On Some New Varieties of Binary Relation
Jayanta Biswas
Assistant Professor of Mathematics, Barasat govt. College,
Barasat, North 24 Parganas, West Bengal, India, Pin- 700124
E-mail – [email protected]
Received: 28th November 2015, Revised: 28th March 2016, Accepted: 29th March 2016.
Abstract
Here we introduce some new types of binary relations on a non-empty set, named, finite ordered reflexive relation,
k-reflexive relation, type-1 and type-2 k-symmetric relation, weak k-symmetric relation which are some kind of generalization of
reflexivity and symmetricity of binary relation. Also we study their properties.
Keywords : k-reflexive relation, type-1 k-symmetric relation, type-2 k-symmetric relation, weak k-symmetric relation .
1. Introduction : We know many types of binary relations [1- 4 ] and their properties. Viewing these
properties, I am motivated to introduce here some new types of binary relations. Throughout this paper, we shall
mean “non-empty set” by “set “, and “ binary relation” by “relation” until otherwise stated. We shall denote the
equality relation on a set S by , i.e.,
. Firstly we remind some related definitions of binary
relations.
Definition ( 1.1) A relation on a set S is said to be
(i) reflexive if
.
(ii) right-shift reflexive if
.
(iii) left-shift reflexive if
.
(iv) Co-reflexive if
, i.e., if
.
(v) quasi-reflexive if
.
(vi) irreflexive if
.
(vii) symmetric if
.
(viii) asymmetric if
.
(ix) transitive if
.
(x) anti-transitive if
.
(xi) left total if for every
such that
.
(xii) injective if
, where for
(xiii) [3] injective of type-2 if
, where for
(xiv) surjective if for every
such that
.
(xv) right-Euclidean if
.
(xvi) left-Euclidean if
.
(xvii) right unique or functional if
.
.
.
Note (1.1) It is clear that a right-Euclidean ( left-Euclidean) relation on a set S is right-shift reflexive ( leftshift reflexive ).
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MAIN RESULTS
2. k-reflexive relation
Definition(2.1) A binary relation on a set S is said to be of finite ordered reflexive relation if there exists
, such that
is reflexive; otherwise is called an infinite ordered reflexive relation.
If be of finite ordered reflexive relation, then its reflexive order is defined to be the smallest positive integer k
such that
is reflexive, and is denoted by
. If
, a finite positive integer, then is called a
k-reflexive relation.
Thus is a k-reflexive relation on a set S iff k is the smallest positive integer such that for every
for which
, where
for k = 1.
Example (2.1) Let = {1,2,3,4} and cons er the relations
and
on S. is not reflexive but
is reflexive. Thus
is 2-reflexive.
Again is not reflexive,
is not reflexive and
and so is not of
finite ordered reflexive relation.
Again let A ={1,2,3} and consider the relation
which is not reflexive.
Now,
is not reflexive.
and hence is not finite ordered reflexive relation on A.
Note (2.1) Clearly every finite ordered reflexive relation on a set is left total and surjective. Also it is clear
that every reflexive relation on a set is 1-reflexive.
Theorem (2.1) If
Proof : Trivial.
be a k-reflexive relation on a set S, then
Theorem (2.2) If be a k-reflexive relation on a set S, then
Proof : Since
is reflexive, hence
is reflexive, i.e.,
Therefore
is a finite ordered reflexive relation. Let
i.e.,
is reflexive and so
is reflexive. Therefore
Theorem (2.3) If
, then
Proof : Trivial.
is a reflexive relation on S.
is a k-reflexive relation on S.
is reflexive. ( since
).
. Then
. Now
is reflexive,
. Therefore
and so
is k-reflexive.
be two relations on a set S, such that is k-reflexive and
is finite ordered reflexive relation on S.
Theorem(2.4) Let
be two relations on a set S, such that is k-reflexive and
ordered reflexive relations on S.
Proof :
Remaining part of the proof is trivial.
is m-reflexive and
. Then
is a finite
Corollary(2.4) If one of two relations on a set is finite ordered reflexive, then their union is a finite ordered
reflexive relation.
Problem (2.1) Let
be two relations on a set S, such that is k-reflexive
finite ordered reflexive relations on S.
Solution : Let = {1,2,3,4} and cons er the relations
on S. is not reflexive. Now
is not reflexive.
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. Then
and
may not be a
But
is reflexive. Thus is 3-reflexive.
Again is not reflexive, Now
is not reflexive and
not of finite ordered reflexive relation, though
.
and so
is
Problem(2.2) Intersection of two finite ordered reflexive relations on a set may not be of finite ordered
reflexive relation.
Solution : Let = {1,2,3,4} and cons er the relations
and
on S . Then
. is not reflexive.
Now
is not reflexive. But
is reflexive.
Thus is 3-reflexive.
Again is not reflexive, Now
is not reflexive.
But
is reflexive. Therefore is 3-reflexive.
Now
is not reflexive.
is not reflexive and
and so
is not of finite ordered reflexive relation.
Note (2.2) From Problem(2.2), we see that intersection of two finite ordered reflexive relations on a set may
not be finite ordered reflexive relation. But intersection of two reflexive relations on a set is a reflexive relation.
Again, if intersection of two relations be finite ordered reflexive then both the relations are finite ordered
reflexive ( from problem(2.2) and theorem(2.4) ).
Also it can be easily proved that, if two relations and on a set S be k-reflexive, for some
and
, then
is finite ordered reflexive relation.
Problem(2.3) (i) Complement of a finite ordered reflexive relation on a set may not be a finite ordered
reflexive relation.
(ii) Difference of two finite ordered reflexive relations on a set may not be a finite ordered reflexive relation.
(iii) Symmetric difference of two finite ordered reflexive relations on a set may not be a finite ordered reflexive
relation.
Theorem(2.5) Let
and
is non-empty non-reflexive for
be a relation on a set S such that is not a proper co-reflexive relation and
Then
is -reflexive, where
and
for all relations on S for which
is a -reflexive
are strictly increasing (
)
relation, provided
Proof : Since
Therefore
Therefore
Let
and so
, hence
. Let
be arbitrary. If
. If
, then
and so
.
so that
is reflexive.
is finite ordered reflexive relation.
. Then
such that
. We claim that
If possible, let
. Then
. Therefore,
are strictly increasing for
( since
Therefore,
Therefore,
(since
Therefore,
, then
.
and
).
), which contradicts the fact that
.
is -reflexive.
Let be any binary relation on S such that
in this case and hence
.
is -reflexive. If
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, then
is -reflexive and so
,
Let
. If
Define
, then there is nothing to prove. Let
by,
Since for any
with
.
.
,
and
are strictly increasing for
. Again
. Therefore, if
. Hence the function f is well-defined and injective so that
, then
, hence
, for some
. Hence the proof.
Theorem (2.6) Let be a -reflexive relation on a set S, such that is injective.
(i) Then is bijective and
.
(ii) If be a -reflexive relation on S and
be left total, then
is -reflexive.
Proof : (i) Since is -reflexive, hence is surjective ( by Note(2.1) ). Again is injective. Therefore,
bijective. Since is injective, hence
is injective. Again
is reflexive. Hence
.
is
(ii) Since
are -reflexive and is injective, hence
( by (i) of this theorem ) and
is reflexive
and so
. Now
…………………..……………. (1)
Again, since
be left total, hence
is left total and so
…..…..…(2) ( by (1) ).
Therefore
is finite ordered reflexive relation on S. Let
. Then
( by (2) ).
Let
be arbitrary. Then
. Therefore, there exists
such that
, This implies that
. But
was arbitrary.
Therefore,
is reflexive. Therefore,
. Therefore
so that
is -reflexive.
Note (2.3) The result of theorem(2.6) does not hold for a type-2 injective relation. For example, let
. Consider the relation
. Clearly is a type-2
injective. Now is not reflexive.
is
reflexive. Therefore is 2-reflexive. But
.
Theorem (2.7) Let
reflexive.
Proof : Trivial.
be a -reflexive relation on a set S, such that
is right-shift reflexive . Then
Theorem (2.8) Let
reflexive.
Proof : Trivial.
be a -reflexive relation on a set S, such that
is left-shift reflexive. Then
Note (2.4) From Note(1.1), theorem(2.7) and theorem(2.8), it is clear that, if
right-Euclidean ( left-Euclidean) relation on a set S, then is reflexive.
is
is
be a -reflexive as well as
Theorem (2.9) Let
Proof : Trivial.
be a -reflexive relation on a set S, such that
is quasi-reflexive. Then
Theorem(2.10) Let
be a -reflexive relation on a set S, such that
such that
, for some
is irreflexive. Then for every
.
Proof : Trivial.
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is reflexive.
Theorem (2.11) Let
Proof : Trivial.
be a -reflexive relation on a set S, such that
is transitive. Then
Theorem (2.12) Let
mapping with
.
Proof : Trivial.
be a -reflexive relation on a set S, such that
is functional. Then
is reflexive.
is a bijective
Theorem (2.13) Let be a -reflexive relation on a set S, such that is injective. Then
is a bijective
mapping with
.
Proof : Since is -reflexive, it is surjective and so
is left total. Again is injective. Hence
is a
mapping. Again since is -reflexive,
is -reflexive ( by theorem(2.2) ). Therefore
and so
is bijective. Therefore
is a bijective mapping. Now,
.
Theorem (2.14) Let
Proof : Trivial.
be a -reflexive relation on a set S. Then
.
Theorem(2.15) A relation on a set S is finite ordered reflexive relation iff there exists
having the
property that k is the smallest positive integer such that for every
with
and
such that
, where
for
= 1.
Proof : Let
is finite ordered reflexive relation on a set S. Let
. From this fact it can be said that, for every
such that
. Then
, where
is reflexive and so
with
and
for
= 1.
Consider
the set of all such ’s. Since
, hence A has the maximum element k (say). And
it is clear that k is the smallest positive integer such that for every
with
and
such that
, where
for
= 1. In this case
actually
.
Conversely, let there exists
every
with
having the property that k is the smallest positive integer such that for
and
such that
,
where
for
= 1.
Then clearly A the set of all such ’s, is a finite set. If m be the least common multiple of all the elements of
A, then, since for every
with
such that
and
divides m,
.
Therefore
is reflexive, i.e., is finite ordered reflexive relation on S.
Note (2.5) From the fact of the converse part of theorem (2.15) it can be easily established that
.
Example (2.2) : Consider the set
Now,
and the relation
,
,
on S.
,
.
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and
Here,
for which
Definition (2.2) A relation
integer such that for every
so that k = 3 is the smallest positive integer such that for every
. Again
on a set S is said to be -reflexively regular if
with
and
, where
for
= 1, i.e.
is the smallest positive
for which
and
and at least one
.
Definition (2.3) A relation
irreflexive for
on a set S is said to be -reflexively perfect
.
Example (2.3) : Consider the set
if
is
and the relation
.Then
Clearly
and
.
is 2-reflexively regular.
Again, let
Then
.
. Clearly
is 2-reflexively perfect.
Theorem (2.16) Let be a non-reflexive relation on a finite set S having at least three elements such that
are strictly increasing . Then
such that is -reflexively regular.
Proof : Trivial.
Theorem (2.17) Let
-reflexively regular relation and be a -reflexively regular relation on a set S.
If
, then
is a -reflexively regular relation on S.
Proof : Since be a -reflexively regular relation, hence
and hence,
implies that
is finite ordered reflexive relation ( by theorem(2.4) ) and clearly
.
Without loss of generality, let
so that
.
Since is
for every
be a
-reflexively regular relation on the set S, hence
is the smallest positive integer such that
with
for which
and
and at least one
. This shows that
is a -reflexively regular relation
on S.
Corollary(2.17)(a) Let
such that
, then
be a -reflexively regular relation on a set S and
is a m-reflexively regular relation on S.
where
is a relation on S
Corollary(2.17)(b) Union of two k-reflexively regular relations on a set S is k-reflexively regular.
Theorem (2.18) Let be a relation on a set S such that is irreflexive for
with
. Then is be -reflexively perfect iff for every
.
Proof : Trivial.
Corollary(2.18) For any set S having at least two elements,
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, for some
is 2-reflexively perfect relation
on S.
3. k-symmetric relation.
Definition (3.1) (a) A relation on a set S is said to be type-1 finite ordered symmetric relation if there
exists
such that
. In this case type-1 symmetric order of is
defined by the smallest positive integer k for which
.
The type-1 symmetric order of is denoted by
.
For a relation on a set S, if
, we call as “ type-1 k – symmetric relation on S”.
(b) A relation on a set S is said to be type-2 finite ordered symmetric relation if there exists
such
that
is symmetric. In this case type-2 symmetric order of is defined by the smallest positive integer k for
which
is symmetric. The type-2 symmetric order of is denoted by
.
For a relation on a set S, if
, we call as “ type-2 k – symmetric relation on S”.
Example (3.1) Let
Then
It can be easily verified that
Theorem (3.1) If and
symmetric relation on S.
Proof : Trivial.
. Consider two relations and on S, given by
and
.
and
.
is a type-1 2-symmetric relation on S and is a type-2 2-symmetric relation on S.
be type-1 k – symmetric relations on a set S, then
Corollary (3.1) If be a type-1 k – symmetric relation on a set S and
type-1 k – symmetric, then
is type-1 finite ordered symmetric.
is a type-1 finite ordered
is a relation on S such that
is
Problem(3.1) Intersection of two type-1 finite ordered symmetric relations on a set may not be type-1 finite
ordered symmetric .
Solution : Let = {1,2,3,4} and cons er the relations
and
on S . Then
. is not type-1 1-symmetric.
Now
from which it is clear that is type-1 2-symmetric.
Again, is not type-1 1-symmetric. Now
from which it is clear that is
type-1 2-symmetric.
Now,
which is not type-1 1-symmetric. Now,
from
which it is clear that
is not type-1 finite ordered symmetric relation.
Note (3.1) It can be easily proved that, if two relations and on a set S be type-1 k-symmetric, for some
and
, then
is type-1 finite ordered symmetric relation.
Theorem (3.2) If
on S.
Proof : Let
be a type-1 k – symmetric relation on a set S then
be arbitrary. Then
and so
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is a type-1 k – symmetric relation
. This implies that,
Then
that
. Therefore
is a type-1 finite ordered symmetric relation on S. Let
.
. Let
be arbitrary. Then
and so
. This implies
. Therefore
. Therefore
. Therefore,
is a type-1 k – symmetric relation on S.
Theorem (3.3) If
be type-1 k – symmetric relations on a set S such that
type-1 finite ordered symmetric relation on S.
Proof : Let
be arbitrary. Then there exists
such that
This implies that
and
. Therefore
Hence the proof.
Theorem (3.4) If
Proof : Trivial.
then
and
(
be a transitive and type-1 k – symmetric relation on a set S, then
is a
.
).
is symmetric.
Theorem (3.5) If be a type-1 k – symmetric relation on a set S, then
; and in addition, if S be a
finite set then there exists
with
such that
.
Proof : Let
be arbitrary. Then
. Therefore, there exists
such that
. This implies that
.
Therefore
so that
.
Now,
implies that
………………. (3)
If
with
, then from (3) it is clear that
is an infinite set and hence S is an
infinite set, in this case. Hence the proof.
Theorem (3.6) If
reflexive.
Proof : Trivial.
be a right shift-reflexive and type-1 k – symmetric relation on a set S, then
is left shift
Theorem (3.7) If be a left total ( or surjective) and type-1 k – symmetric relation on a set S, then
surjective ( or left total). Also is finite ordered reflexive.
Proof : Trivial.
Theorem (3.8) If
Proof : Let
Therefore
be an asymmetric and type-1 k – symmetric relation on a set S, then
be arbitrary. Then
. Therefore, there exists
. This implies that
so that Therefore
.
is
.
such that
Theorem (3.9) If
be type-2 k – symmetric relation on a set S, then
is type-2 finite ordered
symmetric relation on S, provided
.
Proof : Since
, hence
……….…………….. (4)
Let
be arbitrary. Then
( by (4) ) and so there exists
such that
. Now
and
are symmetric relations and so
. This implies
that
(by (4)). Therefore
is type-2 finite ordered symmetric relation on S.
Problem(3.2) Union of two type-2 finite ordered symmetric relations on a set may not be type-2 finite
ordered symmetric .
Solution : Let = {1,2,3,4,5,6,7} and cons er the relations
and
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.
on S .
Then
and is not type-2 1-symmetric. Now
is clear that is type-2 2-symmetric. We see that
. Again since
Therefore
is not symmetric, so that
.
is type-2 1-symmetric
from which it
. Therefore
, hence
.
is not type-2 finite ordered symmetric relation.
Problem(3.3) Intersection of two type-2 finite ordered symmetric relations on a set may not be type-2 finite
ordered symmetric .
Solution : Let = {1,2,3,4} and cons er the relations
and
on S . Then
. is not type-2 1-symmetric. Now
from which it is clear that is type-2 2-symmetric.
Again, is not type-2 1-symmetric. Now
from which
it is clear that is type-2 2-symmetric. Clearly,
is not type-1 1-symmetric. Now,
from which it is clear that
is not type-2 finite ordered symmetric
relation.
Note (3.2) It can be easily proved that, if two relations and on a set S be type-2 k-symmetric, for some
and
(i)
, then
is type-2 finite ordered symmetric relation.
(ii)
, then
is type-2 finite ordered symmetric relation.
Theorem (3.10) If be a type-2 k – symmetric relation on a set S, then
is type-2 k – symmetric relation
on S.
Proof : Let
be arbitrary. Then
( since
). Since
is symmetric,
hence
so that
. This shows that
is type-2 finite ordered symmetric
relation on S and
is symmetric. Let
. Then
. Since inverse of inverse of a relation
on a set is the relation itself, by similar argument it can be proved that
. Therefore
.
This completes the proof.
Theorem (3.11) If
Proof : Trivial.
be an asymmetric and type-2 k – symmetric relation on a set S, then
.
Theorem (3.12) If be a transitive and type-2 k – symmetric relation on a set S, then
(i)
, where
is the symmetric part of .
(ii)
is right-shift as well as left-shift reflexive.
Proof : (i) Clearly
is the largest symmetric relation on S contained in . Since is type-2 k – symmetric
and transitive, hence
is a symmetric relation on S contained in . Therefore
,
(ii) Let
be arbitrary. Then
. Therefore, there exists
such that
. This shows that
(since
). Hence the result.
Theorem (3.13) If
(i)
be an anti-transitive and type-2 k – symmetric relation on a set S, then
.
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(ii)
(iii)
.
Proof : (i) Anti-transitivity of implies that
Let
be arbitrary. Then
. Therefore, there exists
such that
From (5) and (6), we have the result.
…………..(5)
………………………. (6)
(ii) & (iii) : Applying the result of (i) of this theorem, these results will come immediately.
Definition (3.2) (a) A relation on a set S is said to be weak finite ordered symmetric if there exists a
positive integer m such that for every
with
for which
. In this
case weak symmetric order of
with
is defined by the smallest positive integer k for which, for every
such that
. The weak symmetric order of
by
.
For a relation
on a set S, if
(b) A relation
for
on a set S is said to be perfect k – symmetric (
.
, we call
is denoted
as “ weak k – symmetric relation on S”.
) if
(c) A relation on a set S is said to be strongly finite ordered symmetric if there exists a positive integer m such
that for any
. In this case strong symmetric order of is
defined by the smallest positive integer such that for any
.
The strong symmetric order of is denoted by
. For a relation on a set S, if
, we call as “
strongly k – symmetric relation on S”.
Note (3.3) It is clear that every perfect k – symmetric (
) relation on a set is type-1 k – symmetric and
every strongly k – symmetric relation on a set is type-1 finite ordered symmetric.
Example (3.2) (a) Let
. Consider the relation
. Then
on S, given by
. It is easy to check that
is a
weak 2-symmetric.
(b) Let
Then
3-symmetric.
(c) Let
Then
. Consider the relation
,
. Consider the relation
on S, given by
.
. It is easy to check that is a perfect
on S, given by
.
,
,
,
. It is easy to check that
is a strongly
5-symmetric.
Theorem (3.14) If
Proof : Let
be a weak k – symmetric relation on a set S, then
be arbitrary. Then
and so there exists
. This implies that
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. Therefore
is weak k – symmetric.
with
such that
is weak finite ordered
symmetric relation. Let
. Then clearly
above part of the proof of this theorem, we shall get
. Now interchanging &
and m & k in the
. Therefore
. Hence the result.
Theorem (3.15) If
be a weak
– symmetric and
be a weak
– symmetric relation on a set S, then
is weak finite ordered symmetric relation on S.
Proof : Let
be arbitrary. If
then there exists
with
such that
; and if
then there exists
with
. Thus, in any case, there exists
that
. Therefore
such that
with
such
is weak finite ordered symmetric relation on S.
Theorem (3.16) If be a left total and weak k – symmetric relation on a set S, then is finite ordered
reflexive relation.
Proof : Let
be arbitrary. Left totality of implies that
, for some
. Therefore, there exists
with
such that
. Hence
.
But
. Therefore, by theorem(2.15),
is finite ordered reflexive relation.
Theorem (3.17) If be a right (or left) shift-reflexive and weak k – symmetric relation on a set S, then
type-1 k – symmetric.
Proof : Let
be arbitrary. Then
(or
)…………….………….. (7).
Also there exists
with
such that
……………….……….(8).
From (7), (8) and definition(3.2)(a), we can say that k is the smallest positive integer such that
. Therefore is type-1 k – symmetric.
Corollary (3.17)(a) If
k – symmetric.
be a quasi-reflexive and weak k – symmetric relation on a set S, then
Corollary (3.17)(b) If
type-1 k – symmetric.
be a right ( or left ) Euclidean and weak k – symmetric relation on a set S, then
is type-1
Theorem (3.18) If
Proof : Trivial.
be a transitive and weak k – symmetric relation on a set S, then
Theorem (3.19) If
Proof : Trivial.
be a perfect k – symmetric relation on a set S, then
Theorem (3.20) If
Proof : Trivial.
be a perfect k – symmetric relation on a set S such that
, then
is asymmetric.
Corollary (3.20) If
be a perfect k – symmetric relation on a set S such that
, then
is irreflexive.
is symmetric.
is perfect k – symmetric.
Theorem (3.21) If be a relation on a set S such that is irreflexive for
with
then is k-reflexively perfect iff is perfect (k - 1)-symmetric.
Proof : Follows from theorem(2.18).
, for some
Theorem (3.22) If be a strongly k – symmetric relation on a set S , then for any divisor d of ,
strongly finite ordered symmetric.
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is
is
is
Proof : Trivial.
Theorem (3.23) If
Proof : Trivial.
be a strongly k – symmetric relation on a set S , then
is strongly k – symmetric.
Theorem (3.24) If
be strongly k – symmetric relations on a set S such that
strongly finite ordered symmetric relation.
Proof : Trivial.
, then
is
Conclusion : Further study may be done to get some more properties of the relations, introduced here. Also
we may search for some algebras from which these relations may be obtained; or on the basis of these relations
some algebras may be introduced.
References
[1] Edmund Woronowicz : Relations and their basic properties, Formalized Mathematics, 1, 73-83, 1990.
[2] Edmund Woronowicz and Anna Zalewska : Properties of Binary Relations, Formalized Mathematics, 1,
85 – 89, 1990.
[3] Jayanta Biswas : Factorization of Mapping and Binary Relation, Seminar Proceedings (RAMA-2015),
Department of Pure Mathematics, University of Calcutta, pp 52, March 12, 2015.
[4] Krzysztof Hryniewiecki : Relations of Tolerance , Formalized Mathematics, Vol. 2, No. 1 : 105 – 109,
January-February 1991.
ISSN-0976-9625@Acad. J. Aureole 6 &7(1)PP-15-30 (2015-16).