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

By Stankey Burris, H. P. Sankappanavar

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We will think of B as the center of S, and measure the distance from B using the “rings” Cnk+1(B) − Cnk (B). 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 (∗).

5. 6. 7. M. Cohn [9] G. G. I. H. S. Pierce [28] W. Taylor [35] Exercises §1 1. An algebra A, F is the reduct of an algebra A, F ∗ to F if F ⊆ F ∗ , and F is the restriction of F ∗ to F. , every element of S is such that its order divides n). 2. Two elements a, b of a bounded lattice L, ∨, ∧, 0, 1 are complements if a∨b = 1, a∧b = 0. In this case each of a, b is the complement of the other. A complemented lattice is a bounded lattice in which every element has a complement. (a) Show that in a bounded distributive lattice an element can have at most one complement.

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).