UDC 513.831

MATHEMATICS

V. P. ZOLOTAREV

ON THE INTERSECTION OF TOPOLOGIES

(Presented by Academician P. S. Aleksandrov on 15 IV 1970)

It is well known that under refinements the dimension of a space may both increase and decrease. Therefore it is of interest to know how dimension behaves under refinements of a special kind: let \(\dim (X,\tau_1)=0\) and \(\dim (X,\tau_2)=0\), where \(\tau_1,\tau_2\) are different topologies on the set \(X\), and \(\tau=\tau_1\cap\tau_2\) (a set \(U\) is open in the topology \(\tau\) if and only if \(U\) is open both in the topology \(\tau_1\) and in the topology \(\tau_2\)). What, then, can be said about the dimension of the space \((X,\tau)\)? It is clear, for example, that the topology of an interval can be represented as the intersection of two 0-dimensional topologies. For this it is enough to take, as a base of the topology \(\tau_1\), the family \(\{[r,a)\}\), and as a base of the topology \(\tau_2\), the family \(\{(b,r]\}\), where \(a,b,r\) are arbitrary rational numbers, \([r,a)=\{x\mid r\le x<a\}\) and \((b,r]=\{x\mid b<x\le r\}\).

In A. V. Arkhangel’skii’s seminar of P. S. Aleksandrov in Moscow the following hypothesis was formulated: \(\dim (X,\tau)\le n\) if and only if there exist topologies \(\tau_1,\ldots,\tau_{n+1}\) for which \(\dim (X,\tau_i)\le 0\) and

\[ \tau=\bigcap_{i=1}^{n+1}\tau_i. \]

They set the problem of verifying this hypothesis in various classes of completely regular spaces, in particular in the class of metrizable spaces.

The reader may judge how close the following result is to the solution of the latter problem.

Theorem. Let \((X,\tau)\) be a metrizable space. Then \(\dim (X,\tau)\le n\) if and only if there exist topologies \(\tau_1,\ldots,\tau_{n+1}\) such that the following conditions are satisfied:

a) \((X,\tau_i)\) is a metrizable space, \(i=1,\ldots,n+1\);

b) \((X,\bigcap_{i=1}^{k}\tau_i)\) is a metrizable space for each \(k=1,\ldots,n+1\);

c) \(\dim (X,\tau_i)\le 0\);

d) \(\tau=\bigcap_{i=1}^{n+1}\tau_i\).

Proof. Necessity:

Let \(\dim (X,\tau)\le n\). Then there exist such \(A_i\subset X\), \(i=1,\ldots,n+1\),

\[ X=\bigcup_{i=1}^{n+1}A_i, \]

and \(\dim (A_i\tau)\le 0\). Moreover, one may assume that the \(A_i\) are of type \(G_\delta\), \(i=1,\ldots,n+1\) (see \((^2)\)).

Denote by \(\tau_i\) the topology whose base is the union of the family \(\tau\) and the family of all one-point subsets of the set \(X\setminus A_i\). Since \(\tau\) forms a base of the topology \(\tau_i\) at the points of the set \(A_i\) and \(\tau\subset\tau_i\), \(i=1,\ldots,n+1\), we conclude that

\[ \tau=\bigcap_{i=1}^{n+1}\tau_i. \]

Fix the value of the index \(i\) and put \(\tau_i=\tau'\), \(A_i=A'\).

Lemma. \((X,\tau')\) is a normal space.

Proof. We shall prove this without using the fact that there exists a \(\sigma\)-locally finite base in \((X,\tau')\) (it is obvious that the space \((X,\tau')\) is regular).

Let \(D\) and \(C\) be disjoint \(\tau'\)-closed sets; then
\[ D=D_1\setminus D_2,\qquad C=C_1\setminus C_2, \]
where \(D_1,C_1\) are \(\tau\)-closed and \(D_2,C_2\subset X\setminus A'\).
\[ D\cap C=\Lambda=(D_1\setminus D_2)\cap(C_1\setminus C_2) =(D_1\cap C_1)\setminus(D_2\cup C_2). \]
It follows from this equality that
\[ D_1\cap C_1\subset D_2\cup C_2\subset X\setminus A'. \]
Thus, \(D_1\cap C_1\) is \(\tau'\)-open and \(\tau'\)-closed.

Let
\[ L=X\setminus(D_1\cap C_1). \]
Then \(L\cap D_1\), \(L\cap C_1\) are disjoint \(\tau\)-closed sets in \(L\), and, consequently, there exist \(\tau\)-open sets \(U_1,V_1\) in \(L\) such that
\[ L\cap D_1\subset U_1,\qquad L\cap C_1\subset V_1,\qquad \overline{U}_1\cap\overline{V}_1=\Lambda,\qquad U_1\cup V_1\subset L. \]
Since \(L\) is \(\tau'\)-open in \(X\), the sets \(U_1\) and \(V_1\) are \(\tau'\)-open. And because the discrete topology is introduced on the set \(D_1\cap C_1\), the sets \(U\) and \(V\) are \(\tau'\)-open and
\[ U\supset D,\qquad V\supset C,\qquad U\cap V=\Lambda, \]
where
\[ U=U_1\cup\bigl((D_1\cap C_1)\cap D\bigr),\qquad V=V_1\cup\bigl((D_1\cap C_1)\cap C\bigr). \]
The lemma is proved.

a) \((X,\tau')\) is a metrizable space.

Proof. \(X\setminus A'\) has type \(F_\sigma\) in the topology \(\tau\):
\[ X\setminus A'=\bigcup_{j=1}^{\infty}F_j. \]
Thus a base \(B'\) of the topology \(\tau'\) can be represented in the form
\[ B'=B_\tau\cup\left(\bigcup_{j=1}^{\infty}\{\{x\}\mid x\in F_j\}\right), \]
where \(B_\tau\) is a \(\sigma\)-locally finite base of the topology \(\tau\). Clearly, \(B'\) is \(\sigma\)-locally finite.

b) \(\left(X,\bigcap_{i=1}^{k}\tau_i\right)\) is a metrizable space.

This assertion follows immediately from the definition of the topologies \(\tau_i\) and property a) (the intersection of a finite family of sets of type \(F_\sigma\) has type \(F_\sigma\)).

c) \(\dim (X,\tau')\le 0\).

Proof. Let \(\dim(A',\tau)\le k\), and let \(\{\Gamma_j\}\) be a finite covering of the set \(X\) by \(\tau'\)-open sets. Then in the covering \(\{A'\cap\Gamma_j\}\) of the \(\tau'\)-closed set \(A'\) we combinatorially inscribe a \(\tau'\)-closed covering \(\{K_j\}\) of multiplicity \(k+1\) (this is possible, since on the set \(A'\) the topologies \(\tau\) and \(\tau'\) coincide). Since \((X,\tau')\) is a normal space, one can construct \(\tau'\)-open sets \(\{V_j\}\) such that
\[ K_j\subset V_j\subset\Gamma_j \]
and the multiplicity of the family \(\{V_j\}\) does not exceed \(k+1\). Let \(\{U_j\}\) be an arbitrary covering of multiplicity \(1\) of the set
\[ X\setminus\left(\bigcup_j V_j\right)\subset X\setminus A', \]
combinatorially inscribed in
\[ \left\{\Gamma_j\cap\left(X\setminus\left(\bigcup_j V_j\right)\right)\right\}. \]
Then \(\{U_j\cup V_j\}\) is a \(\tau'\)-open covering of \(X\), combinatorially inscribed in \(\{\Gamma_j\}\), of the required multiplicity.

Remark. If one dispenses with property b), then one can ensure that the weight of each \((X,\tau_i)\) does not exceed the weight of the space. Indeed (see \((^1)\)), take, for some \(i\), a \(\sigma\)-locally finite base \(B_i=\{U_\alpha\}\) such that
\[ \operatorname{Fr}(U_\alpha)\cap A_i=\Lambda. \]
It is clear that the family
\[ B_i'=\{U_\alpha\}\cup\{\operatorname{Fr} U_\alpha\} \]
is \(\sigma\)-locally finite. Moreover, the family \(B_i''\), obtained from \(B_i'\) by finite intersections of elements, is also \(\sigma\)-locally finite. Declare \(B_i''\) to be a base of some topology \(\tau_i\). From the construction it is seen that \(B_i''\) consists of \(\tau_i\)-open-closed sets. Hence it is clear that \((X,\tau_i)\) is regular, metrizable, and zero-dimensional; moreover, at the points of \(A_i\) there exists a base consisting of sets open in the topology \(\tau\), i.e.
\[ \bigcap_{i=1}^{n+1}\tau_i=\tau. \]
It is clear that the weight of the space has not changed.

Sufficiency. We shall prove a more general assertion: let \((X,\tau)\) be a hereditarily normal Fréchet–Urysohn space and
\[ \dim(X,\tau)\ge n+1. \]
Suppose also that topologies \(\tau_1,\ldots,\tau_{n+1}\) are given on the set \(X\) such that:

a′) \((X,\tau_i)\) is a hereditarily normal Fréchet–Urysohn space;

b′) \(\left(X,\bigcap_{i=1}^{k}\tau_i\right)\) is a hereditarily normal Fréchet–Urysohn space for each \(k=1,\ldots,n+1\);

c′) \(\dim (X,\tau_i)\leqslant 0\), then
\[ \tau\ne \bigcap_{i=1}^{n+1}\tau_i . \]

We shall prove this assertion by induction: for \(n=0\) the assertion is obvious. Suppose that we have proved that \(\dim (X,\tau')\leqslant n-1\), where
\[ \tau'=\bigcap_{i=1}^{n}\tau_i . \]
Assume that \(\tau'\cap\tau_{n+1}=\tau\).

Lemma 1. Let \((X,\tau_1),\ldots,(X,\tau_k)\) and \((X,\bigcap_{i=1}^{k}\tau_i)\) be Fréchet–Urysohn spaces; then
\[ [A]_{\bigcap_{i=1}^{k}\tau_i}=\bigcup_{i=1}^{k}[A]_{\tau_i}, \]
where \(A\subset X\) and \([A]_\tau\) denotes the closure of the set in the topology \(\tau\).

Proof. Let
\[ x\in [A]_{\bigcap_{i=1}^{k}\tau_i} \quad\text{and}\quad x\notin \bigcup_{i=1}^{k}[A]_{\tau_i}. \]
Since \((X,\bigcap_{i=1}^{k}\tau_i)\) is a Fréchet–Urysohn space, there exists a sequence \(\{x_n\}\subset A\) such that \(\{x_n\}\) converges to \(x\) in the topology \(\bigcap_{i=1}^{k}\tau_i\), but for each \(i\) the set \(\{x_n\}\) is closed in the topology \(\tau_i\) and, consequently, closed in the topology \(\bigcap_{i=1}^{k}\tau_i\)—a contradiction (the reverse inclusion is obvious).

Lemma 2. Let \((X,\tau_1)\), \((X,\tau_2)\), and \((X,\tau_1\cap\tau_2)\) be hereditarily normal Fréchet–Urysohn spaces, where \(\dim (X,\tau_1)\leqslant n\) and \(\dim (X,\tau_2)\leqslant m\). Then
\[ \dim (X,\tau_1\cap\tau_2)\leqslant n+m+1. \]

Proof. Suppose the contrary. Then there exist (see (1)) pairs of \(\tau_1\cap\tau_2\)-closed sets
\[ \{C_i,C_i'\},\quad i=1,\ldots,n+m+2, \]
such that
\[ C_i\cap C_i'=\Lambda \]
and for every collection \(\{B_i\}\), \(i=1,\ldots,n+m+2\), where \(B_i\) is a partition between \(C_i\) and \(C_i'\),
\[ \bigcap_{i=1}^{n+m+2} B_i\ne \Lambda \]
(\(B_i\) is \(\tau\)-closed, where \(\tau=\tau_1\cap\tau_2\)).

Since \(\dim (X,\tau_1)\leqslant n\), there exist \(n+1\) pairs of \(\tau_1\)-open sets
\[ \{V_i,V_i'\},\quad i=1,\ldots,n+1, \]
such that
\[ [V_i]_{\tau_1}\cap [V_i']_{\tau_1} = X\setminus (V_i\cup V_i'),\qquad V_i\supset [\Gamma_i]_\tau\supset \Gamma_i\supset C_i,\quad V_i'\supset [\Gamma_i']_\tau\supset \Gamma_i'\supset C_i', \]
and \(\{V_i,V_i'\}\) is a cover (here \(\Gamma_i\) and \(\Gamma_i'\) are \(\tau\)-open sets).

Consider also a \(\tau_2\)-open cover
\[ \{U_i,U_i'\},\quad i=n+2,\ldots,n+m+2, \]
having analogous properties.

Introduce the notation
\[ B_i=[V_i]_\tau\cap [V_i']_\tau,\quad i=1,\ldots,n+1. \]
It is clear that \(B_i\) is a partition between \(C_i\) and \(C_i'\) in the topology \(\tau\).

Assertion. Let
\[ D=\bigcap_{i=1}^{n+1} B_i. \]
Then there is a pair
\[ U,U'\in \{U_i,U_i'\},\quad i=n+2,\ldots,n+m+2, \]
such that \(D\cap U\) contains some point \(x\) which is a \(\tau\)-limit point for \(D\cap U'\) (or conversely).

Proof. If this were not so, then there would exist a partition \(B_i\), closed in the topology \(\tau\), between
\[ C_i\cup (D\cap U_i) \quad\text{and}\quad C_i'\cup (D\cap U_i') \]
for each \(i=n+2,\ldots,n+m+2\) (by the assumption: the space \((X,\tau)\) is hereditarily normal and \(C_i\) does not meet \([V_i']_\tau\)). Note that
\[ D\cap B_i\subset X\setminus (U_i\cup U_i'), \]
whence follows the equality
\[ \bigcap_{i=n+2}^{n+m+2}(D\cap B_i)=\Lambda, \]
which contradicts the choice of the sets \(\{C_i,C_i'\}\).

Thus there exists
\[ x\in (D\cap U)\cap [D\cap U']_\tau. \]
Since the family \(\{V_i,V_i'\}\) is a cover, we may assume that \(x\in V\) for some pair
\[ V,V'\in \{V_i,V_i'\}. \]
It follows from Lemma 1 that \(U\cap V\) contains no points that are \(\tau\)-limit points for \(U'\cap V'\). Knowing this, we arrive at a contradiction.

Since
\[ x\in (U\cap D)\cap [U'\cap D]_\tau, \]
there exists a sequence \(\{x_n\}\subset D\cap U'\) converging to \(x\) in the topology \(\tau\); then, by Lemma 1, \(\{x_n\}\) converges to \(x\) either in the topology \(\tau_1\) or in the topology \(\tau_2\), but \(U\) is \(\tau_2\)-open and, consequently, \(\{x_n\}\) converges to \(x\) in the topology \(\tau_1\).

Since \(\{x_n\}\subset D\subset [V']_\tau\), for each \(n\) there exists a sequence \(\{y_n^i\}\subset V'\), converging with respect to \(i\) to \(x_n\) in the topology \(\tau\), and \(\{y_n^i\}\) cannot converge to \(x_n\) in the topology \(\tau_1\) for any subsequence of the indices \(n\) cofinal in the natural numbers; otherwise
\[ x\in \bigl[\bigcup_{i,n} y_n^i\bigr]_{\tau_1}, \]
whereas we have \(x\in V=X\setminus [V']_{\tau_1}\).

Thus \(\{y_n^i\}\) converges with respect to \(i\) to \(x_n\) in the topology \(\tau_2\) (at least starting from some \(n\)), and therefore it may be assumed that \(\{y_n^i\}\subset U'\cap V'\); but since \((X,\tau)\) is a Fréchet–Urysohn space, it follows that
\[ x\in \bigl[\bigcup_{i,n} y_n^i\bigr]_\tau, \]
a contradiction (this is true because \(U'\) is \(\tau_2\)-open and \(\{x_n\}\subset U'\)).

At first glance it seems that property \((\delta')\) is superfluous, automatically satisfied. I do not know whether one can dispense with it, but in any case, as the example of A. V. Arhangel’skii shows, it is not automatic.

Example. Consider the set of points in the plane
\[ X=(0,0)\cup \left\{\left(\frac1n,0\right)\right\}\cup \left\{\left(\frac1n,\frac1m\right)\right\}, \]
where \(n,m\) are natural numbers. In addition to the usual open sets induced by the topology of the plane, declare open:

1) in the topology \(\tau_1\), the point \((0,0)\);

2) in the topology \(\tau_2\), the set
\[ (0,0)\cup \left\{\left(\frac1n,0\right)\right\}. \]

It is clear that in the topology \(\tau_1\cap\tau_2\) the point \((0,0)\) has no countable base, and the space \((X,\tau_1\cap\tau_2)\) is not a Fréchet–Urysohn space (from the set
\[ \left\{\left(\frac1n,\frac1m\right)\right\} \]
one cannot reach the point \((0,0)\) by any countable sequence, although
\[ \left[\left\{\left(\frac1n,\frac1m\right)\right\}\right]_{\tau_1\cap\tau_2}=X \]
).

Moscow State University
named after M. V. Lomonosov

Received
3 IV 1970

REFERENCES

1. R. Engelking, Outline of General Topology, Amsterdam, p. 291, 1968.
2. Yu. M. Smirnov, Izv. AN SSSR, Ser. Mat., 20, No. 2 (1956).