UDC 513.83

V. ZAITSEV

ON BICOMPACT SEMIREGULAR AND HAUSDORFF EXTENSIONS

(Presented by Academician P. S. Aleksandrov, 4 I 1968)

The purpose of this paper is to obtain all semiregular \((= T_b)\)-bicompact extensions of \(T_\lambda\)-spaces * (Theorem 1) and all Hausdorff bicompact extensions of Tikhonov \((=\) completely regular) spaces (Theorem 2) in the form of limits of certain projection spectra, namely spectra that are so-called refinements (maximal finite ones) of the spectrum \(S_X\) of the given space \(X\).

Definition 1. A projection spectrum \(S^*=\{K_\alpha^*, \mathfrak{F}_\alpha^{\alpha'}\}\) is called a refinement of the spectrum \(S=\{K_\alpha,\mathfrak{F}_\alpha^{\alpha'}\}\) if both spectra are directed by one and the same set of indices \(\{\alpha\}\), and for each \(\alpha\) a one-to-one mapping \(f_\alpha\) is given from the set \(\dot K_\alpha\) of all vertices of the complex \(K_\alpha\) onto the set \(\dot K_\alpha^*\) of all vertices of the complex \(K_\alpha^*\), inducing a simplicial mapping \(f_\alpha\) of the complex \(K_\alpha\) into the complex \(K_\alpha^*\), commuting with the projections in the sense that, for \(\alpha' \ge \alpha\), we have
\[ f_\alpha \mathfrak{F}_\alpha^{\alpha'}=\mathfrak{F}_\alpha^{*\alpha'} f_{\alpha'}; \]
if, moreover, for all \(\alpha\) one has \(f_\alpha K_\alpha=K_\alpha^*\), then the spectra \(S\) and \(S^*\) are isomorphic to one another.

Remark 1. Let the spectrum \(S^*=\{K_\alpha^*,\mathfrak{F}_\alpha^{*\alpha'}\}\) be a refinement of the spectrum \(S=\{K_\alpha,\mathfrak{F}_\alpha^{\alpha'}\}\). Identify each vertex \(e_\alpha\) of the complex \(K_\alpha\) with the corresponding vertex \(f_\alpha e_\alpha\) of the complex \(K_\alpha^*\); then the complexes \(K_\alpha\) and \(K_\alpha^*\) have one and the same set of vertices, hence \(K_\alpha \subseteq K_\alpha^*\) and \(\mathfrak{F}_\alpha^{*\alpha'}=\mathfrak{F}_\alpha^{\alpha'}\).

Let the spectrum \(S^{(1)}=\{K_\alpha^{(1)},\mathfrak{F}_\alpha^{\alpha'}\}\) be a refinement of the spectrum \(S^{(2)}=\{K_\alpha^{(2)},\mathfrak{F}_\alpha^{\alpha'}\}\). Then every maximal thread of the spectrum \(S^{(2)}\) is a thread of the spectrum \(S^{(1)}\); if it is also maximal in the spectrum \(S^{(1)}\), then we call it invariant under the refinement \(S^{(1)}\) of the spectrum \(S^{(2)}\).

Let \(S_X=\{N_\alpha,\mathfrak{F}_\alpha^{\alpha'}\}\) be the spectrum of the \(T_\lambda\)-space \(X\). If \(x\in X\), then by \(t_\alpha(x)\) we denote the simplex of the nerve \(N_\alpha\) whose support corresponds to the set of all elements of the covering \(\alpha\) containing the point \(x\). It is shown that \(\{t_\alpha(x)\}\) is a maximal thread \(\xi(x)=\{t_\alpha(x)\}\) of the spectrum \(S_X\).

Definition 2. A refinement \(S=\{N_\alpha',\mathfrak{F}_\alpha^{\alpha'}\}\) of the spectrum \(S_X=\{N_\alpha,\mathfrak{F}_\alpha^{\alpha'}\}\) of a given \(T_\lambda\)-space \(X\) is called correct if, under this refinement, every thread of the form \(\xi=\{t_\alpha(x)\}\) of the spectrum \(S_X\) is invariant.

Let \(S^{(1)}=\{K_\alpha^{(1)},\mathfrak{F}_\alpha^{\alpha'}\}\) and \(S^{(2)}=\{K_\alpha^{(2)},\mathfrak{F}_\alpha^{\alpha'}\}\) be two refinements of the spectrum \(S=\{K_\alpha,\mathfrak{F}_\alpha^{\alpha'}\}\). We shall say that \(S^{(1)}\) is a natural refinement of the spectrum \(S^{(2)}\) if, for all \(\alpha\), we have \(K_\alpha^{(2)}\subseteq K_\alpha^{(1)}\).

Remark 2. The set of all correct refinements of the spectrum \(S_X\) is partially ordered: \(S^{(2)}\) follows \(S^{(1)}\) if \(S^{(1)}\) is a natural refinement of the spectrum \(S^{(2)}\). The set of all bicompact semiregular extensions of a given \(T_\lambda\)-space \(X\) is partially ordered in the usual way: set \(b_2X \ge b_1X\) if there exists a continuous mapping of the space \(b_2X\) onto \(b_1X\) with fixed \(x\in X\).

* For terminology and notation see (²). We consider only finite \((=\) consisting of finite complexes) spectra. Every finite spectrum is bounded \((=\) every one of its threads is contained in a maximal thread) (³), and, consequently, complete (¹).

1. Let \(S=\{K_\alpha,\mathfrak G_\alpha^{\alpha'}\}\) be an arbitrary spectrum and let the simplex \(t_{\alpha_\sigma}\in K_{\alpha_\sigma}\). Recall that by \(Ot_{\alpha_\sigma}\), respectively \(\Phi t_{\alpha_\sigma}\), we denote the set of all maximal threads \(\xi'=\{t_\alpha'\}\) satisfying the condition \(t_\alpha'\leq t_{\alpha_\sigma}\), respectively \(t_\alpha'\geq t_{\alpha_\sigma}\). By the definition of the topology in the inverse-limit space \(\tilde S\) of the spectrum \(S\), the sets \(Ot_\alpha\) form an open base of the space \(\tilde S\). We shall call the spectrum \(S\) symmetric if the sets \(\Phi t_\alpha\) form a closed base of the space; here it is enough to restrict oneself to the zero-dimensional simplexes \(t_\alpha=e_\alpha\).

Remark 3. If the spectrum \(S_X\) for a given space \(X\) exists, then it is symmetric (([^2], Theorem 1).

Definition 3. A spectrum \(S\) is called a \(\lambda\)-strengthening of the spectrum \(S_X\) of a given \(T_\lambda\)-space \(X\) if it is symmetric and is a correct strengthening of the spectrum \(S_X\).

Theorem 1. Let \(X\) be an arbitrary \(T_\lambda\)-space. For every bicompact semiregular extension \(\bar X=bX\) of the space \(X\), the spectrum \(S_{\bar X}\) is a \(\lambda\)-strengthening of the spectrum \(S_X\). Conversely, the limit of every \(\lambda\)-strengthening \(S\) of the spectrum \(S_X\) is a bicompact semiregular extension of the space \(X\).

The correspondence \(bX\to S_{bX}\) between all bicompact semiregular extensions \(bX\) of the space \(X\) and all \(\lambda\)-strengthenings of the spectrum \(S_X\) of this space, obtained by virtue of what has just been said, is one-to-one and preserves the order in the sense that from \(b_2X\geq b_1X\) it follows that \(S_{b_2X}\geq S_{b_1X}\).

We prove the first assertion of Theorem 1; we take the following proposition as known:

(A) Let \(\bar X\) be an arbitrary extension of the space \(X\). Then, denoting by \(A\) any canonical closed (we write briefly \(\varkappa a\)) set of the space \(X\), and by \(\bar A\) its closure in \(\bar X\), we have \(X\cap \bar A=A\), and the correspondence \(A\to \bar A\) is a one-to-one correspondence between all \(\varkappa a\)-sets in \(X\) and all \(\varkappa a\)-sets in \(\bar X\), and hence also a one-to-one correspondence between all decompositions \(\alpha=\{A_\alpha^1,\ldots,A_\alpha^{s_\alpha}\}\) of the space \(X\) and all decompositions \(\bar\alpha=\{\bar A_\alpha^1,\ldots,\bar A_\alpha^{s_\alpha}\}\) of the space \(\bar X\), which is an order isomorphism (the relations \(\alpha'>\alpha\) and \(\bar\alpha'>\bar\alpha\) are equivalent). Hence it follows at once that

Lemma 1. If the spectrum \(S_X\) exists, then the spectrum \(S_{\bar X}\) also exists, and it is a strengthening of the spectrum \(S_X\).

Lemma 2. If \(\bar X\) is a \(T_\lambda\)-extension of the space \(X\), then \(S_{\bar X}\) is a correct strengthening of the spectrum \(S_X\).

This lemma follows easily from the following assertion:

(B) Suppose that, among the elements of a decomposition \(\alpha=\{A_\alpha^1,\ldots,A_\alpha^{s_\alpha}\}\) of the space \(X\), a given point \(x\in X\) is contained in the elements \(A_\alpha^{i_0},\ldots,A_\alpha^{i_r}\), and only in them. Then, among the elements \(\bar A_i\) of the corresponding decomposition \(\bar\alpha\), only \(\bar A_\alpha^{i_0},\ldots,\bar A_\alpha^{i_r}\) contain the point \(x\).

Indeed, from \(x\in \bar A_\alpha^i\) it follows that \(x\in X\cap \bar A_\alpha^i=A_\alpha^i\).

Let \(\bar X\) be a bicompact \(T_\xi\) (hence \(T_\lambda\))-extension of the space \(X\). Relying on Lemma 2 and Remark 3, we conclude that the spectrum \(S_{\bar X}\) is a \(\lambda\)-strengthening of the spectrum \(S_X\).

The correspondence \(bX\to S_{bX}\) is one-to-one. Indeed, let the spectra \(S_{b_1X}\) and \(S_{b_2X}\) be naturally isomorphic. Since \(\tilde S_{b_1X}=b_1X\), \(\tilde S_{b_2X}=b_2X\) (see [^2]), it follows that \(b_1X=b_2X\).

We pass to the second part of Theorem 1.

Lemma 3. Let \(X\) be an arbitrary \(T_\lambda\)-space and let \(S=\{N_\alpha,\mathfrak G_\alpha^{\alpha'}\}\) be a correct strengthening of the spectrum \(S_X\). Then the limit space \(\tilde S\) is a bicompact \((T_1)\)-extension of the space \(X\), semiregular if \(S\) is a \(\lambda\)-strengthening.

By Theorems 1 and 4 of [2], the space \(\hat S_X\) is a bicompact \(T_\zeta\)-extension of the space \(X\). Denoting by \(\hat S\) the full limiting space of the (arbitrary) spectrum \(S\) (consisting of all, and not only maximal, threads of the spectrum \(S\), with the same topology as in \(\hat S\)), we have the natural mapping \(f:\hat S_X\to \hat S\). From the correctness of the refinement \(S\) it follows easily that under the mapping \(f\) the set \(X\subseteq \tilde S_X\subseteq \hat S_X\) is mapped topologically onto the set \(fX\subseteq \tilde S\subseteq \hat S\). It remains to show that \(fX\) is dense in \(\tilde S\).

Let \(Ost_\alpha\) be an arbitrary element of the open base of the space \(\tilde S\). Take some vertex \(e_\alpha\) of the simplex \(t_\alpha\in N_\alpha'\). Since \(X\) is dense in \(\tilde S_X\), there exists a point \(x\equiv \xi(x)\in Oe_\alpha\subseteq \tilde S_X\). Then \(t_\alpha(x)=e_\alpha\le t_\alpha\), so that (in view of the correctness of the refinement \(S\)) we have \(\tilde S\ni x\equiv \xi(x)\in O_{\hat S}t_\alpha\), and \(fX\) is dense in \(\tilde S\). Let \(S\) be a \(\lambda\)-refinement of the spectrum \(S_X\). We shall prove that then the extension \(\tilde S\) is semiregular. For this it is enough to prove that all \(\Phi_S e_\alpha\) are \(ж\alpha\)-sets. This in turn follows from the equality

\[ \Phi_S e_\alpha^i=[A_\alpha^i]_{\tilde S}, \tag{1} \]

valid for every correct refinement \(S\) of the spectrum \(S_X\) (by \(e_\alpha^i\) one always denotes the vertex of the nerve \(N_\alpha\) corresponding to the element \(A_\alpha^i\) of the cover \(\alpha\)). Fix \(\alpha=\alpha_0\). The inclusion \([A_{\alpha_0}^i]_{\tilde S}\subseteq \Phi_S e_{\alpha^i}\) is checked directly. We prove the reverse inclusion. Let \(\xi=\{t_\alpha\}\in \Phi_S e_{\alpha_0}^i\subseteq \tilde S\). It is required, for arbitrary \(\alpha=\alpha_1\), to find a point \(x\equiv \xi(x)\equiv \{t_\alpha(x)\}\in A_{\alpha_0}^i\) satisfying the condition \(t_{\alpha_1}(x)\le t_{\alpha_1}\in \xi\). For this we take \(\alpha'\), \(\alpha'\ge \alpha_0,\alpha_1\). Then \(\mathfrak D_{\alpha}^{\alpha'}\cdot t_{\alpha'}=t_{\alpha_0}\), \(\mathfrak D_{\alpha_1}^{\alpha'}t_{\alpha'}=t_{\alpha_1}\).

Let \(e_{\alpha'}^k\le t_{\alpha'}\) be such a vertex that \(\mathfrak D_{\alpha}^{\alpha'}e_{\alpha'}^k=e_{\alpha_0}^i\le t_{\alpha_0}\); then \(\mathfrak D_{\alpha_1}^{\alpha'}e_{\alpha'}^k=e_{\alpha_1}^j\le t_{\alpha_1}\), and, consequently,

\[ A_{\alpha'}^k\subseteq A_{\alpha_0}^i,\qquad A_{\alpha'}^k\subseteq A_{\alpha_1}^j,\qquad A_{\alpha_0}^i\cap I A_{\alpha_1}^j\ne \Lambda. \]

Take \(x\in A_{\alpha_0}^i\cap I A_{\alpha_1}^j\), \(\xi(x)=\{t_\alpha(x)\}\). Then \(\xi(x)\equiv x\in A_{\alpha_0}^i\) and \(t_{\alpha_1}(x)=e_{\alpha_1}^j\le t_{\alpha_1}\), and the point \(x\equiv \xi(x)\) we need has been found.

We shall show that the one-to-one mapping \(bX\to S_{bX}\) is a mapping “onto.” For this we prove the formula

\[ S_{\tilde S}=S, \tag{2} \]

valid for every \(\lambda\)-refinement \(S=\{N_\alpha',\mathfrak D_\alpha^{\alpha'}\}\) of the spectrum \(S_X=\{N_\alpha,\mathfrak D_\alpha^{\alpha'}\}\) of the given \(T_\lambda\)-space \(X\). Since, by Lemma 3, \(\tilde S\) is a bicompact \(T_\zeta\)-extension of the space \(X\), the spectrum \(S_{\tilde S}=\{N_\alpha'',\mathfrak D_\alpha^{\alpha'}\}\) is a \(\lambda\)-refinement of the spectrum \(S_X\). It remains to prove that the spectrum \(S_{\tilde S}=\{N_\alpha'',\mathfrak D_\alpha^{\alpha'}\}\) is naturally isomorphic to \(S\), which follows from the assertion

(B) The vertices \(e_\alpha^{i_0},\ldots,e_\alpha^{i_r}\) of the nerve \(N_\alpha\) then and only then are the vertices of some simplex \(t_\alpha\) in the complex \(N_\alpha'\), when
\([A_\alpha^{i_0}]_{\tilde S}\cap\cdots\cap[A_\alpha^{i_r}]_{\tilde S}\ne\Lambda\).

Indeed, let \(\lvert e_\alpha^{i_0},\ldots,e_\alpha^{i_r}\rvert=t_\alpha\in N_\alpha'\). Since the spectrum \(S\) is complete [4], \(\Lambda\ne \Phi_S t_\alpha=\Phi_S e_\alpha^{i_0}\cap\cdots\cap\Phi_S e_\alpha^{i_r}\) and \([A_\alpha^{i_0}]_{\tilde S}\cap\cdots\cap[A_\alpha^{i_r}]_{\tilde S}\ne\Lambda\) by virtue of formula (1).

Conversely, let \(\Lambda\ne [A_\alpha^{i_0}]_{\tilde S}\cap\cdots\cap[A_\alpha^{i_r}]_{\tilde S}=\Phi_S e_\alpha^{i_0}\cap\cdots\cap\Phi_S e_\alpha^{i_r}\ni \xi=\{t_\alpha\}\). Then the vertices \(e_\alpha^{i_0},\ldots,e_\alpha^{i_r}\) are vertices of the simplex \(t_\alpha\), and the simplex \(\lvert e_\alpha^{i_0}\ldots e_\alpha^{i_r}\rvert\in N_\alpha'\) exists.

We prove that from \(b_2X\ge b_1X\) it follows that \(S_{b_2X}\ge S_{b_1X}\). But if \(b_2X\ge b_1X\), then from \([A_\alpha^{i_0}]_{b_2X}\cap\cdots\cap[A_\alpha^{i_r}]_{b_2X}\ne\Lambda\) it follows that \([A_\alpha^{i_0}]_{b_1X}\cap\cdots\cap[A_\alpha^{i_r}]_{b_1X}\ne\Lambda\), and hence \(S_{b_1X}\) is a natural refinement of the spectrum \(S_{b_2X}\).

Definition 4*. A thread \(\xi=\{t_\alpha\}\) of a spectrum \(S=\{K_\alpha,\mathfrak D_\alpha^{\alpha'}\}\) is called regular if for every \(\alpha\) there exist \(\alpha'\), \(\alpha'\ge\alpha\), such that

* In [4], V. I. Ponomarev gave another definition of regularity, equivalent to ours for all complete spectra.

that every simplex \(t'_{\alpha'} \in K_{\alpha'}\) having a common face with the simplex \(t_\alpha \in \xi\) is projected into a face of the simplex \(t_\alpha \in \xi\): \(\delta_{\alpha}^{\alpha'} t_{\alpha'} \leq t_\alpha\).

A spectrum all of whose maximal threads are regular is called regular; its limit space is regular (Ponomarev).

A regular spectrum \(S=\{N_\alpha,\delta_\alpha^{\alpha'}\}\) which is a correct strengthening of the spectrum \(S_X\) of a (Tychonoff) space \(X\) will be called an \(H\)-strengthening of the spectrum \(S_X\). Its limit space \(\tilde S\) is a bicompact \(T_2\)-extension of the space \(X\). Conversely, the spectrum of every bicompact \(T_2\)-extension \(bX\) is regular \((^4)\); hence, being, by the preceding results, a correct strengthening of the spectrum \(S_X\), it is an \(H\)-strengthening. Just as in the case of bicompact \(T_\xi\)-extensions, one proves the mutual one-to-one character of the mapping \(bX \to S_{bX}\) from the set of all bicompact \(T_2\)-extensions of the space \(X\) to the set of all \(H\)-strengthenings of the spectrum \(S_X\).

This mapping is a “mapping onto,” which follows from equality (2), proved for \(H\)-strengthenings in the same way as for \(\lambda\)-strengthenings (formula (2) is valid even for any correct strengthening).

Let us prove that the mapping under consideration is an order isomorphism. We shall call a thread of a spectrum proper if there exists a unique maximal thread containing it. A spectrum is proper if all its threads are proper.

Lemma 4. Every \(H\)-strengthening \(S=\{N'_\alpha,\delta_\alpha^{\alpha'}\}\) of the spectrum \(S_X\) of a Tychonoff space \(X\) is proper.

Indeed, \(\tilde S\) is a \(T_2\)-space; hence, for any two distinct points \(\xi^1=\{t_\alpha^1\}\in \tilde S\) and \(\xi^2=\{t_\alpha^2\}\in \tilde S\) there is an \(\alpha=\alpha_0\) such that \(O_{\alpha_0}\xi^1 \cap O_{\alpha_0}\xi^2=\Lambda\). Lemma 4 follows from the fact that \(t_{\alpha_0}^1\in \xi^1\) and \(t_{\alpha_0}^2\in \xi^2\) do not have a single common vertex \(e_{\alpha_0}^i\) (if \(e_{\alpha_0}^i\) were such a vertex, then \(O_{S e_{\alpha_0}^i}\subseteq O_{\alpha_0}\xi^1 \cap O_{\alpha_0}\xi^2\)). But \(S\) is a correct strengthening of the spectrum \(S_X\), and therefore \(O_{S e_{\alpha_0}^i}\ne \Lambda\). A contradiction.

Lemma 5. Let a proper spectrum \(S_i=\{C_\alpha,\delta_\alpha^{\alpha'}\}\) be a strengthening of a (bounded) spectrum \(S=\{K_\alpha,\delta_\alpha^{\alpha'}\}\). Then, assigning to each maximal thread \(\xi=\{t_\alpha\}\in \tilde S\) the unique maximal thread containing it, \(f\xi=\{t_\alpha^{(c)}\}\), we obtain a natural mapping \(f:\tilde S\to \tilde S_c\) onto \(\tilde S_c\). If the spectrum \(S_c\) is regular, then \(f\) is continuous.

Let us prove that \(f:\tilde S\to \tilde S_c\) is a mapping “onto.” Take arbitrarily \(\eta=\{t_\alpha^{(c)}\}\in \tilde S_c\) and in the spectrum \([\eta]=\{[t_\alpha^{(c)}],\delta_\alpha^{\alpha'}\}\), where by \([t_\alpha^{(c)}]\) is denoted the combinatorial closure of the simplex \(t_\alpha^{(c)}\), take some zero-dimensional thread \(\eta^0=\{e_\alpha\}\). This thread is also a thread of the spectrum \(S\), and it is contained in it in the maximal thread \(\xi=\{t_\alpha\}\). Then \(f\xi=\eta\) (otherwise the thread \(\eta^0\) of the proper spectrum \(S_c\) would have two maximal threads containing it: \(\eta\) and \(f\xi\)). We leave the proof of the continuity of the mapping \(f\) to the reader. Thus, we have

Theorem 2. Let \(X\) be an arbitrary Tychonoff space. To every bicompact Hausdorff extension \(bX\) of the space \(X\) there corresponds an \(H\)-strengthening \(S_{bX}\) of the spectrum \(S_X\). Moreover, every \(H\)-strengthening \(S_{bX}\) of the spectrum \(S_X\) is put in correspondence with a unique Hausdorff bicompact extension \(\tilde S=bX\) of the space \(X\).

The one-to-one correspondence \(bX\to S_{bX}\) thus obtained between the two partially ordered sets—the set of all Hausdorff bicompact extensions of the Tychonoff space \(X\) and the set of all \(H\)-strengthenings of its spectrum \(S_X\)—is an order isomorphism.

The work was written under the supervision of P. S. Aleksandrov, to whom I express my sincere gratitude.

Mechanical-Mathematical Faculty
of Moscow State University
named after M. V. Lomonosov

Received
2 I 1968

REFERENCES

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4. V. Ponomarev, Mat. sborn., 60, 1, 79 (1963).