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1、1关于完备布尔代数的一点注解【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In1, 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 in2. As the same time, the question of we
2、ther a simple complete Boolean algebra exists is raised firstly in2. In3, 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 alg
3、ebras 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, especially4.2Recall that a frame or a locale is a complete lattice L, satisfying the infinite
4、distributive law:aL, SLaS= as: sSBy Pt(L) we mean the set of prime elements of a frame L; a frame L is said to be spatial, if for any aL ,a=pPt(A)|pa . 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 clos
5、ed 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=bL|ba . All more terminology and notation of locale theory which is not explained here i
6、s taken from5; for general background of complete Boolean algebra, we refer to6.Main ResultDefinition 1 A complete lattice L is said to be generated by a set, if there exists a subset B of L, satisfying 3the following conditions: any two elements of L is not compatible(a,bL , we have not ab) for any
7、 aL , a=bB|ba for any aL ,b0aB, abB|ba,bb0Lemma 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 SP(B), let f( S)=S. By the definition 1(2), f is surjective. In the following
8、we show that f is injective. Assuming there exist S1,S2P(B) with a=f(S1)=f(S2). If S1S2 , it is no problem to suppose that there exists an element b0S1 such that b0S2 and b0a, then a=bB|ba,bb0. It is contradictive to condition (3) of definition 1.Lemma 3 Let L be a complete Boolean algebra. For any
9、aL , if a=pPt(L)|p, then for any p0aPt(L),apPt(L)|pa,pp0 .Proof: Let a=pPt(L)|pa,pp0. we have p0a=p0pPt(L)|pa=p0p|pPt(L),pa=pPt(L)|pa,pp0=b .Since L is a Boolean algebra, p0a=(p00)a. If b=a, then (p00)a=a , so p00a . But p0a , so p0(p00) , contradictorily.4Lemma 4 If L is a spatial complete Boolean
10、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 coatom, 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 algebr
11、a, 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 in6, here we give an alternative proof of it. Partic
12、ularly, we pay attention to the property (3). It discovers the essential relation between two elements a and p in L with pPt(L) and pa , which is called essence prime in4. In the following, it plays an important role.Lemma 6 A subframe of spatial frame is spatial.Proof: Suppose L is a spatial frame,
13、 N is a subframe of L. For any pPt(L) , let p=bN|bp . Note that pPt(N) , for assuming that there exist two elements x,yN with xyp, also xyp , so xp or yp , moreover, xp or yp . For any 5aN , we havea=LpPt(L)|pa=NpPt(N)|pPt(L),pa, p=bN|bpi.e. N is spatial.Corollary 7 An atomic complete Boolean algebr
14、a has no atomicless complete subalgebra.Lemma 8 Let L be an complete Boolean algebra, each subalgebra of which is atomic. For any aL , if apPt(L)|pa, then a is an atom.Proof: Suppose that there exists an element a in L with apPt(L)|pa,pp0, a is not atomic, then we can find an element bL with 0 b a.
15、Let b=ax0|xa , it is easy to verify that B is a proper complete subalgebra of L. Since Pt(B)=pPt(L)|paa0 so B is not spatial. By theorem 5, B is atomicless, contradictorily.Lemma 9 Let L be a complete Boolean algebra. For any aL , if apPt(L)|pa implies that a is an atom, then L is spatial.Proof: We only need to show that for any atom aL,apPt(L)|pa . Let bpPt(L)|pa. If ab , since apPt(L)|pa implies that a is an atom, there no exist 6element c in L with a Theorem 10 A complete Boolean algebra is atomic if only if each subalgebra
