Abstract: This study concerns with the concepts of dense element, pseudocomplemented semilattice, *semilattice, sectionally *semilattice and also the concept of *semilattice to each section (0, a) in a semilattice. It imparts an account on the connections of sectionally *semilattice with pseudocomplemented semilattices. We establish that sectionally pseudocomplemented semilattice is a sectionally *semilattice, but not conversely.
INTRODUCTION
Speed (1969) has introduced the concept of distributive *-lattice, the same condition was applied on a distributive lattice to become a *-lattice was placed on any meet semilattice and such semilattice are called as *semilattice and showed that almost all the results of Speed can be extended to a more general class of distributive *semilattices by Krishnamurthy (1980).
In pseudocomplemented semilattices and distributive semilattice, the set of annihilators of an element is an ideal in the sense of Gratzer (1976). But it is not so in general and thus a weakly distributive semilattice S is defined.
The concept of *semilattice has been extended to semilattice by Krishnamurthy (1980) and that of pseudocomplement by Katrinak (1968), Nawar (1974), Birkhoff (1967) and Krishnamurthy (1980).
Venkateswara Rao and Ravi-Kumar (2010a) studied about Distributive Convex Sub lattices. Venkateswara Rao and Ravi-Kumar (2010b) investigated the Characterization of super modular semilattices. Venkateswara Rao and Ravi-Kumar (2010c) introduced the concept of modular and classic ideals of directed below join semi lattice. Venkateswara Rao and Ravi-Kumar (2010d) studied extensively about Characterization of standard and distributive ideals in semi lattice.
In this study we investigate about weakly distributive semilattice and the necessary and sufficient condition for a modular *semilattice to be a weakly distributive semilattice. We extend the concept of *semilattice to each section (0, a) in a semilattice. We also establish its connections with pseudocomplemented semilattices.
PRELIMINARIES
Definition 1
A meet semilattice (S, ∧) is a set with an idempotent, commutative and
associative binary operations on S.
Definition 2
A partial order may be defined on S by b = a if and only if a ∧ b =
b where a, b ∈ S.
Definition 3
A meet semilattice S is said to be directed above if and only if for x,
y in S, there exists a in S such that x = a, y = a.
Definition 4
A non empty subset I of a meet semilattice S is called an Ideal if (i) for
x, y ∈ I implies x ∧ y ∈ I, (ii) for x ∈ I, t ∈ S such
that x = t implies t ∈ I.
Definition 5
A non empty subset D of a meet semilattice S is called a filter of S if
and only if (i) x ∈ D and x = a implies a ∈ D, (ii) x, y ∈ D
implies there exists z ∈ D such that x = z and y = z.
Definition 6
A meet semilattice S with 0 is called a weakly distributive semilattice,
if (a)* is an ideal for any a ∈ S, where (a)* = { x ∈ S/ x ∧ a
= 0, for a ∈ S}.
Definition 7
A meet semilattice S with 0 is said to be a *semilattice if and only if
for any a in S, there exists a1 in S, such that (a)* = (a1)**,
where a ∧ a1 = 0.
Definition 8
If S is a meet semilattice with 0 then for any subset A of S, A*stands for
{x in S/x ∧ a = 0, for all a in A}. If A = {a} then A* = (a)*.
Definition 9
An element a in S is called a dense element of S if and only if (a)* = {0}.
Remark 1
If D denotes the set of all dense elements of S, then D is a filter.
Proof
Let D = {x/x is dense element of S} = {x/(x)* = {0}, for x ∈ S}. (i)
Let x ∈ D and a ∈ S such that a = x. (a)* ⊆ (x)* = {0}. Also
{0}⊆ (a)* as 0 is least element of S. Therefore (a)* = {0}. Hence a ∈
D. (ii) Let x, y ∈ D. Then x and y are dense elements, (x)* = {0}; (y)*
= {0}. Let z∈ S such that x = z, y = z. Then (z)* ⊆ (x)* and (z)*
⊆ (y)* ⇒ (z)* ⊆ {0}. Also {0} ⊆ (z)* as 0 is least element.
Therefore (z)* = {0}. Hence D is a filter.
WEAKLY DISTRIBUTIVE SEMILATTICE
Now we produce a condition for a *semilattice with modularity property to be a weakly distributive semilattice.
Theorem 1
In a weakly distributive semilattice S, the following are equivalent. (i)
S is a *semilattice (ii) For any x ∈ S, there exists x1 ∈
S such that x ∧ x1 = 0 and (x) ∩ (x1) ⊆ D,
where D is a filter.
Proof: Let S be a weakly distributive semilattice, then for every a ∈ S, (a)* = {x ∈ S / x ∧ a = 0} is an ideal.
(i)⇒(ii)
Let S be a *semilattice, then for every x in S, there exists x1
in S, such that (x)* = (x1)** where x ∧ x1 = 0. Implies
x1 ∈ (x)* = (x1)** implies x1 ∈ (x1)**.
Let t ∈ S such that t ∧ x = 0 = t ∧ x1 implies t ∈
(x)*, since (x)* is an ideal for x ∈ S such that x ≤ t, then t ∈
(x), also t ∈ (x1)*.And since (x1)* is an ideal for
x1 ∈ S such that x1 ≤ t implies t ∈ (x1).Thus
t ∈ (x) ∩ (x1) and as D is filter of S for x, x1
in D there exists t in d such that x ≤ t and x1 ≤ t. Hence
(x) ∩ (x1) ⊆ D.
(ii)⇒(i)
Assume that for any x ∈ S, there exists x1 ∈ S such
that x ∧ x1 = 0 and (x) ∩ (x1) ⊆ D. Since
x ∧ x1 = 0 implies x1 ∈ (x)*. Let s ∈ (x1)**,
then s ∧ (x1)* = 0 which implies s ∧ x = 0 for x ∈ (x1)*.
So s ∈ (x)*. Thus (x1)** ⊆ (x)* --- (i).
Let t ∈ S be such that t ∧ x = 0 and r∈ S be such that r ∧ x1 = 0, then t ∧ x ∧ r = 0 ∧ r = 0 which implies x ∈ (t ∧ r)*. Similarly t ∧ r ∧ x1 = t ∧ 0 = 0, then x1 ∈ ( t ∧ r)*. Since for t, r ∈ S, we have t ∧ r ∈ S and (t ∧ r)* is an ideal of S (since S is a weakly distributive). It follows that there exists z ∈ S such that z ≥ x, x1 as z∈ (t ∧ r)* which implies z ∧ t ∧ r = 0. Since z ≥ x, x1, we have z ∈ (x) and z ∈ (x1) which implies z ∈ (x) ∩ (x1) ⊆ D. Thus z ∈ D and hence t ∧ r = 0, thus t ∈ (r)* which shows (x)* ⊆ (x1)** --- (ii) As t ∧ x =0, t ∈ (x)* and r ∧ x1 = 0, r ∈ (x1)*. Therefore, from (i) and (ii), we have (x)* = (x1)** for x ∧ x1 = 0. Hence S is a * semilattice.
Note
For any filter F of a semilattice S, θF = {(x, y) ∈
S X S / x ∧ f = y ∧ f for some f ∈ F} is a congruence on S.
Definition 10
A meet semilattice S is called modular if and only if a ∧ b ≤ w ≤
a implies there exists y in S such that y ≥ b and w = a ∧ y.
Theorem 2
In a modular *semilattice S, we have θD = R, where R = {
(x, y) ∈ S X S / (x)* = (y)*}.
Proof
Let S be a modular * semilattice. For any filter D of S, θD
= {(x, y) ∈ S X S / x ∧ d = y ∧ d for d ∈ D} is a congruence
on S. Given R = {(x, y) ∈ S X S / (x)* = (y)*}. Clearly θD ⊆
R. Let (x)* = (y)* for x, y ∈ S then (x ∧ x)* = (x ∧ y)* which
implies (x ∧ y)* = (x)*. Since (x)* = (x1)** for x1 ∈
S, then (x ∧ y)* = (x1)**. Since x ∧ x1 = 0, x1
∈ (x)* = (y)* thus y ∧ x1 = 0 ≤ y ∧ x ≤ y. Now,
by using the modularity of S, there exists x2 ≥ x1 such
that y ∧ x = y ∧ x2. Thus (x2)* = 0, which implies
x2 is dense, since x2 ≥ x1 and x2
≥ y ∧ x. Similarly, x ∧ x1 = 0 ≤ y ∧ x ≤ x,
by modularity of S, there exists y2 ≥ x1 such that
y ∧ x = y2 ∧ x. As above y2 is dense. Hence x2
∧ y2 is dense. Now x ∧ x2 ∧ y2
= x ∧ x2 ∧ y = y ∧ x2 ∧ x = y ∧ x2
∧ y2. Therefore, x ∧ x2 ∧ y2
= y ∧ x2 ∧ y2 for x2 ∧ y2
∈ D. Thus R ⊆ θD. Hence θD = R.
In the following theorem, it is provided a necessary and sufficient condition for a modular *semilattice to be weakly distributive.
Theorem 3
If S is a *semilattice which is directed above such that θD =
R, then S is a weakly distributive semilattice if and only if for all x, y in
S ((x) ∧ (y) ∨ D = ((x) ∧ D) ∩ ((y) ∨ D).
Proof
Assume S is weakly distributive. Let t ∈ S be such that t ∈ (
((x) ∨ D) ∧ ((y) ∨ D)) which implies that there exists d1,
d2 ∈ D such that x ∧ d1 ≤ t and y ∧ d2
≤ t. Put d = d1 ∧ d2 ∈ D. It follows that,
there exists d ∈ D such that x ∧ d ≤ t and y ∧ d ≤ t. Since
S is a * semi lattice, there exists t1 ∈ S such that (t)* =
(t1)** where t ∧ t1 = 0. It follows that x ∧ d
∧ t1 = 0 = y ∧ d ∧ t1, then d ∈ D is
a dense element and S is a weakly distributive, there exists z ∈ S such
that z ≥ x, y and z ∧ t1 = 0. Since z ∧ t1 =
0, we have (z)* = (t1)** = (t)* which implies (z)* = (t)* or (z ∧
t)* = (z)*, so that there exists e ∈ D such that z ∧ e = z ∧ t
∧ e which implies that z ∧ e ≤ t. Hence there exists z ∈ (x)
∩ (y) and e ∈ D such that z ∧ e ≤ t. Thus we have t ∈ ((x)
∩ (y)) ∨ D. Thus ((x) ∨ D) ∧((y) ∨ D) ⊆ ((x) ∧ (y))
∨ D. Also clearly ((x) ∧ (y)) ∨ D ⊆ ((x) ∨ D) ∧ ((y) ∨
D). Hence, ((x) ∧ (y)) ∨ D = ((x) ∨ D) ∧ ((y) ∨ D) -------
(i) Conversely, suppose (i) holds and x, y ∈ (a)* in order to prove that
(a)* is an ideal for a ∈ S. Thus x ∧ a = y ∧ a = 0. Let t ∈
(x ∧ a1)*, t ∧ x ∧ a1 = 0 which implies t
∧ x = 0, t ∈ (x)*. Thus (x ∧ a1)* ⊆ (x)*. Similarly,
we obtain (x)* ⊆ (x ∧ a1)*.Therefore (x ∧ a1)*
= (x)*. Similarly (y ∧ a1)* = (y)* where a1 is an
element of S satisfying the condition (a)* = (a1)**. Therefore, x
∧ a1 ∈ (x) ∨ D and y ∧ a1 ∈ (y) ∨
D, since θD = R, we get that a1 ∈ (((x) ∨
D) ∧ ((y) ∨ D). By assumption a1 ∈ (((x) ∧ (y)) ∨
D), it follows that there exists e ∈ D such that z ∧ e ≤ a1,
since z ≥ x, y. Now z ∧ e ∧ a ≤ a1 ∧ a = 0, hence
we have z ∧ a = 0 thus z ∈ (a)* for e ∈ D. Hence (a)* becomes
an ideal. Hence S is a weakly distributive.
SECTIONALLY *SEMILATTICE
Now we define a *semilattice as sectionally and we give a relation with pseudocomplemented and sectionally pseudocomplemented semilattices.
Definition 11
A meet semilattice S with 0 is said to be a sectionally *semilattice if
and only if for every a in S, the interval (o,a) is a *semilattice.
Theorem 4
Every *semilattice is a sectionally *semilattice.
Proof
Let (S,∧) be a semilattice and a in S be arbitrary. Since S is a *semilattice,
for x in S, there exists a in S such that (x)* = (a)** where x ∧ a = 0.Now
we show that (S, ∧) is sectionally *semilattice. Consider (0,a) in S and
Put X = (0,a). To Show that (X, ∧) is a *semilattice. Clearly 0 is in X,
let x be in X. Since X ⊂ S and x is in X, x is in S. Thus there exists a1
in S such that (x)* = (a1)** where x ∧ a1 = 0.Put
x1 = x ∧ a1 = a ∧ a1 = 0 (take x =
a)⇒ x1 = 0 is in X. Therefore x ∧ x1 = x ∧
(a ∧ a1) = x ∧ 0 = 0. Therefore x ∧ (a ∧ a1)
= 0. Hence x is in (a ∧ a1)*.
To show that (X)[0, a] = (a ∧a1)[0, a]
Let y be in (X)[0, a] ⇒ y ∧ (a ∧ a1)
= 0
⇒ y ∧ x = 0 for x is in (a ∧ a1) * ⇒ y is in (X)[0,
a] ⇒(a ∧a1)[0, a] ⊆ (X)*[0, a].
Let y is in (X)*[0, a] be arbitrary, then y ∧ x =0.
⇒ y is in (x)* for all x in (0,a) ⊆ S, y is in (x)* for x is in S
⇒ y is (a1)**
⇒ y ∧ (a1)* = 0 ⇒ y ∧ z = 0 for all z in (a1)*
= {x ∈ S/ x ∧ a1 = 0}
⇒ y ∧ z = 0, for every z in (0,a) with z ∧ a ∧ a1
= 0 ⇒ z ∈ (a ∧ a1)*(0,a)
⇒ y ∧ z = 0 for every z in (a ∧a1)[0, a]
⇒ y is in (a ∧a1)[0, a].
Therefore, (x)*⊆ (a ∧a1)[0, a].
Therefore,(x)*[0, a] = (a ∧a1)[0, a]
Therefore, (x, ∧) is a *semilattice Therefore (0,a) is a *semilattice.
Hence, (S, ∧) is a sectionally *semilattice. Hence, every *semilattice is
a sectionally *semilattice.
Theorem 5
Every sectionally *semi lattice is not necessarily a *semilattice.
Proof
Let S be a sectionally *semilattice. Then for every a in S, the interval
(0, a) is a *semilattice. We prove S need not be a *semilattice by taking an
example. Let S = {0, a, x, d} be a sectionally *semilattice.
Clearly the intervals (0, a), (0, d) and (0, x) are *semilattices, but for any b in S, (0)* ≠ (b)**. Hence, S is not a * semilattice.
Theorem 6
A directed above semilattice S is a * semilattice if and only if it is a
sectionally *semilattice and S contains a dense element.
Proof
Let S be directed above and a *semilattice. To prove that S contains a dense
element. Since S is a *semilattice, then S is a sectionally *semilattice and
for every x in S there exists x1 in S such that (x)* = (x1)**
where x ∧ x1 = 0, also there exists an element d in S such that
d ≥ x, x1. ⇒ d ≥ x ∧ x1 = 0 ⇒ (d)*
⊂ (0)* = S Now for any a in (d)*, we have a ∧ d = 0 ⇒ a ∧ d
∧ x = 0.
⇒ a ∧ x = 0 for d ≥ x also a ∧ d ∧ x1 = 0 ⇒
a ∧ x1 = 0 for d ≥ x1. Thus, a is (x)* and a is
in (x1)*
⇒ a is in (x)* ∩ (x1)*.
⇒ a is in (x1)** ∩ (x1)* ⇒ a is in {0} ⇒
a = 0.
Therefore, (d)* = {0}. Therefore S contains a dense element. Conversely, Let S be a directed above sectionally *semilattice and it contains a dense element d. To prove that S is a *semilattice.
Let x is in S and choose a such that a ≥ x, d. Then a is a dense element because (a)* ≤ (d)* = {0}. Since S is a sectionally * semilattice, we have for every x in the interval (0,a) there exists x1 in the interval (0, a) such that (x)* (0,a) = (x1)** (0,a), where x ∧ x1 = 0.Now, it is enough to show that (x)* = (x1)**. Since x ∧ x1 = 0 We have x is in (x1)* ⇒ (x1)** ⊆ (x)* (i)
Let y is in (x)* and z is in (x1)* ⇒ y ∧ x = 0 and z ∧ x1 = 0.
| ⇒ y ∧ x ∧ z = 0 and y ∧ z ∧ x1 = 0 ⇒
y ∧ z ∧ a ∧ x = 0 and y ∧ z ∧ a ∧ x1
= 0 ⇒ y ∧ z ∧ a is in (x)* (0,a) and y ∧ z ∧ a is in (x1)* (0,a) ⇒ y ∧ z ∧ a is in (x)* (0,a) ∧ (x1)* (0,a) = {0} ⇒ y ∧ z ∧ a = 0 ⇒ y ∧ z is in (a)* ={0} since a is dense element ⇒ y ∧ z = 0 |
| ⇒ y is in (z)* for z is in (x1)* ⇒ y is in (x1)**. Therefore (x)* ⊆ (x1)** | (ii) |
Therefore, from (i) and (ii) we have (x)* = (x1)** where x ∧ x1 = 0. Hence S is a * semilattice.
Definition
A meet semilattice S with 0 is said to be pseudocomplemented if and only
if for a in S, a1 is a pseudocomplement of a in S, that is x ∧
a = 0 in S if and only if x ≤ a1. 3.6 Definition A meet semilattice
S with 0 is called a sectionally pseudocomplemented semilattice if and only
if for every a in S, the interval (0, a) is a pseudocomplemented semilattice.
Remark 2
Every pseudocomplemented semilattice is a sectionally pseudocomplemented
semilattice.
Proof
Let S be the pseudocomplemented semilattice, then for a ∈ S, a* is
a pseudocomplement of a in S that is x ∧ a = 0 in S if and only if x = a*.
Since a* is a pseudocomplement of a, then a ∧ a* = 0.Let y ∈ (0, a)
⇒ 0 = y = a ⇒ y ∧ a = y also y ∧ 0 = 0 ⇒ y ∧ a =
0. y ∧ a = y ∧ 0 = y ∧ a ∧ a* ⇒ y ∧ a = y ∧ a*
∧ a ⇒ y = y ∧ a* ⇒ y = a*. Let y = a*, then y ∧ a* = y
⇒ y ∧ a* ∧ a = y ∧ a ⇒ y ∧ 0= y ∧ a⇒ 0 =
y ∧ a.
Therefore, (0, a) is pseudocomplemented semilattice. Hence every pseudocomplemented semilattice is sectionally pseudocomplemented semilatice.
Theorem 7
Every sectionally pseudocomplemented semilattice is sectionally *semilattice.
Proof
Let S be a sectionally pseudocomplemented semilattice. To prove that for
every a in S, the interval (0, a) is a *semilattice. Let x in the interval (0,
a) which is a pseudocomplemented semilattice. Then there exists a pseudocomplement
x1 of x in the interval (0, a) that is y ∧ x = 0 in the interval
(0, a) if and only if y ≤ x1. Then (x)* = (x1) ⇒
(x)** = ((x1))* = (x1)*. ⇒ (x)* = (x1)**
where x1 is a pseudocomplement of x in the interval (0, a). Therefore,
the interval (0, a) is a *semilattice. Hence, S is a sectionally *semilattice.
Theorem 8
Every sectionally *semilattice is not necessarily a sectionally pseudocomplemented
semilattice.
Proof
Let S be a sectionally *semilattice, then for every a in S, the interval
(0, a) is a *semilattice.
For example S = {0, a, x, d} be a meet semilattice with 0, which is a sectionally *semilattice. But the interval (0, d) is not a pseudocomplemented semi lattice, because there does not exists pseudocomplement x1 of x in the interval (0, d) such that y v x = 0 in the interval (0, d) if and only if y ≤ x1. Therefore, S is not sectionally pseudocomplemented semilattice.