Reports of the Academy of Sciences of the USSR
1967. Volume 174, No. 6

UDC 513.831

MATHEMATICS

V. I. PONOMAREV

ON THE METRIZABILITY OF A FINALLY COMPACT \(p\)-SPACE WITH A POINT-COUNTABLE BASE

(Presented by Academician P. S. Aleksandrov on 15 III 1967)

§ 1. A base \(\sigma=\{U_\alpha\}\) of a space \(X\) is called point-countable if each point \(x\in X\) belongs to no more than a countable number of elements of this base. A. S. Miščenko proved \((^3)\) that a point-countable base \(\sigma\) of a bicompact \(X\) is necessarily countable, and the bicompact \(X\) itself, consequently, is metrizable. On the other hand, A. S. Miščenko \((^3)\), and then Michael \((^4)\), constructed an example of a finally compact, hereditarily strongly paracompact space \(X\) with a point-countable but uncountable base (consequently, these spaces are nonmetrizable). A. V. Arhangel’skii introduced the class of \(p\)-spaces (see \((^1,^2)\)) and proved that the class of paracompact \(p\)-spaces coincides with the class of spaces admitting perfect mappings onto metric spaces \((^1,^2)\).

The main result of the present note is

Main theorem 1*. Let \(X\) be a finally compact \(p\)-space, and let \(\sigma=\{U_\alpha\}\) be its point-countable base. Then this base \(\sigma\) is necessarily countable, and the space \(X\) itself is metrizable with a countable base.

The proof of this proposition, in its methods, is close to my paper \((^6)\), and especially important for us will be the following proposition, proved in \((^6)\) (Lemma 2, p. 106):

Lemma 1. Let \(X\) be a finally compact \(p\)-space; \(U\) an open subset of \(X\) of type \(F_\sigma\). Then there exists a metrizable space \(Y\) with a countable base and a perfect mapping \(f:X\to Y\) such that the set \(U\) is marked** under \(f\), i.e. \(f^{-1}fU=U\).

In addition, we shall need the following quite simple proposition (see \((^5)\)).

Lemma 2. Let \(\sigma\) be a point-countable base of the space \(X\); \(F\) a separable*** subspace. Then the system
\[ \sigma_F=\mathcal{E}(U\in\sigma,\ U\cap F\ne\Lambda) \]
is countable.

Definitions (see \((^7)\)). Suppose mappings \(f_\lambda:X\to Y_\lambda\) are given from one and the same space \(X\) onto spaces \(Y_\lambda\). Define the mapping
\[ f:X\to Y\subseteq \prod_\lambda Y_\lambda \]
into the product \(\prod_\lambda Y_\lambda\) as follows:
\[ fx=\{f_\lambda x\}\in \prod_\lambda Y_\lambda. \]
The mapping obtained will be called the product of the mappings \(f_\lambda\). If two mappings \(f:X\to Y\) and \(g:X\to Z\) are given, then we shall write \(g<f\) if for every point \(x\in X\) necessarily \(g^{-1}gx\subseteq f^{-1}fx\).

§ 2. \(\theta_F\)-reconstruction. Let \(X\) be a normal space, and let \(\sigma=\{U_\alpha\}\) be its point-countable base. Suppose, moreover, that \(F\subseteq X\) is a closed—

* The theorem gives a partial answer to a problem of A. V. Arhangel’skii formulated at the end of this note (Problem B).

** A set \(A\subseteq X\) is called marked under a mapping \(f:X\to Y\) if \(f^{-1}fA=A\).

*** By separability we mean the existence of a countable everywhere dense subset.

a closed separable subspace. Consider the countable (by Lemma 2) system \(\sigma_F\). We shall call a pair \((U,U')\), \(U\in\sigma_F\), \(U'\in\sigma_F\), regular if \([U]\subseteq U'\). Then, by the normality of the space \(X\), there exists an open set \(V\) of type \(F_\sigma\) such that \([U]\subseteq V\subseteq V'\).

To each regular pair \((U,U')\), \(U\in\sigma_F\), \(U'\in\sigma_F\), we assign some one completely determined open set
\[ V^F_{(U,U')} \]
of type \(F_\sigma\):
\[ [U]\subseteq V^F_{(U,U')}\subseteq U'. \]
The system of all \(V^F_{(U,U')}\) thus constructed, over all regular pairs \((U,U')\), \(U\in\sigma_F\), \(U'\in\sigma_F\), will be denoted by \(\theta_F\). We shall call the indicated construction the \(\theta_F\)-rearrangement of the system \(\sigma_F\). It is not difficult to verify that \(\theta_F\) is an at most countable system, forms an external base* of the set \(F\), and all its elements intersect \(F\).

Lemma 3. Let \(X\) be a finally compact \(p\)-space; \(F\subseteq X\) a closed separable subspace; \(\sigma\) a point-countable base of \(X\). For the system \(\sigma_F\) consider the system \(\theta_F\) obtained from \(\sigma_F\) by a \(\theta_F\)-rearrangement. Then there exists a metric space \(Y\) with a countable base and a perfect mapping \(f:X\to Y\) such that every \(V^F_{(U,U')}\in\theta_F\) is distinguished under this mapping; moreover, under this mapping the set \(F\) will also be distinguished, and the restriction \(f|F\) to \(F\) is a homeomorphism. In addition, as a consequence, \(F\) has type \(G_\delta\).

3. Proof of the main theorem 1. Let \(X\) be a finally compact \(p\)-space with a point-countable base \(\sigma=\{U_\alpha\}\). Consider a metric space \(Y_1\) with a countable base and a perfect mapping \(f_1:X\to Y_1\). Let \(F_1\subseteq X\) be a subset closed in \(X\) for which \(f_1F_1=Y_1\), but \(f_1:F_1\to Y_1\) is irreducible. Then the set \(F_1\) is separable, and consequently for it the system \(\sigma_{F_1}\) is countable. Consider for the system \(\sigma_{F_1}\) the system \(\theta_{F_1}\) obtained from \(\sigma_{F_1}\) by a \(\theta_{F_1}\)-rearrangement. Then, by Lemma 3, there exists a metric space \(Y_2\) with a countable base and a perfect mapping \(f_2:X\to Y_2\) such that every \(V^{F_1}_{(U,U')}\in\theta_{F_1}\) is distinguished under the mapping \(f_2\). Moreover, it may always be assumed that \(f_2<f_1\). Now consider such a closed set \(F_2\subseteq X\) that \(f_2F_2=Y_2\) and that \(f_2:F_2\to Y_2\) is irreducible. We obtain \(F_1\subseteq F_2\). Consider \(\sigma_{F_2}\) and the system \(\theta_{F_2}=\{V^{F_2}_{(U,U')}\}\), obtained by a \(\theta_{F_2}\)-rearrangement, and again apply Lemma 3 to the system \(\theta_{F_2}\), and so on. As a result we obtain perfect mappings
\[ f_i:X\to Y_i,\qquad f_1>f_2>\cdots>f_i>\cdots, \]
onto metric spaces with a countable base, and closed separable sets
\[ F_i\subseteq X,\qquad F_1\subseteq F_2\subseteq\cdots\subseteq F_i\subseteq\cdots. \]
The following conditions will then be fulfilled:

\(1^\circ.\) \(f_iF_i=Y_i\) and \(f_i:F_i\to Y_i\) is an irreducible perfect mapping.

\(2^\circ.\) All elements \(V^{F_{i-1}}_{(U,U')}\in\theta_{F_{i-1}}\) are distinguished under \(f_i:X\to Y_i\). Here \(\theta_{F_{i-1}}\) is the system obtained by a \(\theta_{F_{i-1}}\)-rearrangement of the system \(\sigma_{F_{i-1}}\).

Since \(\theta_{F_{i-1}}\) is an external base of the set \(F_{i-1}\), the closed set \(F_{i-1}\) is distinguished under the mapping \(f_i:X\to Y_i\), and \(f_i\) is a homeomorphism on the set \(F_{i-1}\).

Now consider the set \(\Phi=\left[\bigcup F_i\right]\) and the system \(\theta=\bigcup_{i=1}^{\infty}\theta_{F_i}\). We shall prove that \(\theta\) forms an external base of the closed set \(\Phi\). Indeed, let \(x_0\in\Phi\), and let \(Ox_0\) be an arbitrary neighborhood of the point \(x_0\) in \(X\). Consider elements \(U\in\sigma\) and \(U'\in\sigma\) such that
\[ x_0\in U\subseteq[U]\subseteq U'\subseteq Ox_0. \]
Since \(x_0\in\left[\bigcup_{i=1}^{\infty}F_i\right]\), there exists an \(i_0\) such that \(U\cap F_{i_0}\ne\Lambda\)

* A system \(\sigma=\{U\}\) of open sets of the space \(X\) is called an external base of the set \(F\subseteq X\) if for any point \(x_0\in F\) and any of its neighborhoods \(Ox_0\) in \(X\) there exists an element \(U\in\sigma\) such that \(x_0\in U\subseteq Ox_0\) (see (8)).

and \(U' \cap F_{i_0}\ne\Lambda\). Then \(U\in\sigma_{F_{i_0}}\) and \(U'\in\sigma_{F_{i_0}}\); consequently, \((U,U')\) is a regular pair for the system
\[ \sigma_{F_{i_0}}=\mathscr E(U\in\sigma,\ U\cap F_{i_0}\ne\Lambda). \]
Consider the system \(\theta_{F_{i_0}}\), obtained by the \(\theta_{F_{i_0}}\)-refinement of the system \(\sigma_{F_{i_0}}\). By the definition of a \(\theta_{F_{i_0}}\)-refinement, for the regular pair \((U,U')\) there is an open set \(V_{(U,U')}^{F_{i_0}}\in\theta_{F_{i_0}}\) of type \(F_\sigma\) for which
\[ [U]\subseteq V_{(U,U')}^{F_{i_0}}\subseteq U'. \]
Then
\[ x_0\in U\subseteq [U]\subseteq V_{(U,U')}^{F_{i_0}}\subseteq U'\subseteq Ox_0. \]

Consequently,
\[ \theta=\bigcup_{i=1}^{\infty}\theta_{F_i} \]
is an external base of the closed set
\[ \Phi=\left[\bigcup_{i=1}^{\infty}F_i\right]. \]
We shall prove that \(\Phi=X\). Suppose the contrary; let
\[ X\setminus\Phi\ne\Lambda. \]

Consider the mapping
\[ f_i:X\to Y\subseteq\prod_{i=1}^{\infty}Y_i, \]
which is the product of the mappings \(f_i:X\to Y_i\). The mapping \(f\) is also perfect (see (6)), and for every point \(x\in X\) necessarily
\[ f^{-1}fx=\bigcap_{i=1}^{\infty} f_i^{-1}f_i x. \]
Moreover, each \(V\in\theta\) is marked under the mapping \(f\) (see (6), p. 108), but, since \(\theta\) is an external base of the closed set \(\Phi\), \(\Phi\) is also marked under the mapping \(f\), i.e. \(f^{-1}f\Phi=\Phi\), and \(f|\Phi\) is a homeomorphism. Then the set \(\Gamma=X\setminus\Phi\) will be marked, i.e., if \(x_0\in X\setminus\Phi\), then
\[ f^{-1}fx_0\subseteq X\setminus\Phi \]
or
\[ f^{-1}fx_0\cap\Phi=\Lambda. \]
On the other hand, \(f_i:F_i\to Y_i\) is irreducible; consequently,
\[ f_i^{-1}f_i x_0\cap F_i\ne\Lambda \]
for every \(i\), and hence, all the more,
\[ f_i^{-1}f_i x_0\cap\Phi\ne\Lambda. \]
Moreover,
\[ \Phi_1\supseteq \Phi_2\supseteq\cdots\supseteq \Phi_i\supseteq\cdots, \]
since
\[ f_1>f_2>\cdots>f_i>\cdots . \]
Consequently, the intersection of the bicompact sets \(\Phi_i\) is nonempty:
\[ \bigcap_{i=1}^{\infty}\Phi_i\ne\Lambda. \]
But then
\[ f^{-1}fx_0\cap\Phi = \bigcap_{i=1}^{\infty}\bigl(f_i^{-1}f_i x_0\cap\Phi\bigr) = \bigcap_{i=1}^{\infty}\Phi_i\ne\Lambda \]
for \(x_0\in X\setminus\Phi\). We have obtained a contradiction. Thus necessarily \(\Phi=X\), and therefore the mapping
\[ f:X\to Y \]
is a homeomorphism into the metric space
\[ \prod_{i=1}^{\infty}Y_i \]
with a countable base, and the base \(\sigma\) is countable.

The main theorem is completely proved.

§ 4. The general case

Theorem 2. Let a normal space \(X\) have pointwise weight\(*\) \(\tau\) and admit a perfect mapping onto a space of weight \(\tau\). Then the weight of the space \(X\) is equal to \(\tau\).

The proof of Theorem 2 is analogous to the proof of Theorem 1 and is based on the following lemmas.

Lemma 4. Let a normal space \(X\) admit a perfect mapping onto a space \(Y\) of weight \(\tau\), and let \(\sigma=\{U\}\) be a system of open subsets of \(X\) of type \(F_\sigma\) and of cardinality \(\leq\tau\). Then there exists a space \(Z\) of weight \(\leq\tau\) and a perfect mapping
\[ g:X\to Z, \]
under which all \(U\in\sigma\) are marked.

Lemma 5. Suppose that in a normal space \(X\) there is a base \(\sigma\) of pointwise cardinality \(\leq\tau\), and \(F\subseteq X\) is a closed set in which there exists a dense set of cardinality \(\leq\tau\). Then the system
\[ \sigma_F=\mathscr E(U\in\sigma,\ U\cap F\ne\Lambda) \]
has cardinality \(\leq\tau\).

Lemma 6. Let
\[ f:X\to Y \]
be a perfect irreducible mapping of the space \(X\) onto a space \(Y\) of weight \(\tau\). Then the space \(X\) has a dense subset of cardinality \(\tau\).

\(*\) A system \(\sigma\) of open subsets of \(X\) has pointwise cardinality \(\leq\tau\) if any point \(x\in X\) is contained in no more than \(\tau\) elements \(U\in\sigma\). The space \(X\) has pointwise weight \(\tau\) if in \(X\) there exists a base \(\sigma\) of pointwise cardinality \(\leq\tau\), but there does not exist a base of pointwise cardinality \(\leq\tau'<\tau\).

What spaces, then, admit perfect mappings onto spaces of weight \(\tau\)? In order to answer this question, let us introduce the following definitions.

Definition 1. A space \(X\) is called \((\tau,\infty)\)-compact if from every open cover \(\omega\) of it one can extract a subcover of cardinality \(\leq \tau\).

Definition 2. Let \(X\) be a completely regular space, and let \(bX\) be its bicompact extension. We shall call the space \(X\) a \(p^\tau\)-space if in \(bX\) there exists a feathering (in the sense of A. V. Arhangel’skii \((^{1,2})\)) of cardinality \(\tau\).

Theorem 3. In order that a space \(X\) admit a perfect mapping onto a space of weight \(\leq \tau\), it is necessary and sufficient that the space \(X\) be a \((\tau,\infty)\)-compact \(p^\tau\)-space.

§ 4. Hereditarily finally compact spaces and statements of problems. At the seminar of P. S. Aleksandrov the following problem was posed.

Problem 1. Will a hereditarily finally compact space \(X\) with a point-countable base \(\sigma\) be metrizable?

This problem, it seems to me, has transcendental difficulties, and we shall now show that it is, at least in one direction, connected with the classical Suslin problem.

Theorem 4. Suppose there exists a nonseparable ordered bicompactum \(X\) satisfying the Suslin condition. Then there exists a nonmetrizable hereditarily finally compact space \(X_0 \subseteq X\) with a point-countable base.

Proof. Indeed, an ordered bicompactum satisfying the Suslin condition is perfectly normal, hence hereditarily finally compact and satisfies the first axiom of countability. To each transfinite number \(\lambda < \omega_1\) assign a point \(x_\lambda \in X\) in such a way that \(x_\lambda \in X \setminus [\{x_{\lambda'}, \lambda' < \lambda\}]\). The points can be chosen in this way, since \(X\) is nonseparable. Now for each point \(x_\lambda\) consider a countable system of neighborhoods \(\{U_i x_\lambda\}=\sigma_\lambda\), under the condition

\[ [U_i x_\lambda]\subseteq X\setminus [\{x_{\lambda'}, \lambda' < \lambda\}], \]

forming a base at the point \(x_\lambda\). Consider the subspace \(X_0=\{x_\lambda,\lambda<\omega_1\}\), the system \(\sigma=\bigcup_\lambda \sigma_\lambda\), and the system \(\sigma^*=\{U\cap X_0,\ U\in\sigma\}\). It is not hard to prove that \(\sigma^*\) is a point-countable base, and since the space \(X\) is nonseparable, it follows that the hereditarily finally compact space \(X_0\) with the point-countable base \(\sigma^*\) is nonmetrizable.

Problem A. Is Problem 1 not equivalent to the Suslin problem?

Problem B. Does there exist a nonseparable perfectly normal bicompactum? Is this problem not equivalent to the Suslin problem?

Problem V. Will a paracompact \(p\)-space with a point-countable base be metrizable? (A. V. Arhangel’skii’s problem.)

In connection with A. V. Arhangel’skii’s problem we formulate a narrower problem.

Problem G. Suppose a space \(X\) admits a perfect irreducible mapping onto a metric space, i.e. it is coabsolute with some metric space and has a point-countable base. Will \(X\) be metrizable?

Moscow State University
named after M. V. Lomonosov

Received
1 III 1966

CITED LITERATURE

1. A. V. Arhangel’skii, DAN, 151, No. 4, 751 (1963).
2. A. V. Arhangel’skii, Matem. sborn., 67 (109), 1, 55 (1965).
3. A. S. Mishchenko, DAN, 144, 985 (1962).
4. E. Michael, Bull. Amer. Math. Soc., 69, No. 3, 375 (1963).
5. V. I. Ponomarev, Bull. Polish Acad. Sci., 8, No. 3, 127 (1960).
6. V. I. Ponomarev, UMN, 21, issue 4 (130), 101 (1966).
7. K. Kuratowski, Topology, 1, 1966.
8. A. V. Arhangel’skii, DAN, 130, No. 3, 495 (1960).