UDC 519.48+513.83

V. V. PASHENKOV

ON STONE’S THEOREM

(Presented by Academician P. S. Aleksandrov on 17 III 1970)

We shall agree to understand by the word space a Hausdorff topological space possessing a prebase of open-and-closed sets. By a topological algebra \(B\) we shall understand a Boolean algebra which is at the same time a space whose topology is given by a prebase consisting of certain ultrafilters \(F\) and their complements \(I=B\setminus F\)* (ultraideals). The signs \(+\), \(\cdot\), \({}'\), \(\Delta\) will be used to denote, respectively, the operations of taking the sum, intersection, complement, and symmetric difference in a Boolean algebra. Set-theoretic operations of union, intersection, and difference will be denoted, respectively, by \(\cup\), \(\cap\), \(\setminus\); \(\varnothing\) denotes the empty set. The zero and unit of a Boolean algebra will be denoted by \(0\) and \(1\), respectively.

Definition 1. A system of subsets \(\{U_\alpha\}\) of a space \(X\) will be called defining if the collection of sets \(\{U_\alpha\}\cup\{X\setminus U_\alpha\}\) forms a prebase for the topology in \(X\). Each of the subsets \(U_\alpha\) and \(X\setminus U_\alpha\) will be called marked with respect to the defining system \(\{U_\alpha\}\).

Lemma 1. Let \(I,F_1,F\) be, respectively, an ideal and filters of a Boolean algebra \(B\). If \(I\cap F_1=\varnothing\) and \(I\cap F_1\subseteq F\), then \(F_1\subseteq F\).

Lemma 2. Let \(F_1,F_2,\ldots,F_n\) be a finite collection of proper filters of a Boolean algebra \(A\), and let \(F\) be an ultrafilter distinct from \(F_i\), \(i=1,2,\ldots,n\). Then
\[ F\not\subseteq\bigcup_{i=1}^{n} F_i . \]
If each of the filters \(F_i\) is an ultrafilter, then, moreover,
\[ F\not\subseteq\bigcap_{i=1}^{n} F_i . \]

Lemma 3. The algebraic operations \(+\), \(\cdot\), \({}'\), \(\Delta\) are continuous in any topological algebra \(B\).

Proof. Let \(a,b,c\in B\), \(a=bc\), and let \(U\) be some open set containing the element \(a\). Then for some open ultrafilters \(F_i\) and open ultraideals \(I_j\) the relations
\[ a\in F_1\cap F_2\cap\ldots\cap F_n\cap I_1\cap I_2\cap\ldots\cap I_k\subseteq U \tag{1} \]
will hold. For each ideal \(I_j\), in view of its maximality, at least one of the relations \(b\in I_j\), \(c\in I_j\) must hold. Let \(I_{j_1},I_{j_2},\ldots,I_{j_s}\) be the collection of all those ideals which occur in the notation (1) and do not contain the element \(b\). Then \(c\in I=I_{j_1}\cap\ldots\cap I_{j_s}\). If \(k=s\), then we take as \(V\) the set \(F=B\cap F_1\cap\ldots\cap F_n\); if \(s<k\), then we put \(V=F\cap I_0\), where \(I_0\) is the intersection of all ideals from (1) distinct from \(I_{j_1},\ldots,I_{j_s}\). Clearly, \(b\in V\), \(c\in W=I\cap F\), and for any elements \(v\in V\) and \(w\in W\) the relation \(v\cdot w\in U\) is valid. If \(a'\) is the complement of the element \(a\) in the Boolean algebra \(B\), then the set
\[ H=(B\setminus F_1)\cap\ldots\cap(B\setminus F_n)\cap(B\setminus I_1)\cap\ldots\cap(B\setminus I_k) \]
will contain the element \(a'\), and for any element \(h\in H\) the relation \(h'\in U\) is valid. Since the sets \(V,W\), and \(H\) are open, we have thereby proved the continuity of the operations of taking intersection and complement. It remains to note that for any elements \(a,b\in B\) the equalities \(a+b=(a'b')'\) and \(a\Delta b=a'b+b'a\) hold.

Lemma 4. Let \(\{F_\alpha\}\) be a defining system of ultrafilters of a topological algebra \(B\), and let \(F\) be some ultrafilter in \(B\). The following conditions are equivalent: 1. \(F\in\{F_\alpha\}\). 2. \(F\) is open-and-closed in \(B\). 3. \(F\) is closed in \(B\). 4. \(F\) is open in \(B\).

* By an ultrafilter we mean a proper maximal filter.

Proof. The implications \(1 \Rightarrow 2 \Rightarrow 3\) are obvious.
\(3 \Rightarrow 4\). By Lemma 3, the topological algebra \(B\) is a continuous group with respect to the operation \(\Delta\). The ideal \(I = B \setminus F\), being an open subgroup in \(B\), is closed. Consequently, \(F\) is open.

\(4 \Rightarrow 1\). We shall show that if \(F \notin \{F_\alpha\}\), then \(F\) cannot contain any nonempty open set at all. Assuming the contrary, we would have \(F \supset F_0 \cap I \ne \varnothing\), where \(F_0\) and \(I\) are intersections of a finite number of marked ultrafilters and ultraideals, respectively. But then, by Lemma 1, we have \(F_0 \subseteq F\), which contradicts Lemma 2.

Definition 2. Let \(B\) be a topological algebra. Denote the set of all its closed ultrafilters by \(T(B)\). For each element \(b \in B\) define the subset \(b_\tau = T(B)\), consisting of all those ultrafilters \(F \in T(B)\) that contain the element \(b\). Let \(b, b_1, b_2,\ldots,b_n\) be arbitrary elements of the topological algebra \(B\). The following relations are easily verified:

\[ (b')_\tau = T(B) \setminus b_\tau, \tag{2} \]

\[ (b_1 \cdot b_2 \cdots b_n)_\tau = (b_1)_\tau \cap (b_2)_\tau \cap \cdots \cap (b_n)_\tau, \tag{3} \]

\[ (b_1 + b_2 + \cdots + b_n)_\tau = (b_1)_\tau \cup (b_2)_\tau \cup \cdots \cup (b_n)_\tau. \tag{4} \]

It is also easy to see that if \(F_1\) and \(F_2\) are distinct ultrafilters from \(T(B)\), then there exists an element \(c \in B\), \(c \in F_1 \setminus F_2\). Then \(F_1 \in c_\tau\) and \(F_2 \in (c')_\tau\). Therefore, taking as a defining system in the set \(T(B)\) the subsets \(b_\tau\) for all \(b \in B\), we turn \(T(B)\) into a space. Note that, by relations (2) and (3), the subsets \(\{b_\tau\}\), \(b \in B\), form a base of open sets of the space \(T(B)\).

Definition 3. Let \(K\) be a subset of the Boolean algebra \(B\). The set
\[ K^*=\{b\in B:\ bc=0\ \text{for all } c\in K\} \]
will be called the disjoint complement of the set \(K\). If \(K^{**}=(K^*)^*=K\), then \(K\) will be called an algebraic component, or simply a component, of the Boolean algebra \(B\). A component \(K\) will be called principal if there exists an element \(c_0 \in K\) such that \(K=\{b\in B;\ b\le c_0\}\). Otherwise the component will be called nonprincipal.

Definition 4. A topological algebra \(B\) will be called closed if, for every nonprincipal component \(K\) of it, there exists a closed ultrafilter \(F\) in \(B\) for which the relation
\[ F\cap K=F\cap K^*=\varnothing \]
holds.

Lemma 5. Let \(B\) be a closed topological algebra and let \(U\) be a clopen subset in \(T(B)\). Then there exists an element \(b_0\in B\) for which \((b_0)_\tau=U\).

Proof. Denote by \(K\) the set of all those elements \(b_\alpha\) of the topological algebra \(B\) for which \((b_\alpha)_\tau \subseteq U\), and by \(P\) the set of all those elements \(b_\beta\) for which \((b_\beta)_\tau \subseteq V = T(B)\setminus U\); \(K=\{b_\alpha\}\), \(P=\{b_\beta\}\). Note that
\[ \bigcup_\alpha (b_\alpha)_\tau = U,\qquad \bigcup_\beta (b_\beta)_\tau = V. \]
We prove the equality

\[ P=K^*. \tag{5} \]

Let \(a\in P\), i.e. \(a_\tau\subseteq V\). Then \(a_\tau\cap U=\varnothing\), hence, for all \(\alpha\),
\[ a_\tau\cap (b_\alpha)_\tau=\varnothing, \]
whence \(ab_\alpha=0\); consequently, \(a\in K^*\). Conversely, let \(a\in K^*\), i.e. \(ab_\alpha=0\) for all \(\alpha\). Then \(a_\tau\cap (b_\alpha)_\tau=\varnothing\), whence \(a_\tau\cap U=\varnothing\), i.e. \(a_\tau\subseteq V\). Consequently, \(a\in P\). Interchanging the roles of \(U\) and \(V\) in this argument, from (5) we obtain \(P^*=K\). Consequently,
\[ K^{**}=(K^*)^*=P^*=K, \]
i.e. \(K\) is a component. Since \(U\cup V=T(B)\), every closed ultrafilter \(F\) of the topological algebra \(B\) belongs either to the set \(U\) or to the set \(V\). But this means that either \(F\cap K\ne\varnothing\) or \(F\cap K^*\ne\varnothing\). Therefore, by the closedness of the topological algebra \(B\), the component \(K\) is principal. Let \(b_0\in K\) and, for all \(\alpha\), \(b_0\ge b_\alpha\). Then it is clear that \((b_0)_\tau=U\).

Definition 5. Let \(X\) be a space. The Boolean algebra of all its clopen subsets, ordered by inclusion, is denoted…

that is, by \(A(X)\). Denote by \(F_x\) the ultrafilter of the Boolean algebra \(A(X)\) consisting of all open-closed subsets of the space \(X\) that contain the point \(x \in X\). We introduce a topology in \(A(X)\) by taking as a defining system the sets \(F_x\) for all \(x \in X\). It is clear that this will be a Hausdorff topology. Therefore the algebra \(A(X)\) will be a topological algebra.

Lemma 6. For any space \(X\), the topological algebra \(A(X)\) is closed.

Proof. Suppose, contrary to the assertion of the lemma, that \(K=\{a_\alpha\}\) is such a nonprincipal component of the topological algebra \(A(X)\) that in \(A(X)\) there is no closed ultrafilter \(F\) for which \(F \cap K = F \cap K^* = \varnothing\). Let \(K^*=\{a_\beta\}\). Considering the elements of the topological algebra as subsets of the space \(X\), form the sets \(U=\bigcup_\alpha a_\alpha\) and \(V=\bigcup_\beta a_\beta\). It is easy to see that \(U \cap V=\varnothing\). Let us show that \(U \cup V=X\). Indeed, any closed ultrafilter \(F\) in \(A(X)\), by Lemma 4, is distinguished, \(F=F_x\). If \(F_x \cap K \ne \varnothing\), then for some \(a_\alpha \in K\) we have \(a_\alpha \in F_x\), whence \(x \in a_\alpha\). If \(F \cap K^* \ne \varnothing\), then analogously we obtain \(x \in a_\beta\). Thus every point \(x \in X\) belongs to the set \(U \cup V\), i.e. \(U \cup V=X\). Since each of the sets \(U\) and \(V\) is open, \(U\) is open-closed, and hence \(U=a\) for some \(a \in A(X)\). Since \(U a_\beta=0\) for all \(\beta\), we have \(U=a \in K^{**}=K\); moreover, \(U=a \ge a_\alpha\) for all \(\alpha\), and hence the component \(K\) is principal, which contradicts our assumption.

Theorem 1. Every space \(X\) is homeomorphic to the space \(T(A(X))\). Every closed topological algebra \(B\) is topologically isomorphic to the topological algebra \(A(T(B))\).

Proof. To each point \(x\) of the space \(X\) assign the closed ultrafilter \(\varphi(x)=F_x\) of the topological algebra \(A(X)\). Since the elements of the topological algebra \(A(X)\) form a base in \(X\), if \(x_1 \ne x_2\) then \(\varphi(x_1) \ne \varphi(x_2)\). Since every closed ultrafilter \(F\) of the topological algebra \(A(X)\), by Lemma 4, is distinguished, \(F=F_x\) for some \(x \in X\), we map this space \(X\) onto the whole space \(T(A(X))\). It is easy to verify that for any open-closed subset \(a \subseteq X\), \(a \in A(X)\), the relation \(\varphi(a)=a_\tau\) holds. Since the subsets \(a \subseteq X\) form a base in the space \(X\), and the subsets \(\varphi(a)=a_\tau\) form a base in the space \(T(A(X))\), the mapping \(\varphi\) is a homeomorphism.

Now let the topological algebra \(B\) be closed. Assign to each element \(b \in B\) the set \(f(b)=b_\tau\), \(b_\tau \subseteq T(B)\). It is clear that if \(b_1 \ne b_2\), then \(f(b_1) \ne f(b_2)\). Since, in view of the closedness of the topological algebra \(B\), by Lemma 5 every open-closed subset of the space \(T(B)\) has the form \(b_\tau\), where \(b \in B\), \(f\) maps \(B\) onto the whole topological algebra \(A(T(B))\). Moreover, by the relations (2), (3), (4), \(f\) is an algebraic isomorphism. Consequently, \(f\) carries ultrafilters of the algebra \(B\) to ultrafilters of the algebra \(A(T(B))\). Let \(F=\{b_\alpha\}\) be a closed ultrafilter of the topological algebra \(B\). Then \(H=f(F)\) consists of the subsets \((b_\alpha)_\tau\), which are open-closed in the space \(T(B)\). From the inclusion \(b_\alpha \in F\) it follows that \(F \in (b_\alpha)_\tau\) for all \(\alpha\). Hence the ultrafilter \(H\) consists of subsets containing the point \(F\) of the space \(T(B)\), i.e. \(H\) is distinguished, and therefore open-closed. Conversely, if \(H=\{(b_\alpha)_\tau\}\) is open-closed in the topological algebra \(A(T(B))\), then, by Lemma 4, \(H\) is a distinguished ultrafilter, i.e. there exists a point \(F_1\) of the space \(T(B)\) (\(F_1\) is a closed ultrafilter in \(B\)) such that \(F_1 \in (b_\alpha)_\tau\) for all \(\alpha\), hence \(F_1 \ni b_\alpha\), and therefore \(F_1 \supseteq F\), whence \(F_1=F\). Thus \(f^{-1}(H)=F\) is an open-closed set in the topological algebra \(B\). This proves the topological character of the isomorphism \(f\).

Corollary 1. To each space \(X\) there corresponds, in view of Lemma 6, the closed topological algebra \(A(X)\). To each closed topological-

Boolean algebra \(B\) corresponds to the space \(T(B)\). These correspondences are mutually inverse and one-to-one.

Corollary 2. The bicompactness of the space \(X\) is equivalent to the fact that in the topological algebra \(A(X)\) all ultrafilters are distinguished. In this case any topological algebra is closed, and our theorem turns into the classical Stone theorem.

Theorem 2. The following conditions are equivalent: 1. The space \(X\) is discrete. 2. The topology of the topological algebra \(A(X)\) coincides with the interval topology. 3. The topology of the topological algebra \(A(X)\) is bicompact.

Proof. \(1 \Rightarrow 2\). We note that the topology of any topological algebra \(A\) is stronger (not weaker) than the interval topology. Let \(X\) be discrete. Then any closed ultrafilter in \(A(X)\) has the form \([p,1]\), where \(p\) is an atom in \(A(X)\), and its complement has the form \([0,p']\), which are, obviously, closed in the interval topology.

\(2 \Rightarrow 1\). It suffices to show that any ultrafilter closed in the interval topology is principal. Let \(F \subseteq A(X)\) be a closed ultrafilter that is not principal. We have \(F=\bigcap_\alpha K_\alpha\), where

\[ K_\alpha=[a_1^\alpha,b_1^\alpha]\cup [a_2^\alpha,b_2^\alpha]\cup\ldots\cup [a_{n(\alpha)}^\alpha,b_{n(\alpha)}^\alpha] \]

and for all \(\alpha\) the number \(n(\alpha)\) is finite. If among the elements \(a_1^\alpha,a_2^\alpha,\ldots,a_{n(\alpha)}^\alpha\) there is no zero element, then all the filters \([a_1^\alpha,1],\ldots,[a_{n(\alpha)}^\alpha,1]\) are proper filters distinct from \(F\). Therefore, in view of Lemma 2, we have

\[ F\not\subseteq [a_1^\alpha,1]\cup\ldots\cup [a_{n(\alpha)}^\alpha,1], \]

whence \(F\not\subseteq K_\alpha\), which contradicts \(F=\bigcap_\alpha K_\alpha\subseteq K_\alpha\). Hence for all \(\alpha\) we obtain \(0\in K_\alpha\), whence \(0\in \bigcap_\alpha K_\alpha=F\), which is impossible.

\(1 \Rightarrow 3\). If the space \(X\) is discrete, then the Boolean algebra \(A(X)\) is complete, and by what has been proved the topology on \(A(X)\) coincides with the interval topology. Hence \((^1)\), \(A(X)\) is bicompact.

\(3 \Rightarrow 1\). We show that any closed ultrafilter of the topological algebra \(A(X)\) is principal. Suppose that \(F\) is a closed ultrafilter that is not principal; let \(\{F_\alpha\}\) be the set of all closed ultrafilters distinct from \(F\). We shall prove that

\[ F\subseteq \bigcup_\alpha F_\alpha . \tag{6} \]

Indeed, for any element \(a\in F\) one can find elements \(b\in A(X)\) and \(c\in A(X)\) such that \(b\ne a\), \(c\ne a\), and \(b+c=a\). Then one of the elements \(b,c\) belongs to \(F\). Suppose, for example, that \(b\in F\). By the Hausdorff property of the topological algebra \(A(X)\), there exists a closed ultrafilter \(F_{\alpha_0}\) containing the element \(c\). Then \(F_{\alpha_0}\ne F\) and \(F_{\alpha_0}\ni a\). In view of the arbitrariness of the element \(a\), relation (6) is proved. However, by Lemma 2, from the covering (6) of the closed set \(F\) one cannot extract a finite subcover, which contradicts the bicompactness of the topological algebra \(A(X)\).

Received
18 II 1970

References

1. G. Birkhoff, Lattice Theory, 1952, p. 97, Theorem 15.