h - Relation and Its Associated Hyperstructures. M K Sen Department of Pure Mathematics, University of Calcutta, 35, Ballygunge Circucar Road, Kolkata-700019, India.
e-mail:
[email protected] Utpal Dasgupta Department of Mathematics, Sree Chaitanya College, Habra-Prafullanagar, 24 Parganas(North), West Bengal, Pin Code : 743268, India.
Abstract An h-relation on a non-empty set S is in fact a relation from S to the power set P (S) of S. Here, in this paper, it is shown that the semigroup of binary relations on a set S can be embedded as a subsemigroup into the semigroup of h-relations on S(with restpect to a suitably defined operation). We have introduced and studied here two types of reflexivities, symmetries and transitivities of h-relations. The conditions imposed on an h-relation R, under which the hyperstructure HR ( induced by the h-relation R on a non-empty set H) can necessarily and sufficiently be a semihypergroup or a hypergroup, have been obtained. We have established further the necessary and sufficient conditions for which a given semihypergroup (H, ◦) can be induced by a specified h-relation R. 2000 Mathematics Subject Classification: 20N20 Keywords and phrases: h-relations, Hypergroupoids, Semihypergroup.
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Preliminaries
The hyperstructure theory was surfaced up in 1934 when Marty defined hypergroups ([1]), started up analysing their properties and applied them to groups, rational algebraic functions. Since then many researchers have studied in this field and developed it, for example see [4]. In this paper we have studied hyper relation. A hyper relation ( h-relation ) on a non-empty set S is in fact a relation from S to the power set P (S) of S. In [7], Dasgupta introduced the notion of h-relation and studied some of its properties, in the name of p-relation. Definition 1.1. An h-relation on a non-empty set S is a subset of S × P (S).
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Remark 1.1. (a) Any mapping f from a non-empty o set S into itself induces a binary n relation of S, such as ρ = (a, b) ∈ S × S : f (a) = f (b) . The inverse f −1 of f may not −1 be a mapping from S to S, but n as a natural phenomenon, weoobserve that f induces an h-relation on S such as R = (a, A) ∈ S × P (S) : A = f −1 (a) .
(b) For a non-empty set S, 1).
i=1
On the other hand, for B, C ∈ P ∗ (S), if B ⊆ CRn , then of course | C |≤ n. Thus clearly for any A, B, C, ∈ P ∗ (S), A ⊆ BRn , B ⊆ CRn ⇒ A ⊆ CRn whence Rn (n > 1) is transitive of type 2. This is to be noted that Rn (n ∈ N) is a type 1 (and hence type 2) reflexive and type 1 symmetric h-relation, but Rn (n ∈ N) is not a symmetry of type 2. Observe that R1 ⊂ R2 ⊂ R3 ⊂ · · · ⊂ Rn ⊂ Rn+1 · · · is a non-terminating increasing sequence of h-relations ∞ [ on S (since S is infinite) and Ri = R is the h-relation as defined in the example 2.7(1), i=1
which is reflexive (of both the types) symmetric (of type 1 but not type 2) and transitive (of both the types). Proposition 2.14.
A type 2 reflexive h-relation R is type 1 reflexive if it is a type 1
symmetric h-relation. Proof.
Let R be a type 2 reflexive as well as a type 1 symmetric h-relation on S. Then
for any x ∈ S, there exists A ∈ P (S) such that x ∈ A and (x, A) ∈ R i.e., x ∈ AR . So, x ∈ {x} ⊆ AR and | {x} |≤| A | (here | A |≥ 1, since x ∈ A). Thus, (a, {x}) ∈ R for all a ∈ A (since R is a type 1 symmetry). But x ∈ A ; so (x, {x}) ∈ R for all x ∈ S, whence R is reflexive of type 1. Corollary 2.15. A type 2 reflexive h-relation R is type 1 reflexive if it is a type 2 symmetric h-relation. Proof. It follows from proposition 2.6 and 2.14. Proposition 2.16. A type 2 transitive h-relation R is a type 1 transitive h-relation if it is a type 2 symmetric h-relation. Proof.
Let R be a type 2 transitive and type 2 symmetric h-relation on a non-empty
set S. Suppose (x, A) ∈ R and (a, Ba ) ∈ R for each a ∈ A. Then, A ⊆ {x}R (∵ R is a type 2 symmetry); i.e., {a} ⊆ {x}R for all a ∈ A. Hence {x} ⊆ {a}R for all a ∈ A. So, {x} ⊆ {a}R , {a} ⊆ (Ba )R ⇒ {x} ⊆ (Ba )R for all a ∈ A (∵ R is a type 2 transitive [ [ h-relation) whence Ba ⊆ {x}R for all a ∈ A . Thus, Ba ⊆ {x}R implying {x} ⊆ ( Ba )R a a∈A [ i.e., x, Ba ∈ R. So, R is a type 1 transitive h-relation. a
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3
Inverses and Invertibility
Definition 3.1.
An h-relation R on a non-empty set S is said to be perfect if for any
A ∈ P (S), AR = φ or else | AR |≥| A |. Examples 3.2. n [
n 1. For a positive integer n, let S = (i − 1, i), R = (x, A) : x ∈ (i − 1, i) and A is a i=1 o countably infinite subset of (i − 1, i) for some i = 1, 2, · · · , n is a perfect h-relation on S. n o 2. Let S = [0, 1]. Then R = (x, A) : x ∈ (0, 1) and A is an open subset of (0, 1) is a perfect h-relation on S. n 3. Let S be an open interval and R = (x, A) : x ∈ S and A is a δ- neighbourhood o (δ > o) of x ∈ S . Then R is not a perfect h-relation on S. In fact, if S = (a, b), then a+b a+b | SR |= 1 0). Thus, 2 2 n o . SR = a+b 2 Proposition 3.3. The number of perfect h-relations on a finite set S having n(∈ N) elements n h n i n Y X n (j ) n(n+1) is 2 . 1+ . i j=2 i=j Proof. Let S be a (non-empty) finite set with | S |= n. An h-relation R on S is perfect if any A ∈ P (S) satisfies the condition(a) | AR |= 0 or else | AR |≥| A |. Now, φ (empty subset of S) satisfies the condition(a) in 2n ways. A singleton A(⊆ S) satisfies condition (a) in 2n ways. There are n number of singleton sets in P (S). So, any number of singleton sets, each satisfying condition(a) can be taken in (2n )n = 2n ways. Any n i h X n set A (with | A |= k > 1) can satisfy condition(a) in 1 + ways. There are nk i i=k number of sets A (with | A |= k > 1). So, any number of sets each having k elements n i n h X n (k ) and satisfying condition(a) can be taken in 1 + ways. Thus, any number of i i=k n h n i n Y X n (j ) n n2 ways. sets in P (S), each satisfying condition(a) can be taken in 2 .2 . 1+ i j=2 i=j n h n i n Y X n (j ) n(n+1) So, there are 2 . 1+ number of perfect h-relations on S (with | S |= n). i j=2 i=j 2
Definition n 3.4. The type 1 inverse of an h-relation R on S is an h-relation o −1 R = (x, A) ∈ S × P ∗ (S) : ∃X ∈ P (S) such that x ∈ X, A ⊆ XR and | A |≤| X | . Proposition 3.5. An h-relation R is perfect and symmetric of type 1 if and only if R = R−1 . 7
Proof.
Let R be a perfect and type 1 symmetric h-relation on S. Suppose (a, X) ∈ R.
Then X 6= φ. Since R is perfect and a ∈ XR (i.e., XR 6= φ), so | XR |≥| X |. Now, we have A ⊆ XR such that a ∈ A and | A |=| X |. Then (x, A) ∈ R for all x ∈ X (since R is symmetric of type 1), whence X ⊆ AR . Then, by definition of type 1 inverse of R, (a, X) ∈ R−1 (since | X |=| A | and a ∈ A). So, R ⊆ R−1 . Conversely, let (a, X) ∈ R−1 . Then X 6= φ and ∃ A ∈ P (S) such that a ∈ A, X ⊆ AR and | X |≤| A |. Now, X ⊆ AR ⇒ (x, A) ∈ R for all x ∈ X ⇒ for any a0 ∈ A, (a0 , Y ) ∈ R whenever x ∈ Y ⊆ AR and | Y |≤| A | ( ∵ R is a symmetric of type) Since, x ∈ X ⊆ AR , | X |≤| A | and a ∈ A, so we have (a, X) ∈ R. Hence, R−1 ⊆ R whence R = R−1 . Now, suppose that R = R−1 and (a, X) ∈ R, x ∈ X. Also, suppose that A ∈ P (S) be such that a ∈ A ⊆ XR and | A |≤| X |. Clearly then (x, A) ∈ R−1 and hence (x, A) ∈ R [∵ R = R−1 ]. So, R is a symmetry of type 1. Let A ∈ P (S). If A = φ, then AR = φ (∵ R is a symmetry of type 1) and thus | AR |=| A |. Suppose A 6= φ and AR 6= φ. Then, ∃ x ∈ S such that (x, A) ∈ R. So, (x, A) ∈ R−1 (∵ R = R−1 ). Then, ∃X ∈ P (S) such that x ∈ X, A ⊆ XR and | A |≤| X | (by definition of R−1 ). Now, A ⊆ XR ⇒ (a, X) ∈ R for all a ∈ A. Since, a ∈ A ⊆ XR and | A |≤| X |, we have (x, A) ∈ R for all x ∈ X whence X ⊆ AR . Thus | A |≤| X |≤| AR | implying that R is perfect. Examples 3.6. n 1. Let S = { n+1 : nn∈ N}. On S, ano h-relation R is defined as follows n : o n k ( n+1 , A) ∈ R ⇔ A ⊆ k+1 / k ∈ I2n+1 where for any m ∈ N, Im = 1, 2, · · · , m . Then,
R is a perfect h-relation but not a symmetry of type 1 (and hence not a symmetry of type 2). In fact, for any infinite subset A of S, AR = φ. When A is a finite subset n of S; there i /i ∈ Ik }. exists mi ∈ N(i = 1, 2, · · · , k) such that m1 < m2 < · · · < mk and A = mmi +1 n o k m Then, A ⊆ k+1 /k ∈ I2m+1 for any m ≥ mk . So clearly, ( m+1 , A) ∈ R for any m ≥ mk n o m i.e., B = m+1 /m ≥ mk ⊆ AR i.e., | B |≤| AR |. But, B is an infinite set, whereas 4 | A |= n k(∈ N).oHence, n | A |
