UDC 519.50+519.54

MATHEMATICS

S. I. Nedev

SYMMETRIZABLE* SPACES AND $\varepsilon$-MAPS

(Presented by Academician P. S. Aleksandrov on 23 V 1966)

Let $X$ be a symmetrizable space with symmetric $d$ and $\varepsilon>0$.

Definition 1. A map $f$ of the space $X$ onto a topological space $Y$ will be called a weak $\varepsilon$-map if for every $y \in Y$ there is a point $x \in X$ such that

\[ f^{-1}y \subseteq O_{\varepsilon}x = E\{z:\ z \in X,\ d(x,z)<\varepsilon\}. \]

Definition 2. A map $f$ of the space $X$ onto $Y$ is called an $\varepsilon$-map if for every $y \in Y$ $\operatorname{diam}(f^{-1}y)\leq \varepsilon$.

Theorem. If $X$ is a Hausdorff symmetrizable space and for every $n$ the map $f_n:X\to Y_n$ is a continuous weak $1/n$-map, then the natural map $f:X\to \prod_{n=1}^{\infty}Y_n$ is an embedding; if for every $n$ $f_n$ is a continuous closed $1/n$-map, then $f$ is a homeomorphism.

Proof. It is necessary to show that $f$ is one-to-one. Suppose that $x\in X$, $y\in X$ and $fx=fy$, i.e. $f_nx=f_ny$ for every $n$. Since $f_n$ is a weak $1/n$-map, for each $n$ there exists a point $x_n$ such that $d(x_n,x)=1/n$, $d(x_n,y)=1/n$, i.e. $x=\lim x_n$, $y=\lim_{n\to\infty}x_n$, which is possible only when $x=y$, since, by assumption, $X$ is Hausdorff. Thus, $f$ is an embedding.

Suppose now that $f_n$ is a closed $1/n$-map for each $n$. Let $F$ be a closed subset of $X$. Put $\Phi=fX\cap\left(\prod_{n=1}^{\infty} f_nF\right)$ and show that $fF=\Phi$. It is easy to see that $fF\subseteq \Phi$, and therefore let $y\in \Phi$. Consequently, $y=fx=\{f_nx\}$ for some $x\in X$, for which $f_n^{-1}(f_nx)\cap F\neq \Lambda$. Let $z_n\in f_n^{-1}(f_nx)\cap F$. Since $f_n$ is a $1/n$-map, $d(x,z_n)\leq 1/n$. Consequently, $x\in F$ ($x=\lim_{n\to\infty}z_n$, $z_n\in F$). Thus, $fF$ is closed in $fX$, since $\Phi$ is closed in $fX$, i.e. $f$ is a homeomorphism of $X$ onto $fX$.

In what follows, let $P$ denote such a property of topological spaces that: a) if $X$ is a space with property $P$, then every subspace $X_0$ is also a space with property $P$, and b) if for every $n$ $X_n$ is a space with property $P$, then $\prod_{n=1}^{\infty}X_n$ is also a space with property $P$.

Corollary 1. If $X$ is a Hausdorff symmetrizable space, then $X$ embeds in a space with property $P$ if and only if for every $\varepsilon>0$ it admits a weak $\varepsilon$-map onto a space with property $P$.

* We consider spaces symmetrizable in the sense of A. V. Arhangel’skii (see (2)): a space $X$ is called symmetrizable with symmetric $d$ if: 1) $d(x,y)\geq 0$, $d(x,y)=d(y,x)$ for any $x\in X$, $y\in X$; 2) $d(x,y)=0$ if and only if $x=y$; 3) if $U$ is an open subset of $X$ and $x\in U$, then there exists $\varepsilon>0$ such that $O_{\varepsilon}x\subseteq U$; 4) if $A\subset X$, $A$ is not closed and is nonempty, then there exists $x\in X\setminus A$ such that $d(x,A)=0$.

In particular, \(X\) is compactified onto a metrizable space if and only if, for every \(\varepsilon>0\), it admits a weak \(\varepsilon\)-mapping onto a metrizable space. Moreover: \(X\) has property \(P\) if and only if, for every \(\varepsilon>0\), it admits a closed \(\varepsilon\)-mapping onto a space with property \(P\).

In particular, \(X\) is metrizable if and only if, for every \(\varepsilon>0\), it admits a closed \(\varepsilon\)-mapping onto a metrizable space.

Corollary 2. (J. Ceder). A symmetrizable paracompact space with the first axiom of countability (a strongly symmetrizable paracompact space) admits a compactification onto a metrizable space.

The last assertion can be strengthened as follows: if \(X\) is a symmetrizable paracompact space with the first axiom of countability and, for each \(n=1,2,\ldots\), \(\omega_n\) denotes an open cover of the space \(X\), then there exists a metrizable space \(R\) and a compactification \(f:X\) onto \(R\) such that, for each \(n\), \(f\) is an \(\omega_n\)-mapping.

Proof. Put \(\eta_n=\{\langle O_{1/n}x\rangle,\ x\in X\}\). Next we construct a sequence of open covers \(\gamma_1,\gamma_2,\ldots\) such that:
1) \(\gamma_1\) is star-refined into \(\omega_1 \wedge \eta_1\);
2) \(\gamma_{k+1}\) is star-refined into \(\gamma_k \wedge \omega_{k+1} \wedge \eta_{k+1}\).

On the set \(X\) we introduce a symmetric function \(D(x,y)\) as follows:

\(1^\circ\). If for two given points \(x\) and \(y\) there exists no element of the cover \(\gamma_1\) containing both these points, set \(D(x,y)=D(y,x)=1\).

\(2^\circ\). Suppose case \(1^\circ\) does not occur; then there exists \(k\ge 1\) such that both points \(x\) and \(y\) are contained in some element of the cover \(\gamma_k\). We show that if \(x\ne y\), then among these \(k\) there is a largest \(k=k(x,y)=k(y,x)\). Indeed, otherwise for every \(k\) there would exist points \(x_k\) such that \(x\in O_{1/k}x_k,\ y\in O_{1/k}x_k\), which is impossible, since \(X\) is a Hausdorff space. Put then \(D(x,y)=1/2k(x,y)\) and \(D(z,z)=0\) for every \(z\in X\). The symmetry of the function \(D\) is obvious; it is also clear that \(D(x,y)=0\) if and only if \(x=y\). Consequently, we may consider the space \(X;D\), whose topology is defined as follows: a subset \(A\subseteq X\) is called closed if and only if from \(x\in X\) and \(x\notin A\) it follows that \(D(x,A)>0\).

The following property of the function \(D(x,y)\) is easily proved: if \(D(x,y)\le 1/2^{k+1}\), \(D(y,z)\le 1/2^{k+1}\), then \(D(x,z)\le 1/2^k\). Hence it follows that the function \(D\) satisfies the following condition.

Limit axiom. If \(\lim\limits_{n\to\infty} D(x,x_n)=0\) and \(\lim\limits_{n\to\infty} D(x_n,y_n)=0\), then \(\lim\limits_{n\to\infty} D(x,y_n)=0\).

Applying a theorem of A. V. Arhangel’skii \((^3)\), we obtain that the space \((X;D)\) is strongly symmetrizable with the symmetric \(D\) and is metrizable. The identity mapping \(f\) of the space \(X\) onto the space \((X;D)\) is a compactification, since the relation

\[ E\{z:\ z\in X,\ D(x,z)<1/2^k\}=\operatorname{St}_{\gamma_{k+1}}x \tag{1} \]

holds for every point \(x\) and for every \(k\), by construction.

It remains to show that the mapping \(f\) is an \(\omega_n\)-mapping for each \(n\). But this follows immediately from the fact that relation (1) holds and that the cover \(\gamma_{k+1}\) is star-refined into \(\omega_{k+1}\).

Thus Corollary 2 is proved. If it is applied in the special case when \(X\) is bicompact, we obtain the theorem of B. Nemitskii \((^5)\): a symmetrizable bicompact space with the first axiom of countability is metrizable. This is so because a compactification of a bicompact space is a homeomorphism. Similarly, if we take into account a theorem of A. V. Arhangel’skii \((^1)\), asserting that a feathered paracompact space which is compactified onto a metrizable space—

…property, is metrizable, we obtain A. V. Arhangel’skii’s theorem \((^2)\): a symmetrizable feathered paracompact space with the first axiom of countability is metrizable.

In connection with the preceding, the following proposition is of interest.

Proposition. If \(X\) is a symmetrizable space whose symmetrix \(d\) is such that for every point \(x\in X\) and for every \(\varepsilon>0\) there is a \(\delta=\delta(x,\varepsilon)>0\) such that from \(d(x,y)<\delta\) it follows that \(d(x,y)<d(y,X\setminus O_\varepsilon x)\), then \(X\) is paracompact with the first axiom of countability.

Proof. We first show that the symmetrix \(d\) agrees with the topology of the space \(X\) in the strong sense. For this we prove the following lemma:

Lemma. If a space \(Z\) is symmetrizable with symmetrix \(\rho\), then \(Z\) is symmetrizable in the strong sense with symmetrix \(\rho\) if and only if \(\rho\) satisfies the following condition

\[ (\alpha)\quad \text{If } \lim_{n\to\infty}\rho(x,x_n)=0 \text{ and for every } n\ \lim_{k\to\infty}\rho(x_n,x_{n,k})=0, \]
\[ \text{then there exist sequences of natural numbers } \{m_n\}_{n=1}^{\infty},\ \{k_n\}_{n=1}^{\infty} \]
\[ \text{such that } \lim_{n\to\infty}\rho(x,x_{m_n,k_n})=0. \]

Indeed, suppose that \(\rho\) satisfies condition \((\alpha)\) and let \(A\subset Z\). Put
\([A]^1=E\{x:\rho(x,A)=0\}\). Since \(\rho\) satisfies condition \((\alpha)\), we have
\([[A]^1]^1=E\{x:\rho(x,[A]^1)=0\}=[A]^1\), and hence \([A]^1\) is closed, i.e. \(\rho\) agrees with the topology of \(Z\) in the strong sense (\(x\in[A]\) if and only if \(\rho(x,A)=0\)).

If \(Z\) is symmetrizable in the strong sense with symmetrix \(\rho\) and
\(\lim_{n\to\infty}\rho(x,x_n)=0\), \(\lim_{k\to\infty}\rho(x_m,x_{m,k})=0\), then there exist \(m_n\) and \(k_n\) such that, for \(m\ge m_n\) and \(k>k_n\), \(x_{m,k}\in (O_{1/n}x)\). Hence
\(\lim_{n\to\infty}\rho(x,x_{m_n,k_n})=0\), i.e. \(\rho\) satisfies condition \((\alpha)\).

It is easy to verify that the symmetrix \(d\) under the assumptions of the proposition satisfies condition \((\alpha)\).

We now show that the space \(X\) is collectively normal. From this, by the Mack–Olin–Bing theorem (see \((^5)\)), our assertion will follow. Let \(\{F_\alpha\}\) be a discrete family of closed subsets of the space \(X\).

Put

\[ U_\alpha=E\{x:\ d(x,F_\alpha)<d(x,\bigcup_{\beta\ne\alpha}F_\beta)\} \]

and show that

\[ V_\alpha=\langle U_\alpha\rangle\supset F_\alpha. \]

This assertion will prove the claim, since the \(V_\alpha\) are pairwise disjoint (because the \(U_\alpha\), obviously, are pairwise disjoint). Let \(x\in F_\alpha\), and choose \(\varepsilon(x)>0\) so that

\[ O_{\varepsilon(x)}x\cap\left(\bigcup_{\beta\ne\alpha}F_\beta\right)=\Lambda. \]

We have

\[ V_\alpha\supset \bigcup_{x\in F_\alpha}\langle O_{\delta(x,\varepsilon(x))}x\rangle \supset F_\alpha, \]

where \(\delta(x,\varepsilon(x))\) is chosen according to the condition of the proposition, as was required to prove.

Moscow State University
named after M. V. Lomonosov

Received
16 IV 1966

REFERENCES

\(^1\) A. V. Arhangel’skii, Matem. sborn., 67 (109), No. 1 (1965).
\(^2\) A. V. Arhangel’skii, DAN, 164, No. 2, 247 (1965).
\(^3\) A. V. Arhangel’skii, UMN, 4 (130) (1966).
\(^4\) V. Ponomarev, DAN, 141, No. 3 (1961).
\(^5\) F. Michael, Proc. Am. Math. Soc., 9, No. 5 (1958).
\(^6\) V. Niemytzki, Math. Ann., 104, H. 5, 666 (1931).
\(^7\) P. S. Aleksandrov, V. V. Nemytskii, Matem. sborn., 3, No. 3, 663 (1938).

* Therefore, if \(d\) is also a Cauchy symmetrix (see \((^7)\)), i.e. if every convergent sequence is fundamental, then \(X\) will be metrizable.