Theorem 2. Let l be a lattice and let h and I be nonempty subsets of L 1. I is an ideal if and only if the following two conditions hold (a)a,b∈ I implies that a b∈I (b)I is a down-set 2.I= id(h)if and only if I={x|x≤hoU…Uhn-1 for some n≥1 and ho,…,hn-1∈H} Fora∈L id(a)={∩alx∈D} Proof. We prove the theorem one by one. 1. Let I be an ideal. Then a, bE I implies that a I, since I is a sublattice, verifying(a) Ifx≤a∈I, then ar=rna∈r,and(b) is verified. Conversely, let I satisfy(a)and(b). Let a,b E I. Then aUb E I by(a), and, since anb≤a∈I, we also have anb∈Iby(b); thus i is a sublattice.Finl,fr∈ L and a∈I, then an z≤a∈I, thus ana∈rby(b), proving that I is an ideal 2. Let Io be the set on the right side of the displayed formula in 2. Using 1, it is clear that Io is an ideal, and obviously H C lo. Finally, if H CJ and J is an ideal, then Io Cj, and thus To is the smallest ideal containing H: that is, I= lo 3. This proof is obviously simple or direct to applying 2 It is proved 3 Special Lattice Before Boolean lattice is given, we first introduce some lattice with some special properties Definition 3. A lattice L is complete if any(finite or infinite) subset A=ailie I where I is a subset of inder set of L, has a least upper bound Uierai and a greatest lower bound nierai We all know Q, rational number, is close under +,-,x,:. But when you consider a special infinite subset in which there is a sequence converging to a irrational number. The irrational number is the bound but it does not in Q. R instead of Q is complete, which is the truth we have already learned in Calculus Definition 4. A lattice L is bounded if it has a greatest element 1 and a least element o Theorem 5. Finite lattice L=(a1,.,an, is bounded Proof. Let 1= Ui=1@i and 0=n1aTheorem 2. Let L be a lattice and let H and I be nonempty subsets of L. 1. I is an ideal if and only if the following two conditions hold: (a) a, b ∈ I implies that a ∪ b ∈ I, (b) I is a down-set. 2. I = id(H) if and only if I = {x|x ≤ h0 ∪ · · · ∪ hn−1 for some n ≥ 1 and h0, . . . , hn−1 ∈ H}. 3. For a ∈ L, id(a) = {x ∩ a|x ∈ L}. Proof. We prove the theorem one by one. 1. Let I be an ideal. Then a, b ∈ I implies that a ∪ b ∈ I, since I is a sublattice, verifying (a). If x ≤ a ∈ I, then x = x ∩ a ∈ I, and (b) is verified. Conversely, let I satisfy (a) and (b). Let a, b ∈ I. Then a ∪ b ∈ I by (a), and, since a∩b ≤ a ∈ I, we also have a∩b ∈ I by (b); thus I is a sublattice. Finally, if x ∈ L and a ∈ I, then a ∩ x ≤ a ∈ I, thus a ∩ x ∈ I by (b), proving that I is an ideal. 2. Let I0 be the set on the right side of the displayed formula in 2. Using 1, it is clear that I0 is an ideal, and obviously H ⊆ I0. Finally, if H ⊆ J and J is an ideal, then I0 ⊆ J, and thus I0 is the smallest ideal containing H; that is, I = I0. 3. This proof is obviously simple or direct to applying 2. It is proved. 3 Special Lattice Before Boolean lattice is given, we first introduce some lattice with some special properties. Definition 3. A lattice L is complete if any(finite or infinite) subset A = {ai |i ∈ I} where I is a subset of index set of L, has a least upper bound ∪i∈Iai and a greatest lower bound ∩i∈Iai. We all know Q, rational number, is close under +, −, ×, ÷. But when you consider a special infinite subset in which there is a sequence converging to a irrational number. The irrational number is the bound but it does not in Q. R instead of Q is complete, which is the truth we have already learned in Calculus. Definition 4. A lattice L is bounded if it has a greatest element 1 and a least element 0. Theorem 5. Finite lattice L = {a1, . . . , an} is bounded. Proof. Let 1 = ∪ n i=1ai and 0 = ∩ n i=1ai . 2