Mathematics

Yu. Flachsmeyer

On a System of Canonical Open (Closed) Sets

(Presented by Academician P. S. Aleksandrov on 8 I 1964)

In this note it is established which topological spaces have isomorphic systems of canonical open (closed) sets.

A set \(G\) of a topological space \(X\) is called canonically open if it is the open kernel of a closed set. Let \(R_0(X)\) denote the system of all canonically open sets of the space \(X\). Obviously:
\[ G \in R_0(X) \Longleftrightarrow G=\operatorname{Int}\overline{G}. \]
By a canonically closed set is meant the complement of an open canonical set, or, equivalently, the closure of an open set. We denote the system of all canonically closed sets of the space \(X\) by \(R_{\mathfrak A}(X)\), i.e.
\[ F \in R_{\mathfrak A}(X) \Longleftrightarrow F=\overline{\operatorname{Int}F}. \]

In what follows we consider \(R_0(X)\) as ordered by the inclusion relation (similarly \(R_{\mathfrak A}(X)\)). The partially ordered sets \((R_0(X),\subset)\) and \((R_{\mathfrak A}(X),\subset)\) form Boolean structures. These structures are even complete (see \((^2)\)). For the structure operations we have:
\[ (R_0(X),\subset)\quad (R_{\mathfrak A}(X),\subset). \]
1. Structural union
\[ G\vee H=\operatorname{Int}\overline{(G\cup H)},\qquad F\vee K=\overline{\operatorname{Int}(F\cup K)}. \]
2. Structural intersection
\[ G\wedge H=\overline{G\cap H},\qquad F\wedge K=\overline{\operatorname{Int}(F\cap K)}. \]
3. Structural complement
\[ G'=\operatorname{Int}CG,\qquad F'=CF. \]

The operator \(G\to \overline{G}\) (conversely \(F\to \operatorname{Int}F\)) establishes a structural isomorphism between \((R_0(X),\subset)\) and \((R_{\mathfrak A}(X),\subset)\).

As preparation for our main result, we note the following known fact about canonical open (closed) sets of an arbitrary extension.

Lemma 1. If \(X\) is any topological space and \(eX\) is any extension of it (i.e. \(X\) is dense in \(eX\)), then \(X\) and \(eX\) possess isomorphic systems of canonical open (closed) sets. Namely:

a) \(G\to G\cap X\) is an isomorphism between \((R_0(eX),\subset)\) and \((R_0(X),\subset)\);

b) \(H\to O(H)^*\) is the inverse isomorphism to a);

c) \(F\to F\cap X\) is an isomorphism between \((R_{\mathfrak A}(eX),\subset)\) and \((R_{\mathfrak A}(X),\subset)\);

d) \(K\to \overline{K}^{\,eX}\) is the inverse isomorphism to c).

Remark. P. S. Aleksandrov \((^1)\) proved that the given operators map one-to-one the systems \(R_0(X)\), respectively \(R_{\mathfrak A}(X)\), of a normal space \(X\) onto the systems \(R_0(\beta X)\), respectively \(R_{\mathfrak A}(\beta X)\), of the Stone–Čech extension \(\beta X\). Yu. M. Smirnov noted \((^7)\) that this is true for arbitrary \(X\) and \(eX\). Theorem 5 of his paper \((^7)\) is equivalent to assertions a) and b) of the lemma, because the operators are, obviously, monotone.

The proof is elementary; it is based on the identities:
\[ \text{For a set } O \text{ open in } eX:\quad O\cap \overline{X}^{\,X}=O^{eX}\cap X. \tag{1} \]
\[ \text{For a set } F \text{ closed in } eX:\quad \operatorname{Int}_X(F\cap X)=X\cap \operatorname{Int}_{eX}F. \tag{2} \]

It is easy to see that formulas (1) and (2) cease to be true if one does not require in them the openness of the set \(O\) or the closedness of the set \(F\).

\[ \rule{0.35\linewidth}{0.4pt} \]

* \(O(H)\) is the largest open set in \(eX\) whose trace in \(X\) is \(H\), i.e.
\[ O(H)=\bigcup G \quad (G \text{ open in } eX), \]
such that \(G\cap eX=H\). Obviously,
\[ O(H)=eX\setminus (X\setminus H)^{eX}. \]
The operator \(O(H)\) was introduced by Shanin (see \((^7)\)).

The operator \(G \to G \cap X\) maps \(R_0(eX)\) into \(R_0X\); namely, from relations (1) and (2) it follows that
\[ \operatorname{Int}_X (G \cap X)^X = \operatorname{Int}_X (X \cap \overline{G}^{\,eX}) = X \cap \operatorname{Int}_{eX}\overline{G}^{\,eX} = X \cap G. \]
On the other hand, this operator maps \(R_0(eX)\) onto \(R_0(X)\), because
\[ H=\operatorname{Int}_X \overline{H}^{\,X} =\operatorname{Int}_X (X\cap \overline{H}^{\,eX}) = X\cap \operatorname{Int}_{eX}(\overline{H}^{\,eX}) \]
for \(H\in R_0(X)\). For \(G\in R_0(eX)\) we have
\[ G=\operatorname{Int}_{eX}\overline{(G\cap X)}^{\,eX}; \]
hence the operator is one-to-one, and therefore (since the operator is obviously monotone) \(G\to G\cap X\) gives an isomorphism between \((R_0(eX),\subset)\) and \((R_0(X),\subset)\).

For the other operators we note only the relations
\[ O(H)=\operatorname{Int}_{eX}(\overline{H}^{\,eX}) \]
for \(H\in R_0(X)\), and
\[ \overline{F\cap X}^{\,eX}=\overline{F} \]
for \(F\in R_{\mathfrak A}(eX)\).

Lemma 2*. Let \(f:X\to Y\) be a mapping of the space \(X\) onto the space \(Y\). If: 1) \(f\) is closed, 2) irreducible, 3) for every neighborhood \(V\) of any point \(f(x)\) there exists a neighborhood \(U\) of the point \(x\) such that \(f(U)\subset V\)*, then the operator \(F\to f(F)\) is a structural isomorphism between \((R_{\mathfrak A}(X),\subset)\) and \((R_{\mathfrak A}(Y),\subset)\).

The proof consists of three steps:
\[ \text{For every }F\in R_{\mathfrak A}(X)\text{ we have }f(F)=C\operatorname{Int} f(F'). \tag{1} \]
\[ \text{For every }H\in R_{\mathfrak A}(Y)\text{ we have }f^{-1}(\operatorname{Int}H)\in R_{\mathfrak A}(X),\quad f(f^{-1}(\operatorname{Int}H))=H. \tag{2} \]
\[ \text{For }F_1,F_2\in R_{\mathfrak A}(X)\text{ we have }f(F_1)=f(F_2)\text{ when }F_1\ne F_2. \tag{3} \]

Step 1. Since \(F\in R_{\mathfrak A}(X)\), \(CF=\operatorname{Int}F'\); consequently, from
\[ Cf(F)\subset f(CF) \]
it follows that
\[ Cf(F)\subseteq f(F'); \]
\(f\) is closed, and therefore \(Cf(F)\) is open, and
\[ Cf(F)\subseteq \operatorname{Int}f(F'), \]
i.e.
\[ f(F)\supseteq C\operatorname{Int}f(F'). \]
Now let \(y\in f(F)\), and suppose that \(y\in \operatorname{Int}f(F')\). For some \(x\in F\) we have \(f(x)=y\). From (3) there follows the existence of an open neighborhood \(V\) of the point \(x\) such that
\[ f(V)\subseteq \overline{\operatorname{Int}f(F')}\subseteq f(F'). \]
The open set \(V\) intersects \(\operatorname{Int}F\). From (2) there follows the existence of a point \(y_1\in Y\) for which
\[ f^{-1}(y_1)\subseteq V\cap \operatorname{Int}F. \]
But the point \(y_1\) must be in \(f(F')\), although
\[ f^{-1}(y_1)\cap F=\varnothing. \]

Step 2. Let \(H\in R_{\mathfrak A}(Y)\) and \(x\in f^{-1}(\operatorname{Int}H)\). From (3) there follows the existence of an open neighborhood \(U\) of the point \(x\) such that
\[ f(U)\subseteq \overline{\operatorname{Int}H}=H. \]
Then
\[ U\subseteq f^{-1}(\operatorname{Int}H), \]
because otherwise the open set
\[ V=U\cap C\overline{f^{-1}(\operatorname{Int}H)} \]
would contain \(f^{-1}(zy)\) for some \(y\in Y\). Then
\[ y\in f\bigl(\overline{f^{-1}(\operatorname{Int}H)}\bigr)\supset \operatorname{Int}H. \]
Further,
\[ f\bigl(\overline{f^{-1}(\operatorname{Int}H)}\bigr) \]
is closed; consequently
\[ f\bigl(\overline{f^{-1}(\operatorname{Int}H)}\bigr)\supseteq \overline{\operatorname{Int}H}=H, \]
and \(y\in H\), although \(y\in f(U)\subseteq H\). Hence
\[ f^{-1}(\operatorname{Int}H)\subseteq \operatorname{Int}\overline{f^{-1}(\operatorname{Int}H)}, \]
therefore
\[ f^{-1}(\operatorname{Int}H)\in R_{\mathfrak A}(X). \]
To prove the equality
\[ f\bigl(\overline{f^{-1}(\operatorname{Int}H)}\bigr)=H \]
it remains only to prove
\[ H\supseteq f\bigl(\overline{f^{-1}(\operatorname{Int}H)}\bigr). \]
Suppose, for some
\[ x\in \overline{f^{-1}(\operatorname{Int}H)}, \]
that \(f(x)\in H\). Then there exists a neighborhood \(U\) of the point \(x\) such that
\[ f(U)\subseteq CH, \]
i.e.
\[ f(U)\cap \operatorname{Int}H=\varnothing. \]
But
\[ U\cap f^{-1}(\operatorname{Int}H)\ne\varnothing. \]

Step 3. If \(F_1,F_2\in R_{\mathfrak A}(X)\) and \(F_1\ne F_2\), then
\[ \operatorname{Int}F_1\cap CF_2\ne\varnothing \]
or
\[ \operatorname{Int}F_2\cap CF_1\ne\varnothing; \]
suppose
\[ \operatorname{Int}F_1\cap CF_2\ne\varnothing. \]
Then there exists a point \(y\in Y\) for which
\[ f^{-1}(y)\subseteq \operatorname{Int}F_1\cap CF_2, \]
therefore \(y\in f(F_1)\), but \(y\in f(F_2)\).

Relations (1) and (3) give
\[ f(R_{\mathfrak A}(X))=R_{\mathfrak A}(Y), \]
and (3) gives the one-to-one character of the operator \(F\to f(F)\). Moreover, this operator is monotone, which completes the proof.

* See V. I. Ponomarev (8), pp. 106–107, propositions A–B, containing Lemma 2 in the assertion of a continuous (irreducibly closed) mapping \(f\).

** \(f\) is called irreducible if, for every closed \(F\) in \(X\), \(F\ne X\), it is true that \(f(F)\ne Y\).

*** For a regular space \(Y\), condition 3) is equivalent to the continuity of \(f\). If in 3) the existence of a closed neighborhood \(U\) is required, then 3) is equivalent to the \(\theta\)-continuity defined by Fomin (see (4)). For a regular source space \(X\), 3) is equivalent to the \(\theta\)-continuity of \(f\).

Using this, we shall now give a purely topological characterization of the class of Hausdorff spaces that possess isomorphic Boolean structures of their canonical open sets.

Theorem. Let \(B\) be a complete Boolean structure, and let \(K(B)\) be the class of all Hausdorff spaces \(X\) for which the Boolean structures \(R_0(X)\) of canonical open sets are isomorphic to \(B\).

A Hausdorff space \(X\) belongs to \(K(B)\) if and only if it is the image of some dense subspace of the representation space \(M(B)\)* of the structure \(B\) under a closed, irreducible, and \(\theta\)-continuous mapping.

Proof. Let \(X \in K(B)\). In paper \((^3)\) we proved that the space \(\overset{0}{X}\), belonging to the space \(X\), of all fixed regular ultrafilters is dense in \(M(B)\), and that the canonical mapping
\[ \varphi:\overset{0}{X}\to X \]
is \(\theta\)-continuous, irreducible, and closed. If, conversely, the space \(X\) is the \(\theta\)-continuous, irreducible, closed image of a dense subspace \(D\) of the space \(M(B)\), then it follows from Lemma 2 that
\[ R_{\mathfrak A}(X)\cong R_{\mathfrak A}(D), \]
and further, by Lemma 1,
\[ R_{\mathfrak A}(D)\cong R_{\mathfrak A}(M(B)). \]
From the completeness of the structure \(B\) it follows that \(M(B)\) is extremally disconnected; consequently every canonical closed set in \(M(B)\) is even open-closed. Therefore
\[ R_{\mathfrak A}(M(B))\cong B \]
Remark. In this connection we refer to Birkhoff’s problem 82 \((^2)\): to find necessary and sufficient conditions for a given Boolean structure \(B\) to be isomorphic to the structure \(R_0(X)\) of a suitable \(T_1\)-space or a suitable metric space.

The first part of Birkhoff’s problem is solved quite simply by establishing that complete Boolean structures, and only they, can be represented in the form \(R_0(X)\), where \(X\) is a \(T_1\)-space. Namely, their representation space gives such a realization (this was already noted in the proof of the theorem).

The next problem—to find necessary and sufficient conditions for a complete Boolean structure to be representable in the form \(R_0(X)\) for a space \(X\) of a given type—receives, in view of our theorem, the following topological analogue:

For compact extremally disconnected \(T_2\)-spaces \(X\), find conditions under which at least one representative of the given type is contained in the class of all \(\theta\)-continuous, irreducible, closed images of its dense subspaces.

In Birkhoff’s original problem, a type of metric spaces is required.

Mathematical Institute
German Academy of Sciences
Berlin, GDR

Received
31 XII 1963

References

\(^1\) P. S. Aleksandrov, UMN, 2, 5 (1947).
\(^2\) G. Birkhoff, Lattice Theory, N. Y., 1948.
\(^3\) J. Flachsmeyer, Math. Nachr., 26, 1—4 (1963).
\(^4\) S. Il’yadis, DAN, 149, No. 1 (1963).
\(^5\) C. Kuratowski, Fund. Math., 3 (1922).
\(^6\) C. Kuratowski, Topologie, 1, Warszawa, 1958.
\(^7\) Yu. M. Smirnov, Scientific Notes of Moscow University, Mathematics, 4, no. 148, 204 (1951).
\(^8\) V. I. Ponomarev, Matem. sborn., 60 (102), 1, 89 (1963).

* The Stone structure space \(M(B)\), or the representation space of the structure \(B\), is the unique zero-dimensional \((\operatorname{ind}=0)\) Hausdorff compact space for which the (elementary) structure of open-closed sets is isomorphic to \(B\). It is constructed as the space of ultrafilters (or maximal ideals) of the structure \(B\).

** With the aid of the space \(\overset{0}{X}\) belonging to the space \(X\), we may say: for Hausdorff spaces \(X\) and \(Y\),
\[ R_0(X)\cong R_0(Y)\Longleftrightarrow \beta\overset{0}{X}\cong \beta\overset{0}{Y} \]
(because
\[ M(R_0(X))\cong \beta\overset{0}{X}. \]
Here it is essential that \(\overset{0}{X}\) admits a purely topological description (see Lemma 2 and Theorem 1 of paper \((^3)\)).