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MATHEMATICS
V. PONOMAREV
NORMAL SPACES AS IMAGES OF ZERO-DIMENSIONAL SPACES
(Presented by Academician P. S. Aleksandrov on 29 III 1960)
In this paper the following is proved.
Main theorem. Every normal space \(X\) of weight \(\tau\) is the image of a zero-dimensional (in the sense of small inductive dimension \(\operatorname{ind}\)) set \(D \subset D^\tau\) under some continuous closed, bicompact, irreducible mapping.
Addendum. Since \(D \subset D^\tau\), the weight of the space \(D\) does not exceed \(\tau\); since \(X\) is normal, \(D\) is also normal.
If \(X\) is locally bicompact, respectively paracompact, countably paracompact, or strongly paracompact, then, by virtue of known results (see, for example, \((^1)\)), \(D\) will automatically possess the same property. Hence, in particular, it follows that in the case of a strongly paracompact \(X\) the set \(D\) will be zero-dimensional also in the sense of \(\dim D = 0\).
- We pass to the proof of the main theorem. A canonically closed set, as is known, is any set \(A\) which is the closure of some open set (for which one may always take the open kernel \(JA\) of the set \(A\)).
A canonical covering of a space is any finite covering whose elements are canonically closed sets with disjoint open kernels.
If a canonical covering \(\alpha'\) follows a canonical covering \(\alpha\), i.e. is inscribed in it, then: a) each element of the covering \(\alpha'\) is contained in a unique element of the covering \(\alpha\), and b) each element of the covering \(\alpha\) is the sum of the elements of the covering \(\alpha'\) contained in it. In other words, the covering \(\alpha'\) is a subdivision of the covering \(\alpha\). Every set of canonical coverings of a given space we shall regard as partially ordered only in the sense of the just mentioned order relation. The set \(\Sigma=\{\alpha\}\) of canonical coverings of the space \(X\) will be called refining if the following condition is satisfied: whatever the point \(x \in X\) and its neighborhood \(Ox\), there is a covering \(\alpha\) such that the star of the point \(x\) in the covering \(\alpha\) (i.e. the sum of the elements of the covering \(\alpha\) containing the point \(x\)) is contained in \(Ox\). A directed refining set of canonical coverings of the space \(X\) is called simply a chain of subdivisions of the space \(X\).
* By \(D^\tau\) is denoted the “generalized Cantor discontinuum of weight \(\tau\),” i.e. a bicompact space which is the topological product of \(\tau\) spaces \(D_\alpha\), each of which consists of a finite number of isolated points; moreover, it is usually assumed (which does not impair generality) that \(D_\alpha\) consists of two isolated points; for us it is more convenient to abandon this restriction.
A mapping \(f: D \to X\) is called bicompact if the preimage \(f^{-1}x\) of every point \(x \in X\) is bicompact. A mapping \(f\) of the space \(D\) onto the space \(X\) is called irreducible if under the mapping \(f\) no closed set \(F \subset D\) distinct from \(D\) is mapped onto the whole of \(X\).
We shall easily prove the following proposition:
Lemma 1. In every normal space \(X\) of weight \(\tau\) there exists a chain of decompositions of cardinality \(\tau\).
First of all: into every open cover \(\omega\) one can inscribe a canonical one (this is proved, for example, in \((^2)\), Ch. 6, § 4). Let \(\mathfrak B\) be a base of cardinality \(\tau\) of the space \(X\). Consider the system \(\Omega\) of all possible “binary” covers of the form \(\omega=\{U, X\setminus [V]\}\), where \(U\) and \(V\), \([V]\subseteq U\), are elements of the base \(\mathfrak B\). The cardinality of the system \(\Omega\) is equal to \(\tau\). Into each \(\omega\in\Omega\) we inscribe a canonical \(\alpha_\omega\). It is easy to see that the resulting set \(\Sigma_\Omega=\{\alpha_\omega\}\) of canonical covers is refining. In order to obtain from it a directed refining set of canonical covers (i.e., the chain of decompositions of cardinality \(\tau\) which we need), it suffices to augment the system \(\Sigma_\Omega\) by all possible finite products of the covers entering into it, understanding by the product \(\alpha_1\cdots\alpha_n\) of canonical covers \(\alpha_1,\ldots,\alpha_n\) the canonical cover \(\alpha\), whose elements are the closures of all possible nonempty sets of the form \(\Gamma_1\cap\cdots\cap\Gamma_n\), where \(\Gamma_i\) is the open kernel of some element of the cover \(\alpha_i\). Lemma 1 is proved.
- Along with the space \(X\) of weight \(\tau\) we shall consider its bicompact extension \(\overline X\) of the same weight \(\tau\). Let in \(\overline X\) there be given a chain of decompositions \(\overline\Sigma=\{\overline\alpha\}\), \(\overline\alpha=\{\overline A_1^\alpha,\ldots,\overline A_{s_\alpha}^\alpha\}\), of cardinality \(\tau\). Then \(\Sigma=\{\alpha\}\), where \(\alpha=\{A_1^\alpha,\ldots,A_{s_\alpha}^\alpha\}\), \(A_i^\alpha=X\cap \overline A_i^\alpha\), is a chain of decompositions of the space \(X\), and moreover \(\overline A_i^\alpha=[A_i^\alpha]\) and the order \(\alpha'>\alpha\) and \(\overline\alpha'>\overline\alpha\) is one and the same.
For each \(\alpha=\{A_1^\alpha,\ldots,A_{s_\alpha}^\alpha\}\) consider the finite set
\[ D_\alpha=\{1^\alpha,2^\alpha,\ldots,s^\alpha\} \]
of natural numbers furnished with the index \(\alpha\), and for \(\alpha'>\alpha\) in \(\Sigma\) define the projection \(\mathfrak D_\alpha^{\alpha'}:D_{\alpha'}\to D_\alpha\), putting \(\mathfrak D_\alpha^{\alpha'}j^{\alpha'}=i^\alpha\), where \(i^\alpha=i\) is the number of the unique element \(A_i^\alpha\in\alpha\) containing the given element \(A_j^{\alpha'}\in\alpha'\).
Thus we obtain an inverse spectrum \(\{D_\alpha,\mathfrak D_\alpha^{\alpha'}\}\) with limit space \(\overline D=\lim(D_\alpha,\mathfrak D_\alpha^{\alpha'})\), which is a bicompactum lying in the topological product \(D^\tau=\prod_{\alpha\in\Sigma}D_\alpha\). Therefore the weight of the bicompactum \(\overline D\) does not exceed \(\tau\). Let \(\xi=\{i_\alpha\}\in\overline D\). The system \(\{\overline A_{i_\alpha}^\alpha\}\) is a centered system of closed sets of the bicompactum \(\overline X\), and therefore their intersection is nonempty. Since the system \(\overline\Sigma=\{\overline\alpha\}\) is refining, the intersection \(\bigcap_\alpha \overline A_{i_\alpha}^\alpha\) consists of only one point \(\overline x=\bigcap_\alpha \overline A_{i_\alpha}^\alpha\in\overline X\), and we put \(\overline f\xi=\overline x\). This defines the mapping \(\overline f:\overline D\to\overline X\).
Denote by \(U_i^\alpha=\mathscr E(\xi\in\overline D,\ i_\alpha=i)\) the totality of those points \(\xi\in\overline D\) for which \(i_\alpha=i\). These \(\overline U_i^\alpha\), \(i=1,2,\ldots,s^\alpha\), are, for a given \(\alpha\), disjoint bicompacta forming the cover \(\widetilde\delta_\alpha\) of the bicompactum \(\overline D\). The totality of all \(\overline U_i^\alpha\) (over all possible \(\alpha\)) forms a base of the bicompactum \(\overline D\). Moreover, obviously,
\[ \overline f\,\overline U_i^\alpha\subseteq \overline A_i^\alpha . \tag{1} \]
Hence, and from the fact that the system \(\overline\Sigma=\{\overline\alpha\}\) is refining, it follows immediately that the mapping \(\overline f:\overline D\to\overline X\) is continuous.
- Let us call a point \(\xi=\{i_\alpha\}\in\overline D\) marked if \(\bigcap_\alpha A_{i_\alpha}^\alpha\ne\Lambda\); then necessarily
\[ \bigcap_\alpha A_{i_\alpha}^\alpha=\bigcap_\alpha \overline A_{i_\alpha}^\alpha=\overline f\xi\in X. \]
Denote by \(D\) the set of all marked points \(\xi\in \overline D\), and put
\[ U_i^\alpha \cap D = D\,\overline U_i^\alpha,\qquad \delta_\alpha=\{U_1^\alpha,\ldots,U_{s_\alpha}^\alpha\}. \]
We shall prove that for every point \(x\in X\) we have \(\overline f^{-1}x\subset D\). Indeed, if for some point \(\xi=\{i_\alpha\}\in \overline D\) we have \(\overline f\xi=\bigcap_\alpha \overline A_{i_\alpha}^\alpha=x\in X\), then
\[ x=\bigcap_\alpha (X\cap \overline A_{i_\alpha}^\alpha)=\bigcap_\alpha A_{i_\alpha}^\alpha, \]
and this means precisely that \(\xi\) is a marked point, i.e. \(\xi\in D\).
It follows at once that, denoting by \(f:D\to X\) the mapping \(\overline f\) considered only on \(D\), we obtain a continuous compact mapping of the space \(D\) into \(X\).
We shall prove the equality
\[ fU_i^\alpha=A_i^\alpha. \tag{2} \]
This will also prove that \(f\) is a mapping of the set \(D\) onto the whole space \(X\) (and, consequently, \(\overline f\) is a mapping of the space \(\overline D\) onto the whole \(\overline X\)); moreover, from (2) it follows that all \(U_i^\alpha\) are nonempty and, hence, \(D\) is everywhere dense in \(\overline D\).
The inclusion \(fU_i^\alpha\subset A_i^\alpha\) follows from (1).
It remains to prove the more difficult assertion: for every point \(x\in A_i^{\alpha_0}\) there is a point \(\xi\in U_i^{\alpha_0}\) such that \(f\xi=x\).
We turn to this proof. Write \(\Sigma=\{\alpha\}\) as a well-ordered set whose first element is \(\alpha_0\):
\[ \Sigma=\{\alpha_0,\alpha_1,\ldots,\alpha_\lambda,\ldots\},\qquad \lambda<\omega(\tau). \]
For brevity write \(A(0)\) instead of \(A_i^{\alpha_0}\), and \(A(\mu)\) instead of \(A_{i_\mu}^{\alpha_\mu}\). Suppose that for all \(\mu<\lambda\) the \(A(\mu)\in\alpha_\mu\) have been chosen so that \(x\in A(\mu)\), and that, whatever ordinal numbers \(\mu_1,\ldots,\mu_r<\lambda\), taken in a finite number, may be, there exists in some \(\alpha_\nu\) following all \(\alpha_{\mu_1},\ldots,\alpha_{\mu_r}\) (i.e. inscribed in them) a set \(A(\nu)\ni x\), contained in all \(A(\mu_1),\ldots,A(\mu_r)\); here it is not at all required that \(\nu<\lambda\). We shall prove that among the elements of the covering \(\alpha_\lambda\) containing the point \(x\) one can find such an \(A(\lambda)\) that, whatever ordinal numbers \(\mu_1,\ldots,\mu_r\), smaller than \(\lambda+1\), may be, there again exists a covering \(\alpha_\nu\) following all \(\alpha_{\mu_1},\ldots,\alpha_{\mu_r}\), and in it an element \(A(\nu)\ni x\), contained in all \(A(\mu_1),\ldots,A(\mu_r)\). Suppose that no such \(A(\nu)\) can be found. This means that for any
\[ A_i\in\alpha_\lambda,\qquad i=1,2,\ldots,s^{\alpha_\lambda}, \]
one can find such a finite collection of sets
\[ (i)\qquad A(\mu_1^i),\ldots,A(\mu_{r(i)}^i),\qquad \mu_1^i,\ldots,\mu_{r(i)}^i<\lambda, \]
that there exists no \(A(\nu)\ni x\) contained in all the sets \((i)\) and, in addition, in \(A_i\). The union of all the systems \((i)\), \(i=1,2,\ldots,s^{\alpha_\lambda}\), is a finite system \(\sigma\) of sets \(A(\mu)\), \(\mu=\mu_1,\ldots,\mu_r<\lambda\); consequently, for it there exists an \(\alpha_{\nu'}\) following all \(\alpha_{\mu_1},\ldots,\alpha_{\mu_r}\), and in it some \(A(\nu')\ni x\), contained in all \(A(\mu)\in\sigma\). Take a covering \(\alpha_\nu\) following \(\alpha_{\nu'}\) in \(\Sigma\) and following \(\alpha_\lambda\). In \(\alpha_\nu\) there exists a set \(A(\nu)\ni x\), contained in \(A(\nu')\) and, hence, in all the sets of each of the systems \((i)\), for \(i=1,2,\ldots,s^{\alpha_\lambda}\). Let \(A_i^{\alpha_\lambda}=A(\lambda)\in\alpha_\lambda\) be the unique element of the covering \(\alpha_\lambda\) containing \(A(\nu)\). Then \(A(\lambda)\) is contained in all the elements of the system \((i)\) and in \(A_i^{\alpha_\lambda}\)—contrary to the supposition.
Thus, by induction, in every \(\alpha_\lambda\) we can choose such an element \(A(\lambda)\ni x\) that for any finite number of the \(A(\lambda_1),\ldots,A(\lambda_r)\) chosen by us
there is an \(\alpha_\nu\), inscribed in all \(\alpha_{\lambda_1},\ldots,\alpha_{\lambda_r}\), and in it an element \(A(\nu)\ni x\), contained in \(A(\lambda_1),\ldots,A(\lambda_r)\). Hence it follows: if \(\alpha_{\lambda'}>\alpha_\lambda\) in \(\Sigma\), then necessarily \(A(\lambda')\subseteq A(\lambda)\). Indeed, for some \(\alpha_\nu\) following in \(\Sigma\) after \(\alpha_{\lambda'}\) (and after \(\alpha_\lambda\)), some \(A(\nu)\in\alpha_\nu\) is contained in \(A(\lambda')\) and in \(A(\lambda)\), i.e. \(A(\lambda')\), respectively \(A(\lambda)\), is the unique element of the cover \(\alpha_{\lambda'}\), respectively \(\alpha_\lambda\), containing the set \(A(\nu)\), whence the assertion \(A(\lambda)\supseteq A(\lambda')\) follows at once. For the given point \(x\) we have selected one element \(A^{\alpha}_{i_\alpha}\) containing it in each cover \(\alpha\) in such a way that, when \(\alpha'>\alpha\) in \(\Sigma\), necessarily \(A^{\alpha'}_{i_{\alpha'}}\subseteq A^{\alpha}_{i_\alpha}\), and for \(\alpha=\alpha_0\) the selected element is precisely \(A^{\alpha_0}_{i^0}\). But then the point \(\xi=\{i_\alpha\}\in\overline D\) is a distinguished point, with \(\xi\in U^{\alpha_0}_{i^0}\) and \(f\xi=x\). Equality (2) is proved.
- The proof of the closedness of the mapping is preceded by
Lemma 2. Let \(\Phi\) be a bicompactum lying in \(D\); denote by \(U_\alpha\Phi\) the star of this bicompactum in the cover \(\delta_\alpha\). Whatever neighborhood \(O\Phi\) of the bicompactum \(\Phi\) in \(D\) may be, there exists an \(\alpha\) such that \(U_\alpha\Phi\subseteq O\Phi\).
Proof. Let \(\overline{O\Phi}\) be such a neighborhood of the bicompactum \(\Phi\) in \(\overline D\) that \(D\cap\overline{O\Phi}=O\Phi\). We shall prove that there exists an \(\alpha\) such that the star \(\overline{U_\alpha\Phi}\) of the bicompactum \(\Phi\) in \(\overline{\delta_\alpha}\) is contained in \(\overline{O\Phi}\): since \(D\cap\overline{U_\alpha\Phi}=U_\alpha\Phi\) and \(D\cap\overline{O\Phi}=O\Phi\), Lemma 2 will thereby be proved.
Suppose that for every \(\alpha\) the bicompactum
\(B_\alpha=\overline{U_\alpha\Phi}\setminus\overline{O\Phi}\ne\Lambda\). Then the system \(\{B_\alpha\}\) is centered and, consequently, \(\bigcap_\alpha B_\alpha\ne\Lambda\); take a point \(\xi\in\bigcap_\alpha B_\alpha\); for every \(\alpha\) it is contained in some \(\overline{U^\alpha_i}\) intersecting \(\Phi\); but the \(\overline{U^\alpha_i}\) containing the point \(\xi\) form a base of this point in \(\overline D\); therefore \(\xi\in[\Phi]=\Phi\). At the same time \(\xi\in\overline D\setminus\overline{O\Phi}\)—we have arrived at a contradiction. Lemma 2 is proved.
- The mapping \(f\) of the space \(D\) onto the space \(X\) is closed. It is required to prove: whatever open set \(U\subseteq D\) may be, the set \(V=X\setminus f(D\setminus U)=\mathcal E(x\in X,\ f^{-1}x\subseteq U)\) of all \(x\in X\) for which \(f^{-1}x\subseteq U\) is open in \(X\). But if \(\Phi=f^{-1}x\subseteq U\), then, by Lemma 2, there exists an \(\alpha\) such that \(U_\alpha\Phi\subseteq U\). Therefore \(V\) is the union of all possible sets of the form
\(V^*_\alpha=\mathcal E(x\in X,\ f^{-1}x\subseteq U^*_\alpha)\), where \(U^*_\alpha\subseteq U\) is some set that is the union of certain elements of the cover \(\delta_\alpha\), and \(\alpha\) is arbitrary. Thus it is enough to prove that any set \(V^*_\alpha\) is open, or that its complement
\[ X\setminus V^*_\alpha=\mathcal E(x\in X,\ f^{-1}x\cap(D\setminus U^*_\alpha)\ne\Lambda)=f(D\setminus U^*_\alpha) \]
is closed. But \(D\setminus U^*_\alpha\) is the union of those sets \(U^\alpha_i\in\delta_\alpha\) which did not enter as summands into the union \(U^*_\alpha\), so that the set \(f(D\setminus U^*_\alpha)\), as the union of a finite number of closed sets \(fU^\alpha_i=A^\alpha_i\), is closed in \(X\). The assertion is proved.
- The mapping \(f:D\to X\) is irreducible. Otherwise there would exist such a \(U^\alpha_i\) that \(f(D\setminus U^\alpha_i)=X\). But \(A^\alpha_i\) is an element of the canonical cover \(\alpha\), hence there exists a point \(x_0\in A^\alpha_i\setminus\supset\bigcup_{j\ne i}A^\alpha_j\). If \(x_0=f\xi_0,\ \xi_0=\{i^0_\alpha\}\), then necessarily \(i^0_\alpha=i\), i.e. \(\xi_0\in U^\alpha_i\), so that \(f(D\setminus U^\alpha_i)\ne X\). Our theorem is completely proved.
Received
21 III 1960
CITED LITERATURE
- M. Henriksen, J. Isbell, Duke Math. J., 25, 1, 83 (1958).
- P. S. Aleksandrov, Combinatorial Topology, Moscow–Leningrad, 1947.