Page 149 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 149
IV: Skew Boolean Algebras
with the two new, unequal atoms being D-related in LSBAn+1. Thus, given any non-0 central
element c = a1 + … + ak in LSBAn with atoms ai of the given form,
xn+1∧c = (xn+1∧a1) + … + (xn+1∧ak) ≠ (a1∧ xn+1) + … + (ak∧ xn+1) = c∧xn+1.
The case for RSBAn and RSBAn+1, or SBAn and SBAn+1, is similar. We thus have:
Theorem 4.4.16. Given a finite free skew Boolean algebra on n generators, whether left-
handed, right-handed or two sided, its center forms a Boolean algebra of order 2n. In the case of
an infinite free algebra, the center is just {0}. !
We turn to a common property of all free algebras. Trivially, finitely generated free
algebras have intersections since they are finite. It turns out that infinitely generated free algebras
also have finite intersections. Thanks to our observations on adjoining free generators and their
effects on atoms, intersections are stable under the inclusion LSBAn ⊂ LSBAn+1. Thus ∩ for
pairs of elements in LSBA3 remains the same for these elements in the bigger, say LSBA7. What
changes is the decomposition of all outcomes into atoms. The pool of atoms that two elements
share in LSBAn, doubles by splitting to give rise to the new pool of atoms in LSBAn+1 that both
share. As a result:
Theorem 4.4.17 Given any set X, the free left-handed [right-handed] skew Boolean
algebra LSBAX [RSBAX] on X has intersections. Given elements x and y in LSBAX, x∩y can be
calculated in any subalgebra LSBAY, where Y is any finite subset of X such that LSBAY contains
both x and y. Similar remarks hold for RSBAX.
Proof. Suppose that x and y are encountered in LSBAY where Y is a finite subset of X and that u
in LSBAX is such that u ≤ both x and y. Then x∩y relative to LSBAY exists. By our remarks, this
x∩y remains the intersection in any LSBAZ where Y ⊂ Z if Z is finite. Now u must be
encountered in some finite subalgebra LSBAU where U ⊆ X. Then both the current x∩y and u
must lie in the larger subalgebra LSBAY∪U. Since Y∪U is finite, x∩y is the intersection here also,
and u ≤ x∩y follows. Thus x∩y remains the intersection of x and y throughout all LSBAX. The
case for RSBAX is similar. !
Since RSBAX ≅ SBAX/L and LSBAX ≅ SBAX/R, by Theorem 4.4.14 we have:
Theorem 4.4.18. Free skew Boolean algebras have finite intersections. (Indeed, they
have arbitrary intersections since a properly infinite subset must have 0-intersection.) !
147
with the two new, unequal atoms being D-related in LSBAn+1. Thus, given any non-0 central
element c = a1 + … + ak in LSBAn with atoms ai of the given form,
xn+1∧c = (xn+1∧a1) + … + (xn+1∧ak) ≠ (a1∧ xn+1) + … + (ak∧ xn+1) = c∧xn+1.
The case for RSBAn and RSBAn+1, or SBAn and SBAn+1, is similar. We thus have:
Theorem 4.4.16. Given a finite free skew Boolean algebra on n generators, whether left-
handed, right-handed or two sided, its center forms a Boolean algebra of order 2n. In the case of
an infinite free algebra, the center is just {0}. !
We turn to a common property of all free algebras. Trivially, finitely generated free
algebras have intersections since they are finite. It turns out that infinitely generated free algebras
also have finite intersections. Thanks to our observations on adjoining free generators and their
effects on atoms, intersections are stable under the inclusion LSBAn ⊂ LSBAn+1. Thus ∩ for
pairs of elements in LSBA3 remains the same for these elements in the bigger, say LSBA7. What
changes is the decomposition of all outcomes into atoms. The pool of atoms that two elements
share in LSBAn, doubles by splitting to give rise to the new pool of atoms in LSBAn+1 that both
share. As a result:
Theorem 4.4.17 Given any set X, the free left-handed [right-handed] skew Boolean
algebra LSBAX [RSBAX] on X has intersections. Given elements x and y in LSBAX, x∩y can be
calculated in any subalgebra LSBAY, where Y is any finite subset of X such that LSBAY contains
both x and y. Similar remarks hold for RSBAX.
Proof. Suppose that x and y are encountered in LSBAY where Y is a finite subset of X and that u
in LSBAX is such that u ≤ both x and y. Then x∩y relative to LSBAY exists. By our remarks, this
x∩y remains the intersection in any LSBAZ where Y ⊂ Z if Z is finite. Now u must be
encountered in some finite subalgebra LSBAU where U ⊆ X. Then both the current x∩y and u
must lie in the larger subalgebra LSBAY∪U. Since Y∪U is finite, x∩y is the intersection here also,
and u ≤ x∩y follows. Thus x∩y remains the intersection of x and y throughout all LSBAX. The
case for RSBAX is similar. !
Since RSBAX ≅ SBAX/L and LSBAX ≅ SBAX/R, by Theorem 4.4.14 we have:
Theorem 4.4.18. Free skew Boolean algebras have finite intersections. (Indeed, they
have arbitrary intersections since a properly infinite subset must have 0-intersection.) !
147
