Definition 12. If s is a sublattice of L, L is an ertension of s In some case, we have the following special sublattice. Definition13. The subset s of the lattice l is called convex if a,b∈S,c∈L,anda≤c≤b imply that c∈S Given two lattices, they sometimes have the same topology structure by means of Hasse diagram Consider the following example Example 6. Given a poset ((1, 2, 3, 6, ), it is isomorphic to((a, b),S) They have the Hasse diagrams as shown in Figure 4. It is easy to construct a bijection between bl {a} (a)<{1,2,3,6},|> (b)<P({a,b}),g> Figure 4: Finite Boolean algebra example (1, 2, 3, 6 and P(a, b)). And the mapping is not unique Lattice is a specail type of a partial order. Generally, we have the assertion Theorem 14. Given two lattice L and L', a bijection f: L-L fromL to L' is an isomorphism f and only if a≤ b in L implies f(a)≤f(b)imL Proof. If f is isomorphic, f is order preserving. Suppose a s b, we have aUb=b. Then we have ∫(a∪b)=f(a)∪f(b)=f(b), which means f(a)≤f(). Conversely,f(a)≤f(b) also implies a≤b Suppose f is an order preserving bijection, f is isomorphic. Given a, b EL, let d=aUb, we h f(a)Uf(b)≤f(d) because a,b≤ d leads f(a),f(b)≤f(d). For any f(e)= f(a),f(b)≤e', we have a,b≤e. So aub=d≤e. Then we have f(d)≤f(e)=e, which means f(aub=f(au f(b). Similarly, we can prove f(anb)=f(anf(b) Exercies 1. Let A be the set of finitely generated subgroups of a group G, ordered by set inclusion. Prove that (A, C) is a join-semilattice, but not necessarily a lattice Show that every chain is a lattice 3. Prove that the absorption identities imply the idempotency of U and nDefinition 12. If S is a sublattice of L, L is an extension of S. In some case, we have the following special sublattice. Definition 13. The subset S of the lattice L is called convex if a, b ∈ S, c ∈ L, and a ≤ c ≤ b imply that c ∈ S. Given two lattices, they sometimes have the same topology structure by means of Hasse diagram. Consider the following example. Example 6. Given a poset ⟨{1, 2, 3, 6}, |⟩, it is isomorphic to ⟨P({a, b}), ⊆⟩. They have the Hasse diagrams as shown in Figure 4. It is easy to construct a bijection between 1 2 3 6 (a) < {1, 2, 3, 6}, | > ∅ {a} {b} {a, b} (b) < P({a, b}), ⊆> Figure 4: Finite Boolean algebra example {1, 2, 3, 6} and P({a, b}). And the mapping is not unique. Lattice is a specail type of a partial order. Generally, we have the assertion. Theorem 14. Given two lattice L and L ′ , a bijection f : L → L ′ from L to L ′ is an isomorphism if and only if a ≤ b in L implies f(a) ≤ f(b) in L ′ . Proof. If f is isomorphic, f is order preserving. Suppose a ≤ b, we have a ∪ b = b. Then we have f(a∪b) = f(a)∪f(b) = f(b), which means f(a) ≤ f(b). Conversely, f(a) ≤ f(b) also implies a ≤ b because f −1 is also a bijection. Suppose f is an order preserving bijection, f is isomorphic. Given a, b ∈ L, let d = a ∪ b, we have f(a) ∪ f(b) ≤ f(d) because a, b ≤ d leads f(a), f(b) ≤ f(d). For any f(e) = e ′ ∈ L ′ such that f(a), f(b) ≤ e ′ , we have a, b ≤ e. So a ∪ b = d ≤ e. Then we have f(d) ≤ f(e) = e ′ , which means f(a ∪ b) = f(a) ∪ f(b). Similarly, we can prove f(a ∩ b) = f(a) ∩ f(b). Exercies 1. Let A be the set of finitely generated subgroups of a group G, ordered by set inclusion. Prove that ⟨A, ⊆⟩ is a join-semilattice, but not necessarily a lattice. 2. Show that every chain is a lattice. 3. Prove that the absorption identities imply the idempotency of ∪ and ∩. 6