By Stankey Burris, H. P. Sankappanavar

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Next we look at a (constructive) description of θ1 ∨ θ2 . 6. If θ1 and θ2 are two equivalence relations on A then θ1 ∨ θ2 = θ1 ∪ (θ1 ◦ θ2 ) ∪ (θ1 ◦ θ2 ◦ θ1 ) ∪ (θ1 ◦ θ2 ◦ θ1 ◦ θ2 ) ∪ · · · , or equivalently, a, b ∈ θ1 ∨ θ2 iff there is a sequence of elements c1 , c2 , . . , cn from A such that ci , ci+1 ∈ θ1 or ci , ci+1 ∈ θ2 for i = 1, . . , n − 1, and a = c1 , b = cn . Proof. It is not difficult to see that the right-hand side of the above equation is indeed an equivalence relation, and also that each of the relational products in parentheses is contained in θ1 ∨ θ2 .

We want to choose a basis A0 in K such that A0 is as close as possible to B, and such that the last ring which contains elements of A0 contains as few elements of A0 as possible. We choose one of the latter elements a0 and replace it by n or fewer closer elements b1 , . . , bm to obtain a new generating set A1 , with |A1 | < i + n. Then A1 contains an irredundant basis A2 . By the ‘minimal distance’ condition on A0 we see that A2 ∈ K, hence |A2 | > i, so |A2 | ≥ j by (∗). Thus j < i + n. Now for the details of this proof, choose A0 ∈ K such that A0 Cnk (B) imples A Cnk (B) for A ∈ K (see Figure 9).

If L is a distributive lattice, show that the set of ideals I(L) of L (see §2 Exercise 5) forms a distributive lattice. 3. Let (X, T ) be a topological space. A subset of X is regular open if it is the interior of its closure. Show that the family of regular open subsets of X with the partial order ⊆ is a distributive lattice. 4. If L is a finite lattice let J(L) be the poset of join irreducible elements of L (see §1 Exercise 10), where a ≤ b in J(L) means a ≤ b in L. Show that if L is a finite distributive lattice then L is isomorphic to L(J(L)) (see §2 Exercise 4), the lattice of nonempty lower segments of J(L).

### A Course in Universal Algebra by Stankey Burris, H. P. Sankappanavar

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