UDC 513.83

MATHEMATICS

Kh. N. INASARIDZE

ON EXTENSIONS AND REMAINDERS OF FINITE ORDER OF COMPLETELY REGULAR SPACES

(Presented by Academician P. S. Aleksandrov on 28 V 1965)

In this note all topological spaces considered are completely regular Hausdorff spaces, and all mappings are continuous.

A sequence

\[ \left(C^0(X),\ b_1C^0(X),\ b_2C^1(X),\ \ldots,\ b_{n-1}C^{n-2}(X),\ b_nC^{n-1}(X)\right) \]

of spaces, where \(C^0(X)=X\); \(b_iC^{i-1}(X)\) is some compact extension of \(C^{i-1}(X)\) and \(C^i(X)=b_iC^{i-1}(X)\setminus C^{i-1}(X)\) for \(1\le i\le n\), is called a compact extension of \(n\)-th order of the space \(X\), and \(C^n(X)\) an \(n\)-th order remainder of the space \(X\). In the case where all \(b_i\) are Čech extensions \(\beta_i\), the extension is called the Čech extension of \(n\)-th order of the space \(X\), and the remainder of \(n\)-th order \(\Gamma^n(X)\) the Čech remainder of \(n\)-th order of the space \(X\).

Let \(M_n^*\) be the class of all compact extensions of \(n\)-th order of the space \(X\). We introduce in \(M_n^*\) a quasi-ordering as follows:

\[ (\widetilde C^0(X),\ \widetilde b_1\widetilde C^0(X),\ \widetilde b_2\widetilde C^1(X),\ldots,\widetilde b_n\widetilde C^{\,n-1}(X)) \ge (C^0(X),\ b_1C^0(X),\ b_2C^1(X),\ldots,\ b_nC^{n-1}(X)), \]

if there exist mappings

\[ f_i:\widetilde b_i\widetilde C^{\,i-1}(X)\to b_iC^{i-1}(X), \quad i=1,\ldots,n, \]

such that \(f_1\) is an extension of the identity mapping \(e:\widetilde C^0(X)\to C^0(X)\), and for \(i=2,\ldots,n\) the mapping \(f_i\) is an extension, on \(\widetilde b_i\widetilde C^{\,i-1}(X)\), of the mapping induced by \(f_{i-1}\). If all \(f_i\) are homeomorphisms, then we shall call such extensions equivalent.

Let \(M_n\) be the set of classes of equivalent extensions. Using Lemma 1.5 of the paper \((^1)\), one can prove the following lemma.

Lemma 1. The set \(M_n\) is a partially ordered set whose largest element is the class of the Čech extension of \(n\)-th order.

It follows from Lemma 1 that if some \(n\)-th order remainder of the space \(X\) has a perfect property (see \((^1)\)), then all its \(n\)-th order remainders have this property.

Denote by \(P_n\) the following property of a space: the \(n\)-th order remainder of the space has the perfect property \(P\). Then from Lemma 1.5 \((^1)\) and from Lemma 1 the following proposition follows immediately.

Lemma 2. The property \(P_n\) is a perfect property.

Denote by \(R^1(X)\) the set of all points of the space \(X\) that have no compact neighborhoods, and let, for \(n>1\),

\[ R^n(X)=R^1[R^{n-1}(X)] \]

(see \((^1)\)). We shall assume that \(X=R^0(X)\).

Let \(K_n\), where \(n\ge0\), denote the class of all spaces whose \(n\)-th order remainder is compact. By a space with compact remainder of zeroth order we mean a compact space. It can be shown that there exists a space belonging to \(K_n\) and not belonging to \(K_{n-1}\), and that if the space \(X\) has compact

if it is a remainder of finite order, then it contains an open everywhere dense locally compact subset \(A\), and \(X\setminus A=R^1(X)\).

Since compactness is an absolute property, Lemma 2 implies the following theorem.

Theorem 1. If \(f\) is a perfect mapping of \(X\) onto \(Y\), then \(X\) belongs to the class \(K_n\) if and only if \(Y\) belongs to the class \(K_n\).

Theorem 2. A space \(X\) belongs to the class \(K_n\) if and only if \(R^{n/2}(X)\) is compact, when \(n\) is even, and \(R^{(n-1)/2}(X)\) is locally compact, when \(n\) is odd.

This theorem follows from the fact that \(R^j(X)\) is a remainder of order \(2j\) of the space \(X\) for \(j\geqslant 1\).

From Theorem 2, in particular, Theorem 3.1 of Henriksen and Isbell \((^1)\) follows: \(X\) is locally compact at infinity (i.e., has a compact remainder of second order) if and only if \(R^1(X)\) is compact. Further, \(X\) has a locally compact remainder of second order if and only if \(R^1(X)\) is locally compact.

It is easy to see that if \(F\) is a closed subset of the space \(X\), then \(R^k(F)\) is a closed subset of \(R^k(X)\). Hence it follows that if \(X\in K_n\), then \(F\in K_n\). Therefore, if \(X\) is a compact space and \(A\) is its subset belonging to \(K_n\), then \(X\setminus A\) belongs to \(K_{n+1}\).

Theorem 3. Let \(A\) and \(B\) be subsets of a space \(X\), with \(A\) compact and \(B\) a space in \(K_n\). Then, for even \(n\), \(A\cup B\in K_n\), and for odd \(n\), \(A\cup B\in K_{n+1}\).

Proof. For \(n=0\) this is known. Let \(bX\) be a compact extension of \(X\). We have
\[ \overline{A\cup B}\setminus A\cup B=(\overline{B}\setminus B)\setminus A. \]
The closure is taken in \(X\). Let \(\overline{B}\setminus B=D\). It is clear that
\[ [A\cup(\overline{D}\setminus D)]\cap(D\setminus A)=\varnothing \quad\text{and}\quad [A\cup(\overline{D}\setminus D)]\cup(D\setminus A)=\overline{D}\cup A. \]
Let us prove the theorem for \(n=1,2\). If \(B\in K_1\), then \(D\in K_0\) and \(\overline{D}\setminus D=\varnothing\). Then \(D\setminus A\in K_1\), and therefore \(A\cup B\in K_2\). If \(B\in K_2\), then \(D\in K_1\) and \(\overline{D}\setminus D\in K_0\). Since \(A\cup(\overline{D}\setminus D)\in K_0\), we obtain that \(D\setminus A\in K_1\), and therefore \(A\cup B\in K_2\). Suppose the theorem is proved for \(n=m, m+1\); we shall prove it for \(n=m+2, m+3\). If \(B\in K_{m+2}\), then \(\overline{D}\setminus D\in K_m\). By induction \(A\cup(\overline{D}\setminus D)\in K_{m+1}\). Then \(D\setminus A\in K_{m+2}\), and therefore \(A\cup B\in K_{m+3}\). If \(B\in K_{m+3}\), then \(\overline{D}\setminus D\in K_{m+1}\). By induction \(A\cup(\overline{D}\setminus D)\in K_{m+1}\). Therefore we obtain that \(D\setminus A\in K_{m+2}\) and, hence, \(A\cup B\in K_{m+3}\). The theorem is proved.

Using Theorem 3, one can prove the following propositions:

1. The direct product of a compact space and a space in \(K_n\) is a space in \(K_n\).

2. The direct product of a locally compact space and a space in \(K_n\) is a space in \(K_{n+1}\) for even \(n\), and a space in \(K_n\) for odd \(n\).

3. If \(X\) is a space in \(K_n\) and \(A\) is its compact subset, then \(X\setminus A\) is a space in \(K_{n+1}\) for even \(n\), and a space in \(K_n\) for odd \(n\).

4. If \(X\) is a locally compact space and \(A\) is its subset belonging to the class \(K_n\), then \(X\setminus A\) is a space in \(K_{n+1}\) for even \(n\), and a space in \(K_{n+2}\) for odd \(n\).

From 4, in particular, it follows that the complement of a locally compact set in a locally compact space is a space for which the set of points at which the space is not locally compact is locally compact.

Let \(f\) be a mapping of \(X\) onto \(Y\). Denote by \(R(f)\) the set of all points \(y\in Y\) such that \(f^{-1}(y)\) is noncompact. By a theorem of Stone \((^2)\), Theorem 88, \(f\) has a unique extension \(f^*:\beta X\to\beta Y\).

Theorem 4. If \(f\) is a closed mapping of \(X\) onto \(Y\), then
\[ f^*[\Gamma^1(X)]\setminus\Gamma^1(Y)=R(f). \]

Proof. It is clear that \(R(f)\subset \bar f^{*}[\Gamma^{1}(X)]\setminus \Gamma^{1}(Y)\). Let \(b\in \Gamma^{1}(X)\) and \(f^{*}(b)=a'\in f^{*}[\Gamma^{1}(X)]\setminus \Gamma^{1}(Y)\). Suppose that \(f^{-1}(a')=A\) is compact. Since \(A\) is closed in \(\beta X\) and \(b\notin A\), there exists a continuous function \(\varphi:\beta X\to I=[0,1]\) such that \(\varphi(a)=0\) for \(a\in A\) and \(\varphi(b)=1\). Let
\[ Q=\{x\in X,\ \varphi(x)\ge 1/2\}. \]
It is clear that \(Q\) is closed in \(X\), and therefore \(f(Q)=Q'\) is closed in \(Y\), i.e. \(Q'=P\cap Y\), where \(P\) is closed in \(\beta Y\). Obviously, \(a'\notin Q'\). Then there exists a continuous function \(\psi:\beta Y\to I\) such that \(\psi(a')=0\) and \(\psi(p)=1\) for \(p\in P\). Let \(k=\psi f^{*}\). Then \(k(b)=0\). Denote by \(D\) the set of points \(x\) of the space \(X\) for which \(\varphi(x)>1/2\) and \(k(x)<1/2\). It is clear that \(b\in D\). The set \(D\) is open, and therefore there exists a point \(s\in X\) such that \(s\in D\), i.e. \(k(s)<1/2\) and \(\varphi(s)>1/2\). Further, we have \(\psi[f^{*}(s)]<1/2\), and therefore \(f^{*}(s)=f(s)\notin Q'\). Hence it follows that \(s\notin Q\), and therefore \(\varphi(s)<1/2\). The contradiction obtained shows that \(f^{-1}(a')\) is noncompact, i.e. \(a'\in R(f)\). The theorem is proved.

Using Theorems 3 and 4, one can prove the following theorem.

Theorem 5. Let \(f\) be a closed mapping of \(X\) onto \(Y\) such that the \(R(f)\)-space is of class \(K_n\). Then, if \(X\in K_1\), then \(Y\in K_{n+2}\), and if \(X\in K_2\), then for even \(n\), \(Y\in K_{n+2}\), while for odd \(n\), \(Y\in K_{n+3}\).

In particular, it follows from Theorem 5 that under a closed mapping of a locally compact space, if the set of points whose full inverse image is noncompact is compact, then the image is a space for which the set of points having no compact neighborhoods is compact.

Let \(f\) be a mapping of \(X\) onto \(Y\), and let \(f^{*}:bX\to bY\) be an extension of \(f\) to their compact extensions \(bX\) and \(bY\). Let \(X_1=bX\setminus X\), \(Y_1=f^{*}(X_1)\), and let \(f_1\) be the mapping of \(X_1\) onto \(Y_1\) induced by the mapping \(f^{*}\). Then the mapping \(f_1:X_1\to Y_1\) will be called a remainder of first order (or simply a remainder) of the mapping \(f\). If \(bX\) and \(bY\) are Čech compactifications, then \(f_1\) is called the Čech remainder of the mapping \(f\). The mapping \(f_n:X_n\to Y_n\) is called a remainder of \(n\)-th order of the mapping \(f\), if \(f_n\) is a remainder of first order of a remainder of \((n-1)\)-st order of the mapping \(f\). It is easy to see that if some remainder of a closed mapping is a perfect mapping, then all remainders are also perfect. Such a mapping \(f\) will be called perfect at infinity.

With the aid of Lemma 1.5 \((^1)\) and Theorem 4 one can prove the following theorem.

Theorem 6. A closed mapping of \(X\) onto \(Y\) is perfect at infinity if and only if, for every point \(x\in X\) having no compact neighborhoods, the full inverse image of its image is compact.

Theorem 7. A closed mapping of \(X\) onto \(Y\) with a closed Čech remainder is perfect at infinity if and only if the full inverse image of every point of \(Y\) is locally compact.

With the help of Theorems 1, 3, and 4 we obtain the following theorem.

Theorem 8. Let \(f\) be a closed mapping of \(X\) onto \(Y\), perfect at infinity, such that \(R(f)\) is compact. Then: 1) if \(X\in K_n\), then \(Y\in K_n\) for even \(n\) and \(Y\in K_{n+1}\) for odd \(n\); 2) if \(Y\in K_n\), then \(X\in K_{n+1}\) for even \(n\) and \(X\in K_n\) for odd \(n\).

Tbilisi Mathematical Institute named after A. M. Razmadze
Academy of Sciences of the Georgian SSR

Received
28 V 1965

References

1. Melvin Henriksen, J. R. Isbell, Duke Math. J., 25, No. 1, 83 (1958).
2. M. H. Stone, Trans. Am. Math. Soc., 41, 375 (1937).