Conditions forcing the existence of relative complements in lattices and posets

Lemma 1.1

Let L=(L,∨,∧,0,1) be a bounded lattice and x ∈ L. Assume that y and z are complements of x and that (x, y, z) and (x, z, y) are distributive triples. Then y=z.

Proof. We have

Theorem 2.1

Let L=(L,∨,∧) be a lattice, a, b ∈ L with a ⩽ b, x ∈ [a, b], y ∈ L, e≔ (a∨ y) ∧ b and f≔ a∨(y ∧ b). Then f ⩽ e and the following are equivalent:

1. e, f ∈ R(a, b, x),

2. (a∨ y) ∧ x=a and x∨(y ∧ b)=b.

Proof. Because of a ⩽ a∨ y, a ⩽ b, y ∧ b ⩽ a∨ y and y ∧ b ⩽ b we have f ⩽ e. Moreover, we have

Since a ⩽ x and a ⩽ f ⩽ e, we have that x ∧ e=a implies x ∧ f=a, and since x ⩽ b and f ⩽ e ⩽ b, we have that x∨ f=b implies x∨ e=b. Hence (i) and (ii) are both equivalent to e ∧ x=a and x∨ f=b. □

It is worth noticing that we do not assume y to be a complement of x.

Example 2.1

Consider the lattice L depicted in Figure 2. Evidently, L is neither modular nor complemented. Further, x ∈ [a, b], and for y ∈ L we have

Hence, the assumptions of Theorem 2.1 are satisfied. Put e≔ (a∨ y) ∧ b and f≔ a∨(y ∧ b). Then e, f ∈ R(a, b, x). It is worth noticing that neither y belongs to [a, b] nor it is a complement of x.

If y is, moreover, a complement of x then we can derive relative complements of x in [a, b] by using modular triples.

Theorem 2.2

Let L=(L,∨,∧,0,1) be a bounded lattice, a, b ∈ L with a ⩽ b, x ∈ [a, b], y a complement of x, e≔ (a∨ y) ∧ b and f≔ a∨(y ∧ b). Then f ⩽ e and the following are equivalent:

1. e, f ∈ R(a, b, x),

2. (a, y, x) and (x, y, b) are modular triples.

Figure 2.

Proof. Since a ⩽ a∨ y, a ⩽ b, y ∧ b ⩽ a∨ y and y ∧ b ⩽ b, we have f ⩽ e. Because of Theorem 2.1, (i) is equivalent to

Since

(1) is equivalent to

Finally, due to the definition of modular triples, (2) is equivalent to (ii). □

An example of the situation described by Theorem 2.2 is the lattice depicted in Figure 3 where y is a complement of x and the elements e and f satisfy the assumptions of Theorem 2.2. Let us note that f<e in this case.

Example 2.2

Consider the lattice L depicted in Figure 3. The lattice L is not modular, y is a complement of x, the triples (a, y, x) and (x, y, b) are modular since

and e=(a∨ y) ∧ b and f=a∨(y ∧ b) are relative complements of x in [a, b] in accordance with Theorem 2.2.

Example 2.3

Unfortunately, the methods presented in Theorems 2.1 and 2.2 do not produce all relative complements of x in [a, b], even if the lattice L is modular. Consider the lattice L visualized in Figure 4. Then (a∨ y) ∧ x=d ∧ x=a and x∨(y ∧ b)=x∨ c=b. Thus the assumptions of Theorem 2.1 are satisfied. Hence, (a∨ y) ∧ b=d ∧ b=z₂ and a∨(y ∧ b)=a∨ c=z₂ belong to R(a, b, x). However, x has in [a, b] also another relative complement z₁ which is not obtained in this way. The reason is that z₁ is both join- and meet-irreducible and hence it cannot be a result of any term function composed by means of join and meet.

Example 2.4

On the other hand, R(a, b, x) is a convex set, thus if we can compute elements e and f as shown by Theorem 2.1 then also every element z in the interval [f, e] is a relative complement of x, see the lattice depicted in Figure 5. The assumptions of Theorem 2.1 are satisfied, i.e., (a∨ y) ∧ x=d ∧ x=a and x∨(y ∧ b)=x∨ c=b. Thus e, f ∈ R(a, b, x). Here f<z<e , thus z ∈ R(a, b, x).

If y is a complement of x and the triple (a, y, b) is modular, we can determine a relative complement of x in [a, b] by using further modular triples. In this case we have e=f, i.e. we compute only one relative complement of x in [a, b].

Figure 5.

Theorem 2.3

Let L=(L,∨,∧,0,1) be a bounded lattice, a, b ∈ L with a ⩽ b, x ∈ [a, b] and y a complement of x, assume (a, y, b) to be a modular triple and put e≔ (a∨ y) ∧ b=a∨(y ∧ b). Then the following are equivalent:

1. e ∈ R(a, b, x),

2. (a, y ∧ b, x) and (x, a∨ y, b) are modular triples.

Proof. Because of Theorem 2.1, (i) is equivalent to

Since

(3) is equivalent to

Finally, due to the definition of modular triples, (4) is equivalent to (ii). □

Remark 2.1

Consider the following well-known result mentioned in the introduction: If L=(L,∨,∧,0,1) is a bounded modular lattice, a, b ∈ L with a ⩽ b, x ∈ [a, b], y a complement of x and e≔ (a∨ y) ∧ b then e=a∨(y ∧ b) ∈ R(a, b, x). In the proof of this result only modularity of the triples (a, y, b), (a, y, x) and (x, y, b) is used. According to Theorem 2.3, modularity of these three triples must imply modularity of the triples (a, y ∧ b, x) and (x, a∨ y, b) which can be easily shown:

Proposition 3.1

Let P=(P,≤) be a distributive poset and x ∈ P. Then x has at most one complement.

Analogously as in Lemma 1.1, distributivity of (P, ⩽) can be replaced by a weaker condition. The following lemma extends Lemma 1.1 to posets.

Lemma 3.2

Let P=(P,≤) be a poset and x ∈ L. Assume that y and z are complements of x and that (x, y, z) and (x, z, y) are distributive triples. Then y=z.

Proof. We have

and hence y=z. □

Relations between complements and relative complements in posets were investigated in [2].

Again, modularity of (P, ⩽) can be replaced by some weaker conditions, see the following result.

Proposition 3.2

([2]) Let P=(P,≤) be a poset, a, b ∈ P with a ⩽ b, x ∈ [a, b] and y ∈ P and assume that e is the greatest element of L(U(a, y), b) and f the smallest element of U(a, L(y, b)). If

then e, f ∈ R(a, b, x).

Let us note that e is the greatest element of L\big(U(a, y), b\big) if and only if L\big(U(a, y), b\big)=L(e), and f is the smallest element of U\big(a, L(y, b)\big) if and only if U\big(a, L(y, b)\big)=U(f).

Now we show that the converse of Proposition 3.2 is also true, i.e. we can state the following theorem which is an extension of Theorem 2.1 to posets.

Theorem 3.1

Let P=(P,≤) be a poset, a, b ∈ P with a ⩽ b, x ∈ [a, b] and y ∈ P and assume that e is the greatest element of L(U(a, y), b) and f the smallest element of U(a, L(y, b)). Then a ⩽ f ⩽ e ⩽ b and the following are equivalent:

1. e, f ∈ R(a, b, x),

2. L(U(a, y), x)=L(a) and U(x, L(y, b))=U(b).

Proof. We have L(U(a, y), b)=L(e) and U(a, L(y, b))=U(f). Because of U(a, y)∪{b}⊆ U(a, L(y, b))=U(f), we conclude

and hence a ⩽ f ⩽ e ⩽ b. Moreover, we have

Since a ⩽ x and a ⩽ f ⩽ e, we have that L(x, e)=L(a) implies L(x, f)=L(a), and since x ⩽ b and f ⩽ e ⩽ b, we have that U(x, f)=U(b) implies U(x, e)=U(b). Hence (i) and (ii) are both equivalent to L(e, x)=L(a) and U(x, f)=U(b). □

Example 3.1

Consider the poset P depicted in Figure 6.

Clearly, P is not modular. Since

Figure 6.

the assumptions of Theorem 3.1 as well as (ii) of this theorem are satisfied and hence e and f are relative complements of x in [a, b].

Similarly as for lattices, if y is a complement of x then the existence of relative complements of x in [a, b] is assured by certain modular triples as follows.

Theorem 3.2

Let P=(P,≤) be a poset, a, b ∈ P with a ⩽ b, x ∈ [a, b] and y a complement of x and assume that e is the greatest element of L(U(a, y), b) and f the smallest element of U(a, L(y, b)). Then a ⩽ f ⩽ e ⩽ b and the following are equivalent:

1. e, f ∈ R(a, b, x),

2. (a, y, x) and (x, y, b) are modular triples.

Proof. Since U(a, y)∪{b}⊆ U(a, L(y, b))=U(f), we have

and hence a ⩽ f ⩽ e ⩽ b. Because of Theorem 3.1, (i) is equivalent to

Clearly, (5) is equivalent to

Since

(6) is equivalent to

Finally, due to the definition of modular triples, (7) is equivalent to (ii). □

If y is not assumed to be a complement of x, but the triple (a, y, b) is modular then we can determine a relative complement of x in [a, b] and, as in Theorem 2.3 for lattices, again e=f.

Theorem 3.3

Let P=(P,≤) be a poset, a, b ∈ L with a ⩽ b, x ∈ [a, b] and y ∈ P and assume (a, y, b) to be a modular triple and e to be the greatest element of L(U(a, y), b)=LU(a, L(y, b)). Then the following are equivalent:

1. e ∈ R(a, b, x),

2. L(U(a, L(y, b)), x)=L(a) and U(x, L(U(a, y), b))=U(b).

Proof. Since

and

(i) and (ii) are both equivalent to L(e, x)=L(a) and U(x, e)=U(b). □

If we add the assumption that y is a complement of x, then we obtain the following result.

Proposition 3.3

[4] Let P=(P,≤) be a modular poset, a, b ∈ P with a ⩽ b, x ∈ [a, b] and y a complement of x and assume e to be the greatest element of L(U(a, y), b)=LU(a, L(y, b)). Then e ∈ R(a, b, x).

The following theorem generalizes this proposition and Theorem 2.3.

Theorem 3.4

Let P=(P,≤) be a poset, a, b ∈ L with a ⩽ b, x ∈ [a, b] and y a complement of x and assume (a, y, b) to be a modular triple and e to be the greatest element of L(U(a, y), b)=LU(a, L(y, b)). Then the following are equivalent:

1. e ∈ R(a, b, x),

2. L(U(a, L(y, b)), x)=LU(a, L(y, b, x)) and L(U(x, a, y), b)=LU(x, L(U(a, y), b)).

Proof. Because of Theorem 3.3, (i) is equivalent to

Clearly, (8) is equivalent to

Finally, since

(9) is equivalent to (ii). □