ISSN 0973-8975
J.Mech.Cont.& Math. Sci., Vol.-9, No.-2, January (2015) Pages 1418-1424
GLIVENKO CONGRUENCE ON A 0-DISTRIBUTIVE MEET
SEMILATTICE
By
Momtaz Begum
Department of ETE, Prime University, Dhaka, Bangladesh.
Abstract:
In This paper the author studies the Glivenko congruence R in a
0-distributive meet semilattice. It is proved that a meet semilattice S with 0 is
0-distributive if and only if the quotient semilattice
S
is distributive. Hence S is
R
0-distributive if and only if (0] is the Kernel of some homomorphism of S onto a
distributive meet semilattice with 0.
Key words and phrases: Glivenko congruence, 0-distributive semilattice, distributive meet
semilattice.
Introduction:
J.C.Varlet [7] first introduced the concept of 0-distributive lattices. Then
many authors including [1,2,3,5] studied them for lattices and semilattices. By [2],
a meet semilattice S with 0 is called 0-distributive if for all a, b, c  S with
ab  0  ac
imply a  d  0 for some d  b, c . A meet semi lattice S is
called directed above if for all a ,b  S , there exists c  S such that c  a ,b . We
know that all modular and distributive semilattices have the directed above
property. Moreover, [3] have shown that every 0-distributive meet semilattice is
directed above.
Let S be a meet semilattice with 0. For a non-empty subset A of S, we
define A   {x  S | x  a  0 for all a  A } .This is clearly a down set, but we
can not prove that this is an ideal even in a distributive meet semilattice, when A is
infinite.
By [2,3] we know that, for any a  S , {a} is an ideal if and only if S is
0-distributive.
We define a relation R on a meet semilattice S by a  b(R ) if and only if
(a ]  (b]  . In other words, a  b(R ) is equivalent to “for each x  S , a  x  0
if and only if b  x  0 ”.

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J.Mech.Cont.& Math. Sci., Vol.-9, No.-2, January (2015) Pages 1418-1424
We will show below that this is a congruence on the meet semilattice S. We
call it Glivenko congruence. In this paper we establish some results on this
congruence in a meet semilattice.
We start with the following result which is due to [3]. We include its proof
for the convenience of the reader.
Lemma 1: Let S be a meet-semilattice with 0. Again let A, B  S and
a, b  S then we have the followings:
(i) If A  B  (0] , then B  A 
(ii) A  A   (0] ,
(iii) A  B
imply that B   A 
(iv) If a  b imply that {b}  {a} and {a}  {b}
(v) {a}  {a}  (0]
(vi) {a  b}  {a}  {b}
(vii) A  A 
(viii) A   A 
Proof: (i) Let b  B . Then a  b  0 for all a  A , as A  B  {0} . Thus
b  A  . Hence B  A  .
(ii) Let x  A  A  .
= x  A and x  a  0 for all a  A
= xx 0
= x0
(iii) Let A  B
 A  B   B  B   (0]
 A  B   (0]
So, by (i), B   A  .
(iv) Let x  {b} . Then b  x  0 for some x  S . Since a  b ,
then we have a  x  0 for some x  S , which imply that x  {a} .
Hence,
{b}  {a} .
Now let x  {a} . Then y  x  0 for all y  {a} , which implies
that y  x  0 for all y  {b} as
{b}  {a}
Hence,
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Thus x  {b} .
J.Mech.Cont.& Math. Sci., Vol.-9, No.-2, January (2015) Pages 1418-1424
{a}  {b} .
(v) Let x  {a}  {a} . Then x  {a} and x  {a} which
implies that x  a  0 and x  y  0 for all y  {a} . Thus x  x  0 .
Hence
{a}  {a}  (0] .
(vi) Let x  {a}  {b} and y  {a  b} . Then we get
( y  a )  b  0 , which implies that ( y  a)  {b} . Since x  {b} , we get
( x  y)  a  0 .
Hence x  y  {a} . Since x  {a} , we get x  y  {a} . Thus
x  y  {a}  {a}  (0]. Hence x  y  0 for all y  (a  b)  . Therefore
x  (a  b)  . Thus {a}  {b}  {a  b)} .
Conversely we can write that a  b  a , which implies by (i)
(a  b)   {a} . Similarly {a  b}  {b} . Therefore we have,
{a  b}  {a}  {b} .
(vii) Let x  A , consider any r  A  , then we get x  a  0 for all
a  A which implies that r  x  0 . Since x  r  0 for all r  A  . Thus
x  A  . Hence A  A  .
(viii) Since by (vii) A  A  . So by (iii) ( A  )   A  .
Hence A   A  . Since by (vii) A   ( A  )   A  . Therefore we have
A   A  .
Hence the proof is completed. 
Theorem 2: R is a meet congruence on S.
Proof: It is clearly an equivalent relation.
Let a  b(R ) and t  S
Then (a ]   (b]  , so by using Lemma 1, we have (a  t ]   (a  t ] 
= {( a ]   (t ]  }
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= {(b]   (t ]  }
= (b  t ]  = (b  t ] 
This implies a  t  b  t (R ) , and so R is a meet congruence on S. 
A meet semilattice S with 0 is weakly complemented if for any pair of
distinct elements a, b of S, there exists an element c disjoint from one of these
elements but not from the other. In particular, if a  b , then there exists c  S
such that a  c  0 but b  c  0 .
Theorem 3:
If S is weakly complemented. Then R is an equality relation.
Proof: Suppose a, b  S with a  b . Since S is weakly complemented, so there
exist x  S , a  x  0 but b  x  0 . This implies ( a, b)  R . Hence R is an
equality relation. 
S
Theorem 4: For any meet semilattice S.
is also a meet semilattice. Moreover
R
S
S is directed above if and only if
is directed above.
R
Proof:
For [a ], [b] 
S
S
, define [ a ]R  [b]R  [ a  b]R . Thus
is a meet
R
R
semilattice.
Now let a, b  S . If S is directed above, then there exists d  a, b .
Now, [ a ]R  [ d ]R  [ a  d ]R  [ a ]R and
Implies [ d ]R  [ a ]R, [b]R . Thus,
Conversely suppose
Then [a ], [b] 
[b ] R  [ d ] R  [b  d ] R  [b ] R
S
is also directed above.
R
S
is directed above. Let a, b  S
R
S
S
S
. Since
is directed above, so there exists C 
R
R
R
such that C  [ a ]R, [b]R . Then there exists d  C ,
such that [ d ]  C and d  a, b . So S is directed above. 
A meet semilattice S is called a distributive semilattice if w  a  b implies
that there exist x  a , y  b in S such that w  x  y .
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Following result gives some characterizations of distributive meet
semilattices which are due to [4, Theorem 1.1.6], also see [6].
Lemma 5: For a meet semilattice S, the following conditions are equivalent.
i)
S is distributive.
ii)
w  a  b implies that there exists y  S such that y  b , y  w and
y  a  a  w.
iii)
a  b  b  c implies that there exists y  S such that y  b , y  c and
ya  ac.

Theorem 6: For any meet semilattice S, the quotient meet semilattice
complemented. Furthermore, S is 0-distributive if and only if
S
is weakly
R
S
is distributive.
R
S
, there exists a in A
R
and b in B such that a< b, and by the definition of R, there is an element c such that
S
c  a  0 and c  b  0 . Since the minimum class of
has the only element 0,
R
S
the class C of c satisfies A  C  [0] and C  B  [0] . Therefore,
is weakly
R
complemented.
Proof: First part: For any meet semilattice S, when A< B in
S
. So there exists
R
b  B , a  A , c  C such that b  a  c and B  [b]R , A  [a ]R , C  [c]R .
Suppose a  b  x  0 . Then a  c  x  0 . Since S is 0-distributive, so there
exists d  b, c such that a  d  x  0 . On the other hand, for any d  b, c ,
a  d  x  0 implies a  d  x  b  a  b  x  0 . Therefore, a  b  a  d (R )
for some d  b, c . In other words, A  B  A  D where D  [ d ]  B, C .
S
Therefore by [4, Theorem 1.1.6 (ii)],
is distributive.
R
For second part: Let S be 0-distributive. Suppose B  A  C in
S
is distributive. Let a, b, c  S with
R
a  b  a  c  0 . Then [ a ]  [b]  [ a ]  [c ]  [0]R . Since [0] contains only the
element 0, so A  B  A  C  0 , where A  [a ] , B  [b] , C  [c] . Then
S
B  A  C . Since
is distributive, so B  A1  C1 for some A1  A , C1  C .
R
Moreover, B  A1  C1 implies C1  B . Thus 0  A  B  A  A1  C  A  C1 .
Conversely,
suppose
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J.Mech.Cont.& Math. Sci., Vol.-9, No.-2, January (2015) Pages 1418-1424
Now C1  B, C implies C1  [d ]R for some d  b, c . Therefore, a  d  0 for
some d  b, c and so S is 0-distributive. 
We conclude the paper with the following result.
Theorem 7: Let S be a meet semilattice. Then the following conditions are
equivalent
(i)
(ii)
(iii)
S is 0-distributive.
(0] is the kernel of some homomorphism of S onto a distributive semilattice
with 0.
(0] is the kernel of a homomorphism of S onto a 0-distributive semilattice.
Proof: (i)  (ii). Suppose S is 0-distributive. Then by Theorem 1, the binary


relation R defined by x  y(R) iff ( x]  ( y] is a congruence on S. Moreover
S
by Theorem 5,
is a distributive meet semilattice. Clearly the map a  [ a ]R is
R
a homomorphism. Now let a  0( R ) . Then 0  a  0 implies a  a  a  0 . Here
[0]R contains only 0 of S. That is, (0] is a complete congruence class modulo R.
(ii)  (iii) is obvious as every distributive semilattice with 0 is 0-distributive.
(iii)  (i). Let  be a congruence on S for which (0] is the zero element of the
S /
0-distributive
semilattice
.
Then
imply
x y 0 xz
Thus,
[ x ]  [ y ]  [ x  y ]  [ x  z ]  [ x ]  [ z ] .
S
,
[ x ]  [ y ]  (0]  [ x ]  [ z ] . Hence by the 0-distributivity of

[ x]  [d ]  (0] , for some [ d ]  [ y ] , [ z ] . This implies x  d  (0] and so
x  d  0 , where d  y, z . Therefore, S is 0-distributive. 
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