UDC 513.831

MATHEMATICS

V. K. BELNOV

ON COMPACTIFICATIONS ONTO COMPACTA

(Presented by Academician P. S. Aleksandrov on 12 I 1970)

The following definition is known \((^1)\).

Definition 1. Let \(X\) be a topological space. A point \(x \in X\) is called a point of noncompactness of the space \(X\) if at the point \(x\) there exists no base consisting of open sets with bicompact closures.

The set \(N\) of all points of noncompactness of the space \(X\) is, obviously, closed in \(X\).

Theorem 1. A metric space with a countable base is compactifiable onto a complete metric space if and only if the set of all its points of noncompactness is compactifiable onto some complete metric space.

Unfortunately, the analogous theorem on compactifications onto compacta is false.

Example 1. The space \(X\) is a subset of the plane \(R^2\):

\[ X=\bigcup_{n=1}^{\infty}[0,1]\times\left(\frac{1}{n}\right)\cup(0,0)\cup(1,0). \]

As is known \((^1)\), the space \(X\) is not compactifiable onto any compactum, even though the set of all points of noncompactness \(N\) of the space \(X\) is a compactum:

\[ N=\{(0,0);(1,0)\}. \]

Example 2. The space \(Y\) is a subset of the space \(R^3\):

\[ Y=([0,1]\times[0,1]\times(0,1])\cup X, \]

where \(X\) is the space constructed above. The set of all points of noncompactness of the space \(Y\) coincides exactly with \(X\), and consequently is not compactifiable onto any compactum. Nevertheless, one can show that the space \(Y\) is compactifiable onto some compactum \(Z\). (The proof of this fact almost repeats the proof of the theorem on the “burnt” polyhedron \((^1)\).)

Theorem 2. Let \(X\) be a Hausdorff space and let \(A,B \subseteq X\) be closed subsets of the space \(X\) such that:

1) \(A\) and \(B\) are compactifiable onto bicompacta;
2) \(X=A\cup B\);
3) \(K=A\cap B\) is bicompact.

Then the space \(X\) is compactifiable onto a bicompactum.

Proof. Let \(f_1:A\to A'\) and \(f_2:B\to B'\) be compactifications of the spaces \(A\) and \(B\) onto bicompacta \(A'\) and \(B'\), respectively. We construct a bicompactum \(X'\) onto which the space \(X\) is compactified. The underlying set of the space \(X'\) coincides with the set \(X\). The topology of the space \(X'\) is arranged as follows: if a point \(x\in A\setminus K\subseteq X'\), a base at the point \(x\) is formed by all sets of the form \(f_1^{-1}(V)\), where \(V\) is an open set in \(A'\setminus f(K)\) containing the point \(f_1(x)\). A base at points \(x\in B\setminus K\subseteq X'\) is defined analogously. Let a point \(x\in K\subset X'\). A base at the point \(x\) is formed by all sets of the form \(f_1^{-1}(U)\cup f_2^{-1}(V)\), where \(U\) and \(V\) are neighborhoods of the point \(x\) in the spaces \(A'\) and \(B'\), respectively, such that \(f_1^{-1}(U)\cap K=f_2^{-1}(V)\cap K\). It is not difficult to verify that the space \(X'\) is Hausdorff, and the spaces \(A'\) and \(B'\) can be

identify with the subspaces \(A\) and \(B\) of the space \(X'\). Hence it follows that \(X'\) is bicompact. It is clear that the space \(X\) is compactified onto the space \(X'\). The theorem is proved.

Theorem 3. Let \(X\) be such a completely regular space that: 1) the set \(N\) of all points of noncompactness of the space \(X\) is finite—\(N=\{x_i\}\), \(i=1,2,\ldots,n\). 2) For each point \(x_i\in N\) there exists a neighborhood \(Ox_i\) with bicompact boundary such that \(Ox_i\cap Ox_j=\varnothing\) for any points \(x_i,x_j\in N\), \(i\ne j\). Then the space \(X\) is compactified onto a bicompactum.

Proof. From the conditions of the theorem there follows the existence of such neighborhoods \(\Gamma_i\) of the points \(x_i\in N\), \(i=1,2,\ldots,n\), that \(x_i\in \Gamma_i\subset [\Gamma_i]_X\subset Ox_i\), and the set \(\Gamma_i\) has bicompact boundary for all \(i\). Consider \(\beta X\), the maximal bicompact extension of the space \(X\). As is known \({}^{(2)}\), \(\beta X\) is a perfect extension of the space \(X\); therefore
\[ \operatorname{Fr}_{\beta X}O\langle\Gamma_i\rangle = [\operatorname{Fr}_X\Gamma_i]_{\beta X} = \operatorname{Fr}_X\Gamma_i,\quad i=1,2,\ldots,n, \]
where \(O\langle\ \rangle\) is the operator of maximal cut-out in \(\beta X\). Let
\[ A_i=[\Gamma_i]_{\beta X}\setminus (X\setminus N). \]
These are, obviously, closed subsets of \(\beta X\), and for them we have \(A_i\cap X=x_i\) and
\[ A_i\cap A_j\subset \operatorname{Fr}_{\beta X}O\langle\Gamma_i\rangle\cap \operatorname{Fr}_{\beta X}O\langle\Gamma_j\rangle \subset [\Gamma_i]_X\cap[\Gamma_j]_X=\varnothing, \]
if \(i\ne j\). Put
\[ A_0=(\beta X\setminus X)\setminus \bigcup_{i=1}^{n} A_i. \]
We shall show that \(A_0\) is closed in \(\beta X\). Indeed, since
\[ [\beta X\setminus X]_{\beta X}\cap X=N, \]
we have
\[ [A_0]_{\beta X}\cap X=\varnothing. \]
Further, it is easy to see that
\[ [A_0]_{\beta X}\cap \left(\bigcup_{i=1}^{n} A_i\right) = [A_0]_{\beta X}\cap \left(\bigcup_{i=1}^{n}\operatorname{Fr}_{\beta X}O\langle\Gamma_i\rangle\right) = [A_0]_{\beta X}\cap X=\varnothing. \]
Take an arbitrary point \(y\in X\), \(y\notin N\), and consider the continuous decomposition of the space \(\beta X\) whose elements are: the points of the set \(X\setminus (N\cup y)\), the bicompacts \(A_i\), \(i=1,2,\ldots,n\), and the bicompact \(A_0\cup y\). The space of this continuous decomposition is some bicompactum \(Y\), and the restriction of the natural mapping \(f:\beta X\to Y\) to the space \(X\) is a compactification of this space onto the bicompactum \(Y\): \(f|_X:X\to Y\). The theorem is proved.

In what follows we shall need the following lemma:

Lemma 1. Let \(X\) be a metric space with a countable base such that: 1) \(X\) is an absolute \(F_\sigma\), and 2) \(\dim X=0\), and let \(\rho\) be some metric of the space \(X\). Then \(X\) can be represented in the form
\[ X=\bigcup_{i=1}^{\infty} K_i, \]
where the \(K_i\) are pairwise disjoint compacta and \(\operatorname{diam}K_i\to 0\) as \(i\to\infty\), where \(\operatorname{diam}K_i\) is considered in the metric \(\rho\).

From Lemma 2 the following easily follows.

Theorem 4. Let \(X\) be a metric space with a countable base. In order that the space \(X\) have a compact extension with a countable remainder, it is necessary and sufficient that \(X\) be \(\pi\)-compact and be an absolute \(G_\delta\).

Corollary \({}^{(3)}\). Every \(\pi\)-compact metric space with a countable base that is an absolute \(G_\delta\) is compactified onto a compactum.

Definition 2. A metric space \(X\) with a countable base is called weakly \(\pi\)-compact if there exists a system of sets open in the space \(X\),
\[ \{U_i\},\quad i=1,2,\ldots, \]
with compact boundaries, such that for any points \(x,y\in X\), \(x\ne y\), one can find open sets
\[ U,V\in \{U_i\}, \]
for which
\[ x\in U,\quad y\in V,\quad \text{and}\quad U\cap V=\varnothing. \]

Theorem 5. Let \(X\) be a weakly \(\pi\)-compact space. Then the space \(X\) is compactified onto some \(\pi\)-compact space \(Y\), which is metrizable and has a countable base.

Proof. Since the space \(X\) is weakly \(\pi\)-compact, there exists such a system \(\mathcal U\) of open subsets of \(X\):
\[ \mathcal U=\{U_i\},\quad i=1,2,\ldots, \]
with compact boundaries, that for any points \(x,y\in X\), \(x\ne y\), one can find such \(U,V\in\mathcal U\), for which
\[ x\in U,\quad y\in V,\quad \text{and}\quad U\cap V=\varnothing. \]
A certain system of subsets \(\mathfrak A=\{A_\alpha\}\) of the space \(X\) is called algebraically closed if it satisfies the following

conditions: 1) if \(A_{\alpha_0}\) and \(A_{\alpha_1}\in\mathfrak A\), then also \(A_{\alpha_0}\cap A_{\alpha_1}\in\mathfrak A\); 2) if \(A_{\alpha_0}\in\mathfrak A\), then also \(X\setminus [A_{\alpha_0}]_X\in\mathfrak A\). Using induction, it is not difficult to prove that there exists a minimal algebraically closed system \(\overline{\mathcal U}\) containing the system \(\mathcal U\) and having the following properties: a) \(\overline{\mathcal U}\) consists of a countable number of open sets, \(\overline{\mathcal U}=\{V_j\}\), \(j=1,2,\ldots\); b) every \(V_j\in\overline{\mathcal U}\) has compact boundary. Construct the space \(Y\). The underlying set of the space \(Y\) is the set \(X\), and a base of open sets is formed by the elements of the system \(\overline{\mathcal U}\). It is clear that \(Y\) is a Hausdorff space with a countable base. Let us show that the space \(Y\) is \(\pi\)-compact. Let \(y\in Y\), and let \(Oy\) be an arbitrary neighborhood of the point \(y\). There exists an open set \(U\) of the space \(Y\), \(U\in\overline{\mathcal U}\), such that \(y\in U\subseteq Oy\). Since the set \(X\setminus [U]_X\in\overline{\mathcal U}\), the set \([U]_X\) is closed in the space \(Y\) and \([U]_X=[U]_Y\); therefore \(\operatorname{Fr}_Y U=\operatorname{Fr}_X U\) and is a compactum. Let us prove regularity, and consequently metrizability, of the space \(Y\). Let \(y\in Y\), and let \(Oy\) be an arbitrary neighborhood of the point \(y\). Without loss of generality one may assume that \(Oy\in\overline{\mathcal U}\). If \([Oy]_Y=Oy\), then the assertion is obvious. Otherwise, for every point \(x\in\operatorname{Fr}_Y Oy\), by the Hausdorffness of the space \(Y\), there exists a neighborhood \(Ox\) in the space \(Y\) such that \([Ox]_Y\not\ni y\). The set of all such neighborhoods forms an open cover of the compactum \(\operatorname{Fr}_Y Oy\); therefore there is a finite set of points \(x_i\in\operatorname{Fr}_Y Oy\) such that \(\operatorname{Fr}_Y Oy\subseteq\bigcup_i Ox_i\) and \([\bigcup_i Ox_i]_Y\not\ni y\). Let
\[ U=Oy\cap\bigl(Y\setminus [\bigcup_i Ox_i]_Y\bigr). \]
It is not difficult to see that \(y\in U\subseteq [U]_Y\subseteq Ou\). Since the space \(X\) is condensed onto the space \(Y\), the theorem is proved.

Definition 3. Let \(X\) be a topological space and let \(x\in X\). The point \(x\) is called a point of non-\(\pi\)-compactness of the space \(X\) if there is no base at the point \(x\) in the space \(X\) consisting of open sets with bicompact boundaries. If the space \(X\) is metric, the set \(M\) of all points of non-\(\pi\)-compactness of the space \(X\) is a set of type \(F_\sigma\).

Theorem 6. Let \(X\) be a weakly \(\pi\)-compact complete metric space with a countable base, the set of all points of non-\(\pi\)-compactness \(M\) of which is a compactum. Then the space \(X\) is condensed onto some compactum.

Proof. Since the space \(X\) is weakly \(\pi\)-compact, there exists a system \(\{U_i\}\), \(i=1,2,\ldots\), of open sets with compact boundaries such that, for any points \(x,y\in X\), one can find \(U,V\in\{U_i\}\) for which \(x\in U\), \(y\in V\), and \(U\cap V=\varnothing\). The set \(M\) of points of non-\(\pi\)-compactness of \(X\) is a compactum; therefore the set \(X\setminus M\) is open in the space \(X\), and every point \(x\in X\setminus M\) has, in the space \(X\), a base consisting of open sets with compact boundaries. Consequently, there exists a system \(\{V_j\}\), \(j=1,2,\ldots\), of open sets of the space \(X\) with compact boundaries such that, for every point \(x\in X\setminus M\), the set of all elements of this system containing the point \(x\) forms a base at the point \(x\) in the space \(X\). Let \(\mathcal U=\{U_i\}\cup\{V_j\}\), \(i,j=1,2,\ldots\). As in the proof of Theorem 5, construct a \(\pi\)-compact metric space \(Y\), using the system \(\mathcal U\). If \(f:X\to Y\) is the condensation obtained in the proof of the preceding theorem, then it is not difficult to verify that \(f|_{X\setminus M}:X\setminus M\to f(X\setminus M)\subseteq Y\) is a homeomorphism. Since \(f|_M:M\to f(M)\) is also a homeomorphism, the space \(Y\), as the union of two absolute \(G_\delta\)’s—\(f(X\setminus M)\) and \(f(M)\)—is an absolute \(G_\delta\), and, by the corollary to Theorem 4, is condensed onto some compactum \(Z\). Therefore the space \(X\) is also condensed onto the compactum \(Z\). The theorem is proved.

The following theorem is proved analogously:

Theorem 7. Let \(X\) be a complete metric space with a countable base, the set of all points of non-\(\pi\)-compactness \(M\) of which is a discrete and closed subset of \(X\). Let \(M=\{x_i\}\), \(i=1,2,\ldots\). Suppose

that there exists a disjoint system $\{O x_i\}$ of open sets with compact boundaries such that $x_i \in O x_i$, $i = 1, 2, \ldots$. Then the space $X$ is compactified on a certain compactum $Z$.

Corollary. Let $X$ be a complete metric space with a countable base. If the space $X$ has only one point of non-$\pi$-compactness, then it is compactified on a certain compactum.

Faculty of Mechanics and Mathematics
Moscow State University
named after M. V. Lomonosov

Received
8 I 1970

References

1. A. S. Parkhomenko, Izv. Akad. Nauk SSSR, Ser. Mat., 5, No. 3, 225 (1941).
2. E. G. Sklyarenko, ibid., 26, No. 3, 427 (1962).
3. I. L. Raukhvarger, DAN, 66, No. 1, 13 (1949).