number of adjacent measures should remain the same by we have for some .
Let be defined by . So We show that . Let be any given element of . Let also We should show that or equivalently Let . Since is surjective, there exists a natural number such that . Note that, since then the vertices and are adjacent. Since , the vertices and are adjacent which means that the vertices and are adjacent; therefore .
The previous discussion results in but the two sets have the same cardinality, namely, ; therefore which completes the proof.
Theorem 9. Let be an atomic measure; then .
Proof. For and , two vertices and are not adjacent, since is a common index for both and . Therefore, the set is an independent set. By elementary combinatorial principles of counting we have Thus we have an independent set with cardinality
Observe that , since any vertex can be written as , where is a nonempty proper subset of .
Now suppose that is an independent set of . For every , define , where . Obviously for every , is adjacent to ; hence .
For distinct ’s, relative ’s are distinct; thus we will have . Finally, we can say
4. General Measures
In the previous section, we discussed the graph assigned to atomic measures. In this section we study the general case.
Theorem 10. For a given measure , if and are two vertices of , one has the following. (1) iff and are mutually singular.(2) iff and are not mutually singular, but for some measurable set with positive measure, one has .(3) iff and are not mutually singular, but for no measurable set with positive measure, one has .
Proof. () It is done by definition.
() If the conditions hold, then by first part . Define the measure by for every in . Obviously . Since , then . By the definition of , one can obtain . We also have That is, and . Thus is a path and ; therefore we have
Now suppose that ; thus for some , we may have and . Therefore, for some nonzero -measurable sets and , we will have thus we have If then ; thus , a contradiction. Thus ; hence . It is enough to take .
() If , none of the conditions of part (1) and (2) hold; thus the condition mentioned in this part will occur.
Now if the conditions of part (3) hold, by definition of and , since does not contain (indeed ), for some nonzero -measurable sets and , we have where . Now we can define two measures and by Thus we will have , which shows .
By parts (1) and (2), we can deduce that .
By Result 2, every measure can be decomposed into two mutually singular measures: one is atomic and the other is nonatomic. We call the nonatomic measure the nonatomic part of and the other one the atomic part.
Theorem 11. If the nonatomic part of is nonzero, then
Proof. In light of Result 2, suppose that , where is the atomic and is the nonatomic part of .
Since , for some , we will have and . So, . Set . Since is nonatomic, we can split into two disjoint measurable sets with positive -measure. Let be one of them. By continuing the previous process, there exists a sequence of nonzero -measurable sets, and consequently a sequence with positive -measure and therefore positive -measure.
Now define for all . It is easily seen that ’s are mutually singular and for all , that is, is a clique of .
Corollary 12. If the nonatomic part of is nonzero then
In the next theorem, we consider the domination number as well.
Theorem 13. If the nonatomic part of is nonzero then the domination number of cannot be finite.
Proof. Let , where is the atomic and is the nonatomic part of . Suppose that is a dominating set of .
Let , and . is a poset with inclusion relation. Set as the minimal element of . Suppose that and are two subsets of such that and consequently, .
For every , we can choose a nonzero measurable set with and . Let .
Now define the measure by which is a measure different from all of ’s and adjacent to none of them. A contradiction with the hypothesis is that is a dominating set. Therefore, every dominating set is infinite.
Theorem 14. If the nonatomic part of is nonzero, then .
Proof. By the hypothesis of the theorem, there is some measurable set of positive measure which does not contain any atom and hence there is a sequence of measurable sets such that Now define the sequence of measures by for . By definition of ’s we have . Now suppose that for some we have . Thus for , there is some , where and . But by definition which is a contradiction. Therefore, we conclude that the set is an independent set of , which is not finite and the theorem is proved.
Applying the material presented in this paper, we will have the following corollary.
Corollary 15. Given any number (finite or infinite), there exists a nonregular graph such that
Competing Interests
The authors declare that they have no competing interests.