MATHEMATICS

Z. Frolik

LOCALLY TOPOLOGICALLY COMPLETE SPACES*

(Presented by Academician P. S. Aleksandrov, 14 X 1960)

Definition. We shall call a space** \(P\) locally complete if every point \(x \in P\) is contained in some open complete*** subspace of the space \(P\).

Obviously, every complete space is locally complete. The converse assertion is false:

Example 1. Let \(T\) be the set of all countable ordinal numbers, and let \(K\) be a segment of the number line. Order the set \(R = T \times K\) lexicographically. In the topology determined by this order, \(R\) is a compact, locally bicompact, and locally metrizable space. The set \(R'\) of all \((a,0)\), where \(a\) is an isolated number, is not an \(F_\sigma\)-set in \(R\), because only finite \(M \subset R'\) are closed in \(R\). Thus, \(P\) is not a \(G_\delta\)-set in \(R\), and since \(P\) is dense in \(R\), \(P\) is not complete. Obviously, \(P\) is locally complete and locally metrizable.

Theorem 1. Every paracompact locally complete space \(P\) is complete.

Proof. By definition there exists an open cover \(\mathfrak A\) consisting of complete subspaces. Since \(P\) is paracompact, one can inscribe in \(\mathfrak A\) a locally finite open cover \(\mathfrak B\). For every \(B \in \mathfrak B\) there exists a complete \((^6)\) sequence \(\{\mathfrak A_n(B)\}\) of open covers of the space \(B\) such that \(\mathfrak A_{n+1}(B) \subset \mathfrak A_n(B)\). Denote by \(\mathfrak A_n\) the union of all \(\mathfrak A_n(B)\), \(B \in \mathfrak B\). It is not difficult to prove that the sequence \(\{\mathfrak A_n\}\) is complete. Thus, \(P\) is complete.

Theorem 2. Every locally complete space contains a dense, open and complete space.

Proof. If \(\mathfrak A\) is some maximal disjoint subsystem of the system of all complete open subspaces, then the union of the system \(\mathfrak A\) is complete, dense and open.

Obviously, every locally bicompact space is a \(k\)-space****.

However, a locally complete space need not be a \(k\)-space:

Example 2. Let \(N\) be an infinite countable discrete space. Take a sequence \(\{N_n\}\) of subsets of the set \(N\) such that the sets \(N_n - N_{n+1}\) are infinite. Denote by \(K_n\) the closure

* Example 1 and Theorem 1 of the present paper are also contained in A. Arkhangel’skii’s paper “On topological spaces complete in the sense of Čech” (\(^5\)), submitted for publication on 18 V 1960. P. S. Aleksandrov.

** “Space” always means a completely regular topological space.

*** We call a space \(P\) complete if it is topologically complete in the sense of E. Čech, i.e. if for every space \(R \supset P\) such that \(\overline P = R\), \(P\) is a \(G_\delta\)-set in \(R\).

**** A space \(P\) is called a \(k\)-space if every set \(M \subset P\) whose intersection with every bicompact \(K \supset P\) is closed is itself closed.

sets \(N_n\) in the Čech extension \(\beta(N)\) of the space \(N\). It turns out that the complete subspace \(P = N \cup \bigcap_{n=1}^{\infty} K_n\) of the space \(\beta(N)\) is not a \(k\)-space.

Theorem 3. If an open subset \(U\) of a locally complete space \(P\) contains some point of condensation of a countable \(N \subset P\), then there exists a bicompact set \(K \subset U\) intersecting \(N\) in an infinite set.

Proof is easily carried out using the characterization of complete spaces by means of complete sequences of open coverings.

From this it follows easily:

Theorem 4. The topological product of a locally complete compact space and any compact space is compact.

Theorem 5. If an open subset \(U\) of a locally complete space \(P\) contains some point of condensation of a countable disjoint system \(\mathfrak A\) of open subsets of the space \(F\), then there exists a bicompact \(K \subset U\) intersecting infinitely many elements of the system \(\mathfrak A\).

From this it easily follows:

Theorem 6. The topological product of a pseudocompact locally complete space and any pseudocompact space is pseudocompact.

The character of a point \(x\) of a space \(P\) is called \((^1)\) the least cardinality of a complete system of neighborhoods of the point \(x\). The pseudocharacter of a point \(x\) of a space \(P\) is called \((^1)\) the least cardinality of a system \(\mathfrak A\) of open sets for which \(\bigcap \mathfrak A = (x)\).

Theorem 7. In a locally complete space, the characters and pseudocharacters of points coincide.

A mapping of a space \(P\) into a space \(Q\) will be called open if the images of open subsets of the space \(P\) are open in \(Q\).

Theorem 8. Let \(f\) be a continuous and open mapping of a space \(P\) onto a space \(Q\). If \(P\) is locally bicompact, complete, or locally complete, then \(Q\), respectively, is locally bicompact, complete, or locally complete\(^*\).

The first assertion is obvious; the third is a consequence of the second, which we shall now prove. Let \(F\) be the continuous extension of the mapping \(f\) to the Čech extension \(\beta(P)\). Since \(P\) is complete, \(P = \bigcap_{n=1}^{\infty} U_n\), where \(U_n\) are open subsets of the space \(P\). It is not difficult to prove that the open kernel \(V_n\) of the set \(F[U_n]\) contains \(Q\) and \(Q = \bigcap_{n=1}^{\infty} V_n\).

Theorem 9. Let \(f\) be a perfect (i.e., continuous, closed, and such that the complete inverse images of all points are bicompact) mapping of a space \(P\) onto a space \(Q\). In order that the space \(P\) be locally bicompact, complete, or locally complete, it is necessary and sufficient that the space \(Q\) be, respectively, locally bicompact, complete, or locally complete.

Proof. Let \(F\) be the mapping from the proof of Theorem 8. It is not difficult to prove that \(F^{-1}[Q] = P\). Consequently, if \(Q\) is locally bicompact or complete, then \(P\) also is, respectively, locally bicompact or complete. If \(M \subset Q\), then \(f|f^{-1}[M]\) is a perfect mapping of the space \(f^{-1}[M]\) onto \(M\). Thus, by what has already been proved, if \(Q\) is locally complete, then \(P\) is also locally complete.

\(^*\) This result, even for multivalued perfect mappings, was proved by V. Ponomarev \((^2)\) (for locally bicompact spaces and for complete spaces).

Now suppose that \(P\) is complete. Let \(\{\mathfrak A_n\}\) be a complete sequence of additive open covers of the space \(P\). Denote by \(\mathfrak B_n\) the system of all \(Q-f[P-A]\), where \(A \in \mathfrak A_n\). It is not difficult to prove that \(\{\mathfrak B_n\}\) is a complete sequence of the space \(Q\). Thus, \(Q\) is a complete space.

Finally, suppose that the space \(P\) is locally complete. Every point \(x \in P\) is contained in some open and complete \(U(x) \subset P\). Let \(y \in Q\). Denote the set \(f^{-1}[y]\) by \(K\). Since \(K\) is bicompact, there exists a finite set \(M \subset P\) such that
\[ U=\bigcup\{U(x);\, x \in M\} \supset K. \]
Put \(V=Q-f[P-U]\). Clearly, the set \(V\) is open and
\[ H=f^{-1}[V]\subset U. \]
Since \(U\) is complete, \(H\) is also complete, and, finally, by what was proved above, \(V\) is complete. The assertion on local bicompactness is proved quite analogously.

From Theorems 7 and 9 it follows:

Theorem 10. Let \(f\) be a closed and continuous mapping of a complete metrizable space \(P\) onto a space \(Q\). In order that \(Q\) be metrizable, it is necessary and sufficient that \(Q\) be a complete space.

Proof. Since \(f\) is a closed mapping, the pseudocharacters of the points of the space \(Q\) are countable. Thus, if \(Q\) is complete, then, by Theorem 7, the characters of the points of the space \(Q\) are countable. From the countability of the characters of the points of the space \(Q\) and from the properties of the mapping \(f\), it follows by a well-known theorem of Stone \({}^{(3)}\) that \(Q\) is a metrizable space.

If \(Q\) is a metrizable space, then, by a well-known theorem of I. A. Vainshtein \({}^{(4)}\), the boundaries of the inverse images of points are compact. For every \(y \in Q\), let \(K(y)\) be the boundary of the set \(f^{-1}[y]\), if it is nonempty; if it is empty, let \(K(y)\) be some one-point subset of the set \(f^{-1}[y]\).

Clearly, the subspace
\[ K=\bigcup\{K(y);\, y \in Q\} \]
is complete, and the mapping \(f\), considered on \(K\), is perfect. Thus, by Theorem 9, the space \(Q\) is complete.

Charles University
Prague, Czechoslovakia

Received
14 X 1960

REFERENCES

\({}^{1}\) P. Aleksandrov, P. Urysohn, Verh. Kön. Akad. Amsterdam, 14, No. 1 (1929).
\({}^{2}\) V. Ponomarev, DAN, 124, 268 (1959).
\({}^{3}\) A. H. Stone, Proc. Am. Math. Soc., 7, 690 (1956).
\({}^{4}\) I. A. Vainshtein, DAN, 57, 319 (1947); Uch. zap. Mosk. univ., 155, 3 (1952).
\({}^{5}\) A. Arhangel’skii, Vestn. Mosk. univ., ser. matem. i mekh., No. 1 (1961).
\({}^{6}\) Z. Frolik, DAN, 137, No. 3 (1961).