UDC 513.83

MATHEMATICS

V. Z. POLYAKOV

ON PROXIMALLY OPEN MAPPINGS OF METRIC SPACES

(Presented by Academician P. S. Aleksandrov on 1 XI 1968)

Proximally open mappings \(f:P\to Q\) were introduced in \((^{6})\) as mappings preserving \(\delta\)-neighborhoods of sets*, i.e. satisfying the condition: \(A\subset B\subset P \Rightarrow fA\subset fB\). In the same work the study of such mappings was begun. Here we prove that the equi-open image** of a metrizable proximity space is metrizable; if, moreover, the mapping is compact, the image of a metrizable complete \((^{8})\) space is complete.

The notions given below were apparently first used in \((^{2})\).

Definitions. Sequences \(x_n, y_n\) of points of a proximity space are called equivalent if, for every increasing sequence of indices \(n_k\), one has \(\{x_{n_k}\}\delta\{y_{n_k}\}\). A cover \(\mathfrak M\) is called quasiuniform*** if, whatever equivalent sequences \(x_n\) and \(y_n\) may be, for infinitely many indices \(i\) one has \(x_i\in \operatorname{St}(y_i,\mathfrak M)\).

Lemma 1. Let \(f:M\to P\) be an equi-open mapping of a proximity space, \(\mathfrak U\) a uniform cover of \(M\), \(\mathfrak U^f=\{\operatorname{St}(f^{-1}p,\mathfrak U)\mid p\in P\}\), and \(\mathfrak M=f\mathfrak U^f\). Then \(\mathfrak M\) is a quasiuniform cover of the space \(P\).

Proof. Suppose the contrary. Then there exist equivalent sequences \(x_i\sim y_i\) in the space \(P\) and \(\forall i:\ x_i\notin \operatorname{St}(y_i,\mathfrak M)\). Denote \(R_i=f^{-1}x_i,\ S_i=f^{-1}y_i\), and let \(\mathfrak B\) be a uniform cover star-refining \(\mathfrak U\). We assert that each \(\operatorname{St}(R_i,\mathfrak B)\) intersects only finitely many sets \(S_j\). Indeed, if the set \(J_i=\{j\mid S_j\cap \operatorname{St}(R_i,\mathfrak B)\ne\varnothing\}\) were infinite (for some \(i\)), then, in view of
\[ A_i=\bigcup\{S_j\cap \operatorname{St}(R_i,\mathfrak B)\mid j\in J_i\}\subset \operatorname{St}(R_i,\mathfrak U)=L\in \mathfrak U^f, \]
one would also have \(fA_i=\{y_j\mid j\in J_i\}\subset fL\in\mathfrak M\); but \(\{x_j\mid j\in J_i\}\delta\{y_j\mid j\in J_i\}\), and therefore \(\{x_j\mid j\in J_i\}\cap fL\ne\varnothing\), which contradicts our assumptions about the cover \(\mathfrak M\). Similarly, all sets
\[ J'_i=\{j\mid R_j\cap \operatorname{St}(S_i,\mathfrak B)\ne\varnothing\} \]
are finite.

Now one can—in the same way as this was done in \((^{2})\)—select distant cofinal parts from the sequences \(R_i\) and \(S_i\). To this end define the sequence \(\Theta=\{i_n\}\), putting \(i_0=0\),
\[ i_{n+1}=\min\Bigl(N\setminus \bigcup_{k\le i_n}(J_k\cup J'_k)\Bigr). \]
Put \(R=\bigcup\{R_i\mid i\in\Theta\}\) and \(S=\bigcup\{S_i\mid i\in\Theta\}\); then \(R\bar\delta S\), since, obviously,
\[ R\cap \operatorname{St}(S,\mathfrak B)=\varnothing. \]

* Proximally open, or equi-open, mappings are not analogous to uniformly open mappings in the sense of \((^{4})\), i.e. satisfying the requirement \(\mathfrak U\in\mathfrak U_X:\ f\operatorname{St}(x,\mathfrak U)\supset \operatorname{St}(fx,\mathfrak B)\). The point is that in the case of proximity spaces it is not at all clear whether, for instance in Levin–Raiskov’s sense, the restriction \(F\mid F^{-1}Q\) of an open mapping \(F:X\to \bar Q\) of a compact \(X\) onto a compactification \(Q\) (even if the proximity spaces \(Q\) and \(P=F^{-1}Q\) are metrizable) is open. On the other hand, extensions \(\bar f:\bar P\to\bar Q\) of equi-open mappings \(f:P\to Q\) are open (see \((^{6})\)).

** All mappings under consideration are proximally continuous.

*** For the definition of a uniform cover of a proximity space see, for example, \((^{7})\). It can be shown that every uniform cover is quasiuniform.

Thus, \(R \subset M \setminus S\). Therefore \(fR \subset f(M \setminus S)=P \setminus fS\), i.e. \(fR \delta fS\), or \(\{x_i\mid i\in\Theta\}\delta\{y_i\mid i\in\Theta\}\). This contradicts the fact that the sequences \(x_i\) and \(y_i\) are equivalent.

Theorem 1. If the proximity space \(M\) is metrizable and the mapping \(f:M\to P\) is equi-open, then the space \(P\) is also metrizable.

Proof. Let

\[ \mathfrak D_n^{-1}=\left\{O\left(z,\frac1n\right)\mid z\in M\right\} \]

be the “exact \(1/n\)-covering” of \(M\), and let \(\mathfrak R_n=f\mathfrak D_{n^{-1}}^{\,f}\). By the preceding lemma, all the coverings \(\mathfrak R_n\) are quasi-uniform; obviously,

\[ \mathfrak R_1>\mathfrak R_2>\cdots . \]

In [5] S. Leader proved that a proximity space is metrizable if and only if it has a decreasing system of quasi-uniform coverings separating any two sets precisely when they are far apart. We shall now verify that this condition is satisfied for the sequence \(\mathfrak R_n\).

Let \(A\delta B\); then, in view of the openness of \(f\), \(f^{-1}A\delta f^{-1}B\), and for every \(n\)

\[ f^{-1}A\cap \operatorname{St}\bigl(f^{-1}A,\mathfrak D_{n^{-1}}^{\,f}\bigr)\ne\varnothing, \]

whence it follows that

\[ A\cap \operatorname{St}(B,\mathfrak R_n)\ne\varnothing . \]

Conversely, suppose \(A\bar\delta B\). There exists \(O\subset P\) with \(A\subset O\) and \(O\bar\delta B\); therefore also

\[ f^{-1}A\subset f^{-1}O \quad\text{and}\quad f^{-1}O\bar\delta f^{-1}B. \]

For some \(m\) one has simultaneously

\[ \operatorname{St}\bigl(f^{-1}A,\mathfrak D_{m^{-1}}\bigr)\subset f^{-1}O \quad\text{and}\quad \operatorname{St}\bigl(f^{-1}B,\mathfrak D_{m^{-1}}\bigr)\cap f^{-1}O=\varnothing. \]

Therefore, whatever point \(p\in P\) is chosen, the set

\[ \operatorname{St}\bigl(f^{-1}p,\mathfrak D_{m^{-1}}\bigr) \]

does not meet one of the sets \(f^{-1}A\) or \(f^{-1}B\). Indeed, if

\[ \operatorname{St}\bigl(f^{-1}p,\mathfrak D_{m^{-1}}\bigr)\cap f^{-1}A\ne\varnothing, \]

then

\[ \operatorname{St}\bigl(f^{-1}A,\mathfrak D_{m^{-1}}\bigr)\cap f^{-1}p\ne\varnothing; \]

thus \(f^{-1}p\cap f^{-1}O\ne\varnothing\), hence \(f^{-1}p\subset fO\). Thus,

\[ f^{-1}A\cap \operatorname{St}\bigl(f^{-1}B,\mathfrak D_{m^{-1}}^{\,f}\bigr)=\varnothing, \]

and consequently

\[ A\cap \operatorname{St}\bigl(B,f\mathfrak D_{m^{-1}}^{\,f}\bigr)=\varnothing, \]

i.e.

\[ A\cap \operatorname{St}(B,\mathfrak R_m)=\varnothing . \]

Corollary. If \(f:M\to P\) is an equi-open mapping of metric spaces and \(\mathfrak U\) is a uniform covering of \(M\), then \((f\mathfrak U)^{**}\) is a uniform covering of \(P\).

Proof. From the proof of the cited theorem of Leader it follows that, in metrizable proximity spaces, the star of a quasi-uniform covering is a uniform covering; note also that

\[ f\mathfrak U^{\,f}=(f\mathfrak U)^*. \]

Lemma 2. Let the mapping of proximity spaces \(f:M\to P\) be equi-open and compact, let \(a_i\in M\) be a countable sequence of points, \(x_i=fa_i\in P\) a Cauchy sequence, \(\mathfrak B\) a uniform covering in \(M\), and \(V_i=f\operatorname{St}(a_i,\mathfrak B)\). There exist indices \(i\) and \(j\) for which

\[ U_i\supset \{x_r\mid r>j\}. \]

Proof. Suppose the contrary. Let \(\mathfrak W\) be a uniform covering triply star-inscribed in \(\mathfrak B\), and let

\[ W_i=f\operatorname{St}(a_i,\mathfrak W). \]

First we show (incidentally, only here shall we use the compactness of the mapping!) that every infinite intersection of sets \(W_i\) is empty. Indeed, suppose, to the contrary, that

\[ p\in \bigcap\{W_{i_t}\mid t=1,2,\ldots\}. \]

Then for every \(t\) one can find points

\[ q_t\in f^{-1}p\cap \operatorname{St}(a_{i_t},\mathfrak W). \]

The set \(\{q_t\}\) has a point of contact \(q\), and therefore the set

\[ Q=\{q_t\}\cap \operatorname{St}(q,\mathfrak W) \]

is infinite. It is true that

\[ \operatorname{St}(q,\mathfrak W^*)\supset \operatorname{St}(Q,\mathfrak W); \]

thus the set \(\operatorname{St}(q,\mathfrak W^*)\) contains infinitely many elements \(a_{i_t}\)—denote their totality by \(A\). Now, if \(a\in A\), then

\[ q\in \operatorname{St}(a,\mathfrak W^*) \]

and therefore

\[ \operatorname{St}(a,\mathfrak B)\supset \operatorname{St}\bigl(\operatorname{St}(a,\mathfrak W^*),\mathfrak W^*\bigr) \supset \operatorname{St}(q,\mathfrak W^*)\supset A. \]

Hence

\[ V=f\operatorname{St}(a,\mathfrak B)\supset fA. \]

Choose \(H\) with

\[ V\supset H\supset fA. \]

The set \(\{x_i\}\setminus H\) is far from the subsequence \(fA\), and therefore is finite; hence

\[ V\supset H\supset \{x_r\mid r>\mathrm{const}\}. \]

Now, assuming that the lemma is false, for every \(i\) find an index \(j(i)>i\) with \(x_{j(i)}\notin V_i\); put \(y_i=x_{j(i)}\), and let

\[ R_i=f^{-1}y_i. \]

It is asserted that, whatever \(i\) may be, \(\operatorname{St}(a_i,\mathfrak W)\) meets only finitely many of the sets \(R_k\): indeed, if

\[ R_k\cap \operatorname{St}(a_{i^0},\mathfrak W)\ne\varnothing \]

for all \(k\in\{k\}\), then, in view of

\[ \operatorname{St}(a^{i^0},\mathfrak W)\subset \operatorname{St}(a_{i^0},\mathfrak B), \]

it follows that

\[ \{y_k\}\subset W_{i^0}\subset V_{i^0}, \]

and therefore,

if \(W_{i_0}\subset K\subset V_{i_0}\), in view of \(y_k\sim x_k\sim x\), it follows that \(x_i\in K\) for almost all \(i\)—contrary to our assumption.

We shall show that, analogously to the preceding argument, the set \(\operatorname{St}(R_i,\mathfrak B)\), whatever \(i\) may be, contains only finitely many points \(a_k\). If not, then for infinitely many indices \(k\) (with \(i=\mathrm{const}\)) one has \(\operatorname{St}(a_k,\mathfrak B)\cap R_i\ne\varnothing\), and hence \(\bigcap W_k\ne\varnothing\), which, as we have already established, is impossible.

We now construct the sequences \(a_{i_s}\) and \(R_{i_s}\) in the usual way: \(i_0=0\), and \(i_{s+1}\) is the least index, starting from which \(a_i\notin \operatorname{St}(R_{i_s},\mathfrak B)\) and
\[ R_i\cap \operatorname{St}(a_{i_s},\mathfrak B)=\varnothing . \]
It follows from the construction, evidently, that
\[ \{a_{i_s}\mid s\}\cap \operatorname{St}\bigl(\bigcup\{R_{i_s}\mid s\},\mathfrak B\bigr)=\varnothing \]
and, therefore,
\[ \{a_{i_s}\mid s\}\subset M\setminus \bigcup\{R_{i_s}\mid s\}; \]
further, in view of the proximal openness of \(f\),
\[ \{fa_{i_s}\mid s\}=\{x_{i_s}\mid s\}\subset f\bigl(M\setminus \bigcup\{R_{i_s}\mid s\}\bigr) = P\setminus f\bigl(\bigcup\{R_{i_s}\mid s\}\bigr) = P\setminus \{y_{i_s}\mid s\}, \]
i.e. \(\{x_{i_s}\}\,\delta\,\{y_{i_s}\}\). But both these sequences are subsequences of the fundamental sequence \(x_i\), and therefore cannot be far apart!

Theorem 2. Let \(M\) be a metrizable proximity space, and let the mapping \(f:M\to P\) be compact and equi-open. The space \(P\) is complete if \(M\) is complete.

Proof. Let \(x_i\in P\) be an arbitrary fundamental sequence. Put
\[ S_i=\bigcup\{f^{-1}x_j\mid j\ge i\} \]
and consider the system
\[ \mathfrak Z=\{\operatorname{St}(S_i,\mathcal U)\mid i,\mathcal U\} \]
(the covers \(\mathcal U\) are uniform). We assert that \(\mathfrak Z\) is a stable system in Isbell’s sense \((^3)\), i.e.
\[ \forall \mathcal U:\ \bigcap \{\operatorname{St}(Z,\mathcal U)\mid Z\in\mathfrak Z\}\in \mathfrak Z . \]

We first show that, for any fixed uniform cover \(\mathfrak B\) of the space \(M\), there exists an \(n\) with
\[ S_n\subset \operatorname{St}(S_m,\mathfrak B) \]
for whatever \(m\). Suppose this is not so, and that, in particular, for arbitrarily large \(n\) there are infinitely many \(m\) with
\[ \bigcap_m \bigl(f^{-1}x_n\setminus \operatorname{St}(f^{-1}x_m,\mathfrak B)\bigr)\ne\varnothing . \]
Number the chosen \(n\)’s by indices \(r\) and fix points
\[ a_r\in \bigcap_m \bigl(f^{-1}x_{n_r}\setminus \operatorname{St}(f^{-1}x_m,\mathfrak B)\bigr). \]
Thus \(fa_r=x_{n_r}\), while
\[ \operatorname{St}(a_r,\mathfrak B)\cap f^{-1}x_m=\varnothing, \]
i.e.
\[ x_m\notin V_{n_r}=f\operatorname{St}(a_{n_r},\mathfrak B), \]
and for each \(r\) there are infinitely many such numbers \(m\). This contradicts the preceding lemma: indeed, for some \(r\) and \(r_0\) one has
\[ V_{n_r}\supset \{x_{n_{r'}}\mid r'>r_0\}, \]
and, consequently, it must also hold that
\[ V_{n_r}\supset (x_i\mid i>i_0). \]

Now consider an arbitrary uniform cover \(\mathcal U\) in the space \(M\), and assume that \(\mathfrak B\) is star-refined into it. Let
\[ Z=\operatorname{St}(S_n,\mathfrak B)\in\mathfrak Z, \]
where \(n=n(\mathfrak B)\) is the number found in the preceding paragraph. If \(X\in\mathfrak Z\), then, for some \(m\), \(S_m\subset X\); consequently,
\[ Z=Z(\mathcal U)\subset \operatorname{St}(\operatorname{St}(S_m,\mathfrak B),\mathfrak B) =\operatorname{St}(S_m,\mathfrak B^*)\subset \operatorname{St}(X,\mathcal U). \]
Thus the system \(\mathfrak Z\) is indeed stable.

J. R. Isbell in \((^3)\) showed that every complete metric space satisfies the following ultracompleteness condition: an arbitrary stable regular\(^*\) filter has nonempty intersection. Let
\[ J=\bigcap\mathfrak Z; \]
choose an arbitrary \(u\in J\) and prove that \(u=\lim x_i\). Indeed, let \(O\) be an open neighborhood of \(u\). Put
\[ G=f^{-1}u\cap J; \]
obviously,
\[ G\subset f^{-1}O. \]
We shall now verify that
\[ f^{-1}x_i\cap f^{-1}O\ne\varnothing \]
for almost all \(i\): otherwise one would have
\[ M\setminus f^{-1}O\supset \bigcup\{f^{-1}x_i\mid i\ge r\}=S_r,\qquad M\setminus G\supset M\setminus f^{-1}O, \]
and hence
\[ M\setminus G\in\mathfrak Z. \]
But from this it should follow that \(G\cap J=\varnothing\), i.e. \(G=\varnothing\), which is absurd. Thus
\[ f^{-1}O\cap f^{-1}x_i\ne\varnothing \]
and, consequently, \(x_i\in O\) for almost all \(i\).

We have proved that an arbitrary fundamental sequence

\(^*\) The regularity condition for the system \(\mathfrak Z\) means that for every \(X\in\mathfrak Z\) there is a \(Y\in\mathfrak Z\) with \(X\supset Y\).

converge in \(P\), i.e. \(P\) is sequentially complete. Since, by Theorem 1, this space is metrizable, it is complete.

Remark. In proving Theorem 2, we have established, in essence, the following more general assertion:

The image of an ultracomplete proximity space under countably compact proximity-open mappings is sequentially complete.

Noncompact equi-open mappings of metric spaces do not preserve the property of completeness, as the following simple example shows.

Example. Consider the set of points of the plane

\[ \Psi=\bigcup\left\{\left\{\frac{1}{n}\right\}\times [n,\infty)\mid n=1,2,\ldots\right\} \]

with the induced proximity. Its projection onto the axis of abscissas

\[ p:\Psi\to \left\{\frac{1}{n}\mid n=1,2,\ldots\right\} \]

is a proximity-open mapping.

If, on the contrary, a mapping \(\Phi:P\to Q\) of proximity spaces is “uniformly” open, i.e. if for every uniform cover \(\mathfrak P\) there is a \(\mathfrak Q\) such that

\[ \Phi\,\operatorname{St}(x,\mathfrak P)\supset \operatorname{St}(\Phi x,\mathfrak Q), \]

then it is not difficult to prove the theorem on preservation of completeness (analogous to (4)) in full generality:

Theorem 3. The image of an ultracomplete proximity space under a “uniformly” open mapping is ultracomplete.

Proof. Let \(\mathfrak X\) be a stable regular filter in \(Q\). Consider the trace

\[ \mathfrak Y=\{\Phi^{-1}X\mid X\in \mathfrak X\}; \]

obviously, all the sets

\[ Z_{\mathfrak P}=\bigcap\{\operatorname{St}(Y,\mathfrak P)\mid Y\in\mathfrak Y\} \]

are nonempty, since, by the properties of \(\Phi\) and \(\mathfrak X\), \(\Phi Z_{\mathfrak P}\in\mathfrak X\). It is verified directly (independently of \(\Phi\)) that the filter with trace

\[ \mathfrak Z=\{Z_{\mathfrak P}\} \]

is stable. The set \(\Phi(\cap\mathfrak Z)\) is the limit of the filter \(\mathfrak X\).

The author expresses his gratitude to Yu. M. Smirnov for his attention to the work.

Steklov Mathematical Institute
Academy of Sciences of the USSR

Received
30 X 1968

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