【摘要】  在本研究中，给出了完备布尔代数是原子的等价刻画。同时，证明了一个完备的布尔代数是原子的当且仅当每一个真子代数是原子的。

【关键词】  完备布尔代数 注解

　doi:10.3969/ j.issn. 10044337.2010.06.007 Introduction and preliminaries

　　In[1], an atomless complete Boolean algebra is called simple if it has no proper atomless complete subalgebra, the equivalence of which to rigid and minimal is proved in[2]. As the same time, the question of wether a simple complete Boolean algebra exists is raised firstly in[2]. In[3], a positive answer is given. Similarly, in the note, we will give some results of the complete atomic Boolean algebra as same as the complete atomless Boolean algebra.

　　It is well known that the properties of complete Boolean algebras correspond to properties of generic models obtained by forcing with these algebras, which the majority work on complete Boolean algebra come from. But the ideal of this note comes from locale theory, especially[4].

　　Recall that a frame or a locale is a complete lattice L, satisfying the infinite distributive law:a∈L, SLa∧∨S=∨{ a∧s: s∈S}

　　By Pt(L) we mean the set of prime elements of a frame L; a frame L is said to be spatial, if for any a∈L ,a=∧{p∈Pt(A)|p≥a} . The subframe of Frame L is a subset of it, which is closed under finite meets and arbitrary joins; A subset of complete Boolean algebra that is closed under arbitrary meets and arbitrary joins is called complete subalgebra. A complete Boolean algebra is a frame, the complete subalgebra of which is a subframe. The power set of set B is denoted by P(B); Let ↑a={b∈L|b≥a} . All more terminology and notation of locale theory which is not explained here is taken from[5]; for general background of complete Boolean algebra, we refer to[6].

　　Main Result

　　Definition 1 A complete lattice L is said to be generated by a set, if there exists a subset B of L, satisfying the following conditions:① any two elements of L is not compatible(a,b∈L , we have not a≤b)② for any a∈L , a=∧{b∈B|b≥a}③ for any a∈L ,b0∈↑a⌒B, a≠∧{b∈B|b≥a,b≠b0}Lemma 2 Let L be a complete lattice. L is isomorphic to the power set of a set if and only if it is generated by a set.Proof:  It is trivially. Suppose L is generated by B. For any SP(B), let f( S)=∧S. By the definition 1(2), f is surjective. In the following we show that f is injective. Assuming there exist S1,S2∈P(B) with a=f(S1)=f(S2). If S1≠S2 , it is no problem to suppose that there exists an element b0∈S1 such that b0S2 and b0≥a, then a=∧{b∈B|b≥a,b≠b0}. It is contradictive to condition (3) of definition 1.

　　Lemma 3 Let L be a complete Boolean algebra. For any a∈L , if a=∧{p∈Pt(L)|p≥}, then for any p0∈↑a⌒Pt(L),a≠∧{p∈Pt(L)|p≥a,p≠p0} .Proof: Let a=∧{p∈Pt(L)|p≥a,p≠p0}. we have p0…a=p0→∧{p∈Pt(L)|p≥a}=∧{p0→p|p∈Pt(L),p≥a}=∧{p∈Pt(L)|p≥a,p≠p0}=b .Since L is a Boolean algebra, p0→a=(p0→0)∨a. If b=a, then (p0→0)∨a=a , so p0→0≤a . But p0≥a , so p0≥(p0→0) , contradictorily.

　　Lemma 4 If L is a spatial complete Boolean algebra, L is generated by Pt(L).Proof: L is spatial, the condition 1 of the definition 1 is satisfied. Since L is a Boolean algebra, the element of Pt(L) is a coatom, so any two elements Pt(L) are not compatible. The condition 3 can be obtained by lemma 3.

　　Theorem 5 If L is a complete Boolean algebra, then the following properties are equivalence:　　(1) L is atomic;(2) L is spatial;(3) L is generated by a set;Proof: (1) (2) By the duality of Boolean algebras.(2) (3) By the lemma 4.(3) (1) It is trivially.Remark: Some of the theorem can be found in[6], here we give an alternative proof of it. Particularly, we pay attention to the property (3). It discovers the essential relation between two elements a and p in L with p∈Pt(L) and p≥a , which is called essence prime in[4]. In the following, it plays an important role.

　　Lemma 6 A subframe of spatial frame is spatial.

　　Proof: Suppose L is a spatial frame, N is a subframe of L. For any p∈Pt(L) , let p′=∨{b∈N|b≤p} . Note that p′Pt(N) , for assuming that there exist two elements x,y∈N with x∧y≤p′, also x∧y≤p , so x≤p or y≤p , moreover, x≤p′ or y≤p′ . For any a∈N , we havea=∧L{p∈Pt(L)|p≥a}=∧N{p′∈Pt(N)|p∈Pt(L),p≥a, p′=∨{b∈N|b≤p}}

　　i.e. N is spatial.

　　Corollary 7 An atomic complete Boolean algebra has no atomicless complete subalgebra.Lemma 8 Let L be an complete Boolean algebra, each subalgebra of which is atomic. For any a∈L , if a≠∧{p∈Pt(L)|p≥a}, then a is an atom.Proof: Suppose that there exists an element a in L with a≠∧{p∈Pt(L)|p≥a,p≠p0}, a is not atomic, then we can find an element b∈L with 0< b< a. Let b=↑a∪{x→0|x∈↑a} , it is easy to verify that B is a proper complete subalgebra of L. Since Pt(B)={p∈Pt(L)|p≥a}∪{a→0} so B is not spatial. By theorem 5, B is atomicless, contradictorily.

　　Lemma 9 Let L be a complete Boolean algebra. For any a∈L , if a≠∧{p∈Pt(L)|p≥a} implies that a is an atom, then L is spatial.Proof: We only need to show that for any atom a∈L,a∧{p∈Pt(L)|p≥a} . Let b∧{p∈Pt(L)|p≥a}. If a≠b , since a≠∧{p∈Pt(L)|p≥a} implies that a is an atom, there no exist element c in L with a Theorem 10 A complete Boolean algebra is atomic if only if each subalgebra of it is atomic.

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