Page 145 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 145
IV: Skew Boolean Algebras
having finite intersections imply that both S/L and S/R do also? Do examples exist where S has
finite intersections but not S/L or S/R? Conversely, if S/L and S/R do also:
Theorem 4.4.11. If a skew lattice S with finite intersections has a lattice section, then
both S/L and S/R also have finite intersections.
Proof. If T ⊆ S is a lattice section in S, then a copy of S/R in S is given by T[L] = ∪t ∈TLt. For
all x ∈ S let tx be the unique element in Dx ∩ T and set xL = x∧tx in T[L]. Then xL R x and
T[L] = {xL⎪x ∈ S}. Given a, b ∈ T[L], we claim that its intersection a∩b in S is already in T[L].
We use the fact that x ≥ y in S iff both xL ≥ yL in T[L] and dually, xR ≥ yR in the dual subalgebra
T[L]. (In general, if x and y correspond to (xʹ, xʺ) and (yʹ, yʺ) in S/R ×S/D S/L, then x ≥ y iff both
xʹ ≥ yʹ and xʺ ≥ yʺ.) Since a = aL and b = bL, it follows that a, b ≥ (a∩b)L in T[L]. But since
both a∩b D (a∩b)L and a∩b ≥ (a∩b)L by definition of a∩b, a∩b is (a∩b)L in T[L]. Being
closed under ∩, T[L] and also T[R] have finite intersections and so do their copies, S/R and S/L.
£
Thus for a symmetric skew lattice S with finite intersections, if S/D is countable (so that
S has a lattice section), then both S/R and S/L also have finite intersections. We can do better.
Theorem 4.4.12. If a symmetric skew lattice S has finite intersections, so do S/R and
S/L.
Proof. So let S be a symmetric skew lattice and let xʹ and yʹ in S/R be given with pre-images x
and y in S. Let S1 be the sub-algebra generated from x and y, let T be a lattice section of S1 and
let T[L] be the maximal left-handed subalgebra T[L] = ∪t ∈TLt of S1. If xL = x∧tx and yL = y∧ty
in S1 as in the previous proof, then xL and yL are also pre-images of xʹ and yʹ in S. Let S2 be the
generated result of adjoining xL∩yL in S to S1 and let T2 be a lattice section of S2 extending
T1. Then xL∩yL must lie in T2[L], as above.
Suppose next that xʹ, yʹ ≥ zʹ in S/R. If z is a pre-image of zʹ in S, let S3 be the extension
of S2 generated from S2 and z, and let T3 be a lattice section of S3 extending T2. By what was
seen in the above proof, zL in T3[L] is also a pre-image of zʹ such that both xL ≥ zL and yL ≥ zL
hence xL∩yL ≥ zL. Clearly, if θ is the image of xL∩yL in S/R, then xʹ, yʹ ≥ θ ≥ zʹ. Given that xʹ
and yʹ are fixed and zʹ is arbitrary but subject to the constraint xʹ, yʹ ≥ zʹ, θ must be xʹ∩yʹ in
S/R. Since this must be true for all xʹ, yʹ in S/R, the latter has intersections. Likewise, so does
S/L. £
Indeed, the above arguments essentially show: if a skew lattice S with finite intersections
is such that for each countable sublattice T of S/D, the inverse image S|T of T in S has a lattice
section, then both S/R and S/L also have finite intersections.
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having finite intersections imply that both S/L and S/R do also? Do examples exist where S has
finite intersections but not S/L or S/R? Conversely, if S/L and S/R do also:
Theorem 4.4.11. If a skew lattice S with finite intersections has a lattice section, then
both S/L and S/R also have finite intersections.
Proof. If T ⊆ S is a lattice section in S, then a copy of S/R in S is given by T[L] = ∪t ∈TLt. For
all x ∈ S let tx be the unique element in Dx ∩ T and set xL = x∧tx in T[L]. Then xL R x and
T[L] = {xL⎪x ∈ S}. Given a, b ∈ T[L], we claim that its intersection a∩b in S is already in T[L].
We use the fact that x ≥ y in S iff both xL ≥ yL in T[L] and dually, xR ≥ yR in the dual subalgebra
T[L]. (In general, if x and y correspond to (xʹ, xʺ) and (yʹ, yʺ) in S/R ×S/D S/L, then x ≥ y iff both
xʹ ≥ yʹ and xʺ ≥ yʺ.) Since a = aL and b = bL, it follows that a, b ≥ (a∩b)L in T[L]. But since
both a∩b D (a∩b)L and a∩b ≥ (a∩b)L by definition of a∩b, a∩b is (a∩b)L in T[L]. Being
closed under ∩, T[L] and also T[R] have finite intersections and so do their copies, S/R and S/L.
£
Thus for a symmetric skew lattice S with finite intersections, if S/D is countable (so that
S has a lattice section), then both S/R and S/L also have finite intersections. We can do better.
Theorem 4.4.12. If a symmetric skew lattice S has finite intersections, so do S/R and
S/L.
Proof. So let S be a symmetric skew lattice and let xʹ and yʹ in S/R be given with pre-images x
and y in S. Let S1 be the sub-algebra generated from x and y, let T be a lattice section of S1 and
let T[L] be the maximal left-handed subalgebra T[L] = ∪t ∈TLt of S1. If xL = x∧tx and yL = y∧ty
in S1 as in the previous proof, then xL and yL are also pre-images of xʹ and yʹ in S. Let S2 be the
generated result of adjoining xL∩yL in S to S1 and let T2 be a lattice section of S2 extending
T1. Then xL∩yL must lie in T2[L], as above.
Suppose next that xʹ, yʹ ≥ zʹ in S/R. If z is a pre-image of zʹ in S, let S3 be the extension
of S2 generated from S2 and z, and let T3 be a lattice section of S3 extending T2. By what was
seen in the above proof, zL in T3[L] is also a pre-image of zʹ such that both xL ≥ zL and yL ≥ zL
hence xL∩yL ≥ zL. Clearly, if θ is the image of xL∩yL in S/R, then xʹ, yʹ ≥ θ ≥ zʹ. Given that xʹ
and yʹ are fixed and zʹ is arbitrary but subject to the constraint xʹ, yʹ ≥ zʹ, θ must be xʹ∩yʹ in
S/R. Since this must be true for all xʹ, yʹ in S/R, the latter has intersections. Likewise, so does
S/L. £
Indeed, the above arguments essentially show: if a skew lattice S with finite intersections
is such that for each countable sublattice T of S/D, the inverse image S|T of T in S has a lattice
section, then both S/R and S/L also have finite intersections.
143
