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UDC 513.83
MATHEMATICS
V. V. FILIPPOV
ON FACTORIAL \(s\)-MAPPINGS
(Presented by Academician P. S. Aleksandrov, 18 XII 1967)
The present note is devoted to the study of factorial \(s\)-mappings of spaces with a point-countable base onto spaces of point-countable type. All necessary definitions, as well as some comments, may be found in \((^{1})\).
Theorem 1*. Let \(f: X \to Y\) be a factorial \(s\)-mapping of a space \(X\) with a point-countable base onto a \(T_2\)-space \(Y\) of point-countable type. Then the first axiom of countability holds in the space \(Y\).
We shall carry out the proof in several stages.
I. If \(Y'\) is a set closed in \(Y\), and \(X' = f^{-1}(Y')\), then \(f|_{X'}: X' \to Y'\) is a factorial mapping; therefore the following lemma, proved by A. V. Arhangel’skii for completely regular spaces and by M. Choban for \(T_2\)-spaces, allows us to conduct the proof under the assumption that \(Y\) is bicompact, which we shall do.
Lemma (see \((^{2})\)). If a bicompactum \(B\) has countable character in a \(T_2\)-space \(Y\), and a point \(y \in B \subseteq Y\) has countable character in the subspace \(B\), then the point \(y\) has countable character also in the space \(Y\).
II. Let \(m = \{y_k^i,\, i,k = 1,2,\ldots\} \subseteq Y\) and let
\[ y \in \left[\{y_k^i,\, i=1,2,\ldots,\ k>l\}\right] \setminus \{y_k^i,\, i=1,2,\ldots,\ k>l\} \]
for every \(l\). Then from the set \(m\) one can choose a sequence of points
\[ \{y_{k(n)}^{i(n)},\, n=1,2,\ldots,\ k(n+1)>k(n)\} \]
such that
\[ \left[\{y_{k(n)}^{i(n)}\}\right]=\{y_{k(n)}^{i(n)}\}\cup\{y\}. \]
As the sum of a countable number of separable sets, the set \(f^{-1}(m)\) is separable, and therefore, for any point-countable base \(\Delta\) of the space \(X\), the set \(\Delta^*\) of those of its elements which intersect \(f^{-1}(m)\) is at most countable. Let
\[ g=\{Y\setminus [f(\delta)],\ \delta\in \Delta^*,\ y\in [f(\delta)]\}. \]
By construction, \(g\) is a countable family of open sets in \(Y\) containing the point \(y\). We shall suppose the elements of the family \(g\) enumerated, i.e. let \(g=\{G_n\}\). We construct by induction a family \(\gamma=\{\Gamma_n\}\). As \(\Gamma_1\) take \(G_1\). Suppose open sets \(\Gamma_1,\ldots,\Gamma_{n-1}\) have been constructed. As \(\Gamma_n\) take an arbitrary open set satisfying the relation
\[ y\in \Gamma_n \subseteq [\Gamma_n]\subseteq \Gamma_{n-1}\cap G_n. \]
Let
\[ B=\bigcap_{n=1}^{\infty}\Gamma_n. \]
Since \([\Gamma_n]\subseteq \Gamma_{n-1}\), it follows that
\[ B=\bigcap_{n=1}^{\infty}[\Gamma_n], \]
and consequently \(B\) is a bicompactum. The family \(\gamma=\{\Gamma_n\}\) is a certain base of neighborhoods** of the bicompactum \(B\): for any open set \(U\) containing \(B\), beginning with some number \(N\),
\[ [\Gamma_n]\subseteq U, \]
for otherwise \([\Gamma_n]\setminus U\) would be a decreasing sequence of nonempty bicompacta; their intersection would be nonempty and would lie, on the one hand,
* Note added in proof. Very recently the author obtained a considerably stronger result.
Theorem \(1'\). A \(T_2\)-space of point-countable type which is a factorial \(s\)-image of a space with a point-countable base itself has a point-countable base.
Theorem \(1'\) completely covers almost all results known to the author in this direction, and also gives an answer to some of the questions that had remained open.
** In other words, for any open set \(U\) containing the set \(B\), there is an element \(\Gamma\) of the family \(\gamma\) such that \(B\subseteq \Gamma\subseteq U\).
in \(\bigcap_{n=1}^{\infty}[\Gamma_n]=B\), and on the other—in \(Y\setminus U\), which contradicts the inclusion \(B\subseteq U\).
We construct by induction the sequence \(\{y^i_{k(n)},\ n=1,2,\ldots\}\). As \(y^i_{k(1)}\) take any point of the set \(\{y^i_k\}\). Suppose the points \(y^i_{k(1)},\ldots,y^i_{k(n-1)}\) have been chosen. Since, by assumption, \(y\in[\{y^i_k;\ k>k(n-1)\}]\), it follows that \(\Gamma_n\cap\{y^i_k,\ k>k(n-1)\}\ne\varnothing\). As \(y^i_{k(n)}\) take any point of the set \(\Gamma_n\cap\{y^i_k,\ k>k(n-1)\}\). Since every neighborhood of the set \(B\) contains all the points \(y^i_{k(n)}\), beginning with some one of them, there are in the bicompactum \(B\) (and only there—since a closed set and a point in a bicompactum are separable) points that are limit points for this sequence. We shall show that \(y\) is the unique limit point. For this it suffices to show that the set
\[
f^{-1}(\{y^i_{k(n)}\}\cup\{y\})=Q
\]
is closed. Suppose this is not so. Then there exists a point \(x'\in[Q]\setminus Q\). Let \(y'=f(x')\ne y\). Since \(Y\), by assumption (see I), is a bicompactum, there exists an open set \(V\ni y'\) such that \(y\in[V]\). Take an element \(\delta\) of the base \(\Delta\) with \(x'\in\delta\subseteq f^{-1}(V)\). Since \(x'\in[Q]\subseteq[f^{-1}(m)]\), we have \(\delta\in\Delta^*\). Moreover, \(f(\delta)\subseteq[f(\delta)]\subseteq[V]\) and \(y'\in[V]\); therefore \(Y\setminus[f(\delta)]\) is an element of the family \(g\) that does not contain the point \(y'\). Hence \(y'\in B\). But, as was already noted, the sequence \(\{y^i_{k(n)}\}\) has no limit points outside \(B\). Thus the supposition that there exists a point \(x'\in[Q]\setminus Q\) has led us to a contradiction and, consequently, \(y\) is indeed the unique limit point of the sequence \(\{y^i_{k(n)}\}\), as required.
III. We shall show that the mapping \(f\) is pseudo-open.
Suppose this is not so. Then there exists an open set \(V\subseteq X\) such that \(\operatorname{Int} f(V)\) does not contain all points whose preimages lie entirely in \(V\). Consider \(U=\operatorname{Int} f^{-1}f(V)\). Since, obviously, \(V\subseteq U\) and \(f(V)=f(U)\), the set \(\operatorname{Int} f(U)=\operatorname{Int} f(V)\) also does not contain all points whose preimages lie entirely in \(U\). It follows that there exist points
\[
\begin{aligned}
&y^i_k\in Y\quad (i,k=1,2,\ldots),\qquad f^{-1}(y^i_k)\subseteq X\setminus U,\\
&y_k\in Y,\\
&x^i_k\in f^{-1}(y^i_k),\qquad x_k\in f^{-1}(y_k),\\
&x^i_k\to x_k\quad *\ \text{as } i\to\infty,\\
&x'_k\in f^{-1}(y_k),\\
&y\in Y,\qquad f^{-1}(y)\subseteq U,\\
&x'_k\to x\in f^{-1}(y).
\end{aligned}
\]
Indeed, otherwise we would have that \(f^{-1}f(X\setminus U)\) is a closed set, as is
\[
f\bigl(f^{-1}f(X\setminus U)\bigr)=f(X\setminus U),
\]
which contains not a single point whose preimage lies entirely in \(U\). Then all such points would lie in the open set
\[
Y\setminus f(X\setminus U)\subseteq f(U)=f(V),
\]
which contradicts our supposition.
By II, from the set \(\{y^i_k\}\) we can choose a sequence \(\{y^i_{k(n)},\ n=1,2,\ldots\}\) such that
\[
[\{y^i_{k(n)}\}]=\{y^i_{k(n)}\}\cup\{y\}.
\]
It follows that the set \(f^{-1}(\{y^i_{k(n)}\})\) has no limit points not belonging to it and lying outside \(f^{-1}(y)\). But since \(f^{-1}(\{y^i_{k(n)}\})\subset X\setminus U\), where the set \(X\setminus U\) is closed and
\[
f^{-1}(y)\cap(X\setminus U)=\varnothing,
\]
the set \(f^{-1}(\{y^i_{k(n)}\})\) has no limit points in \(f^{-1}(y)\) either. But then the set \(\{y^i_{k(n)}\}\) is closed, which contradicts the inclusion \(y\in[\{y^i_{k(n)}\}]\). The contradiction obtained proves that the mapping \(f\) is pseudo-open.
IV. Let \(y\in Y\), and let \(\{H_n\}\) be an arbitrary countable covering of the set \(f^{-1}(y)\) by open subsets of \(X\). We shall show that from it one can choose
* \(x_i\to x\) if every neighborhood of the point \(x\) contains all the points \(x_i\), beginning with some one of them.
a finite number of elements \(\{H_1,\ldots,H_n\}\) such that
\[ y \in \operatorname{Int} f(H_1\cup\cdots\cup H_n). \]
Suppose that this is not so; then, since the mapping \(f\) is pseudo-open, for any \(k\) there exist sequences of points
\[ y_k^i\in Y\quad (i=1,2,\ldots),\qquad f^{-1}(y_k^i)\cap (H_1\cup\cdots\cup H_k)\ne\varnothing, \]
\[ x_k^i\in f^{-1}(y_k^i),\qquad x_k^i\to x_k\in f^{-1}(y). \]
In accordance with II, choose from the set \(\{y_k^i\}\) a sequence \(\{y_{k(n)}^{i(n)}\}\) such that
\[ [\{y_{k(n)}^{i(n)}\}]=\{y_{k(n)}^{i(n)}\}\cup\{y\}. \]
Take an arbitrary \(x\in f^{-1}(y)\); then, for some \(N\), \(x\in H_N\), and \(H_N\) is a neighborhood of the point \(x\) which does not meet the sets \(f^{-1}(y_{k(n)}^{i(n)})\) as soon as \(k(n)\ge N\), and consequently
\[ x\in [f^{-1}\{y_{k(n)}^{i(n)}\}]. \]
By II the set \(f^{-1}(\{y_{k(n)}^{i(n)}\})\) has no limit points outside \(f^{-1}(y)\), and therefore the set \(f^{-1}(\{y_{k(n)}^{i(n)}\})\) is closed. Consequently, the set
\[ \{y_{k(n)}^{i(n)}\}=ff^{-1}(\{y_{k(n)}^{i(n)}\}) \]
is closed as well, which contradicts the equality
\[ [\{y_{k(n)}^{i(n)}\}]=\{y_{k(n)}^{i(n)}\}\cup\{y\}. \]
The assertion is proved.
V. The set \(\Delta'\) of elements of the base \(\Delta\) which meet \(f^{-1}(y)\) is at most countable. Therefore the family of all finite subsets of the set \(\Delta'\) is also at most countable, as is the family
\[ \beta=\{\operatorname{Int} f(\delta_1\cup\cdots\cup\delta_n),\ \delta_1,\ldots,\delta_n\in\Delta',\ n=1,2,\ldots\}. \]
If we show that this family is a certain base at the point, then theorem 1 will thereby be proved. Let \(V\) be an open subset of \(Y\) containing the point \(y\). For each point \(x\in f^{-1}(y)\) there is an element \(\delta\) of the base \(\Delta\) such that
\[ x\in\delta\subseteq f^{-1}(V). \]
The set of all such elements is at most countable; they all belong to \(\Delta'\) and together cover the set \(f^{-1}(y)\). By IV, from this family one can choose a finite number of elements \(\delta_1,\ldots,\delta_n\) such that
\[ y\in \operatorname{Int} f(\delta_1\cup\cdots\cup\delta_n), \]
i.e. \(\operatorname{Int} f(\delta_1\cup\cdots\cup\delta_n)\) is an element of the family \(\beta\) containing the point \(y\) and contained in the set \(V\), as was required.
From the corollaries of the theorem just proved we give one (see (2)).
Theorem 2. Let \(f:X\to Y\) be a factor \(s\)-mapping of a metric space \(X\) onto a regular separable space \(Y\) of point-countable type. Then the space \(Y\) is metrizable.
In particular:
Corollary. A separable bicompactum which is a factor \(s\)-image of a metric space is metrizable.
In conclusion I express my gratitude to my scientific adviser A. V. Arhangel’skii, and also to M. Choban for valuable discussions.
Mechanical-Mathematical Faculty
of M. V. Lomonosov Moscow State University
Received
15 XII 1967
References
- A. V. Arhangel’skii, UMN, 21, no. 4, 133 (1966).
- M. Choban, DAN, 166, no. 3, 562 (1966).