MATHEMATICS

D. B. DOICHINOV

ON THE EMBEDDING OF UNIFORM SPACES*

IN HILBERT AND EUCLIDEAN SPACES

(Presented by Academician P. S. Aleksandrov on 7 VII 1960)

P. S. Urysohn, in his famous metrization theorem, proved that every metric space with a countable base is homeomorphically embedded in Hilbert space \((^1)\). If, moreover, a metric space with a countable base has finite dimension \(n\), then it can be homeomorphically embedded in Euclidean space \((^{2–4})\) of dimension \(2n+1\). Yu. M. Smirnov posed the question of finding sufficient conditions for a uniform** (in both directions) embedding of uniform (even just metric spaces with a countable base!) spaces in Hilbert (generalized \((^{10})\), generally speaking) or Euclidean spaces \((^5)\).

Since for compacta every homeomorphism is uniform, and the uniform dimension*** coincides with the ordinary one for them, the two above-mentioned assertions are true for compacta also in the uniform sense. They are also valid in the uniform sense for totally bounded metric spaces \((^5)\).

In the present note, leaving open the question of the uniform embedding of arbitrary metric spaces with a countable base in Hilbert space, we give (see Theorem 2) the following condition, stronger than total boundedness, sufficient for a metric space of arbitrary weight to be uniformly embedded in a (generalized) Hilbert space of the same weight (even in its unit ball):

K. In every uniform covering one can inscribe a finite-multiplicity uniform covering.

This condition is certainly satisfied for finite-dimensional spaces. The assertion of interest to us follows from the following general assertion:

Theorem 1. Every uniform space \(X\) satisfying condition K can be uniformly embedded in the (uniform) product**

* A uniform space will be defined by means of a system (structure) of coverings (see \((^{8,9})\)), which we shall call uniform. A metric space is considered only in the natural uniform structure of all such coverings, into each of which one can inscribe one of the coverings of the form \(\omega_\varepsilon=\{O_\varepsilon x:x\in X\}\).

* A mapping \(f\) is called uniformly continuous* if the inverse image
\(f^{-1}\gamma=\{f^{-1}(\Gamma):\Gamma\in\gamma\}\) of every uniform covering \(\gamma\) of the space \(Y\) is a uniform covering for \(X\).

* Uniform dimension** (see \((^{6,7})\)) is the smallest of the numbers \(n=0,1,2,\ldots\) such that in every uniform covering one can inscribe a uniform covering of multiplicity \(\leq n+1\).

**** Only \(T_1\)-spaces are considered.

** The *product structure of uniform spaces \(Y_\alpha\), by definition, consists of all such coverings into each of which is inscribed a covering of the form
\(\{\Gamma_{\alpha_1}\times\cdots\times\Gamma_{\alpha_s}\times\prod_{\alpha\ne\alpha_i}Y_\alpha\}\), where \(\alpha_1,\ldots,\alpha_s\) is an arbitrarily chosen finite set of indices, \(\Gamma_{\alpha_i}\) is an element of an arbitrarily chosen uniform covering \(\gamma_{\alpha_i}\) of the space \(Y_{\alpha_i}\).

\(\prod_\alpha S_\alpha^\tau\) of unit balls \(S^\tau\), taken from one and the same generalized Hilbert space \(H^\tau\) of the same topological weight \(\tau\) as the space \(X\), whose number is equal to the uniform weight\(^*\) of the space \(X\).

Finally, in connection with the second question, an example is given here of a uniformly zero-dimensional (and even uniformly locally bicompact\(^ {**}\)) metric space with a countable base which cannot be uniformly embedded in any Euclidean space.

Plan of the proof of Theorem 1.

Lemma 1. If \(f_\alpha\) are mappings\(^ {***}\) of the space \(X\) into the corresponding spaces \(Y_\alpha\), then the mapping \(f=\{f_\alpha\}\), which assigns to each point \(x\in X\) the point \(y=\{f_\alpha(x)\}\) of the (uniform) product \(\prod Y_\alpha\), is uniformly continuous.

Indeed, a base of the uniform structure of the product \(\prod Y_\alpha\) consists of all covers of the form
\[ \gamma=\{\Gamma_{\alpha_1}\times\ldots\times\Gamma_{\alpha_s}\times \prod_{\alpha\ne\alpha_i}Y_\alpha\}, \]
where \(\Gamma_{\alpha_i}\in\gamma_{\alpha_i}\), and \(\gamma_{\alpha_i}\) is an arbitrary cover of the space \(Y_{\alpha_i}\). The inverse image
\[ f_{\alpha_i}^{-1}(\gamma_{\alpha_i})=\{f_{\alpha_i}^{-1}(\Gamma):\Gamma\in\gamma_{\alpha_i}\} \]
of the cover \(\gamma_{\alpha_i}\) under the mapping \(f_{\alpha_i}\) is a uniform cover. Since
\[ f^{-1}(\Gamma_{\alpha_1}\times\ldots\times\Gamma_{\alpha_s}\times \prod_{\alpha\ne\alpha_i}Y_\alpha) = f_{\alpha_1}^{-1}(\Gamma_{\alpha_1})\cap\ldots\cap f_{\alpha_s}^{-1}(\Gamma_{\alpha_s}), \]
the inverse image \(f^{-1}(\gamma)\) is the “product”\(^ {****}\) of a finite number of covers \(f_{\alpha_i}^{-1}(\gamma_{\alpha_i})\), and therefore is itself uniform. Hence \(f\) is uniformly continuous.

Lemma 2. If \(f_\alpha\) are arbitrary mappings of the space \(X\) into the corresponding spaces \(Y_\alpha\), and if, moreover, the inverse images \(f_\alpha^{-1}(\gamma)\) of covers \(\gamma\) of the spaces \(Y_\alpha\) constitute a pseudobase\(^ {*****}\) of the uniform structure of the space \(X\), then the mapping \(f=\{f_\alpha\}\) is one-to-one and the inverse mapping \(f^{-1}\) is uniformly continuous.

Indeed, for distinct points \(x\) and \(y\) there exists such a product of covers \(f_{\alpha_i}^{-1}(\gamma_{\alpha_i})\), \(i=1,\ldots,s\), that \(x\in U\), but \(y\notin U\), for some set
\[ U=f_{\alpha_1}^{-1}(\Gamma_{\alpha_1})\cap\ldots\cap f_{\alpha_s}^{-1}(\Gamma_{\alpha_s}), \]
where \(\Gamma_{\alpha_i}\in\gamma_{\alpha_i}\). Then \(f_{\alpha_i}(x)\in\Gamma_{\alpha_i}\), but \(f_{\alpha_i}(y)\notin\Gamma_{\alpha_i}\) for some \(\alpha_i\). Hence \(f(x)\ne f(y)\). Next, let \(\omega=\{U_\lambda\}\) be a cover of the space \(X\). Take covers \(f_{\alpha_i}^{-1}(\gamma_{\alpha_i})\), \(i=1,\ldots,s\), whose product is inscribed in \(\omega\). If
\[ f_{\alpha_1}^{-1}(\Gamma_{\alpha_1})\cap\ldots\cap f_{\alpha_s}^{-1}(\Gamma_{\alpha_s})\subset U_\lambda, \]
then
\[ \Gamma_{\alpha_1}\times\ldots\times\Gamma_{\alpha_s}\times \prod_{\alpha\ne\alpha_i}Y_\alpha\subset f(U_\lambda). \]
Thus the cover
\[ \{\Gamma_{\alpha_1}\times\ldots\times\Gamma_{\alpha_s}\times \prod_{\alpha\ne\alpha_i}Y_\alpha\} \]
is inscribed in \(f(\omega)\). Therefore \(f(\omega)\) is a uniform cover. In view of the arbitrary choice of the cover \(\omega\), the mapping \(f^{-1}\) is uniformly continuous.

Lemma 3. If \(\omega=\{U_\lambda\}\) is a finite-fold cover of cardinality \(\tau\), then there exists a mapping \(f\) of the space \(X\) into the ball \(S^\tau\) of a generalized

\(^*\) A uniform base is any system of uniform covers such that into every uniform cover one can inscribe a cover of this system. The uniform weight of a uniform space is the least of the cardinalities of all its uniform bases.

\(^ {**}\) A uniform space is called (7) uniformly locally bicompact if there exists a uniform cover consisting of closed bicompact sets.

\(^ {***}\) Everywhere below, unless explicitly stated otherwise, mappings are assumed to be uniformly continuous, spaces and covers—uniform.

\(^ {****}\) The “product” of covers \(\gamma_i\) is the cover whose elements are all possible intersections \(\bigcap \Gamma_i\), where \(\Gamma_i\in\gamma_i\).

\(^ {*****}\) A pseudobase of a uniform space is any system of uniform covers such that into every uniform cover one can inscribe the product of a finite number of covers of this system.

Hilbert space of weight \(\tau\) such that the preimage of some space \(Y=f(X)\) is inscribed in \(\omega\).

Indeed, there exists a uniformly continuous bounded pseudometric \(d\) on \(X\) for which the given cover \(\omega\) of multiplicity \(n\) is uniform \((^9)\). Let, for example, \(d(x,y)\leqslant 1/\sqrt n\) for any points \(x,y\) of \(X\). There exists a cover \(\omega_\varepsilon=\{O_\varepsilon x\}\) (\(\varepsilon\)-neighborhoods are taken in the pseudometric \(d\)) inscribed in the cover \(\omega\). Put
\[ f_\lambda(x)=d(x, X\setminus U_\lambda). \]
For any points \(x\) and \(y\) of \(X\) we have: 1) \(f_\lambda(x)\leqslant 1/\sqrt n\); 2) no more than \(n\) of the numbers \(f_\lambda(x)\) are nonzero, and at least one of them is \(\geqslant \varepsilon\); 3) \(|f_\lambda(x)-f_\lambda(y)|\leqslant d(x,y)\).

The mapping \(f=\{f_\lambda\}\) uniformly maps \(X\) into \(S^\tau\), since \(\rho(f(x),0)\leqslant 1\) and \(\rho(f(x),f(y))\leqslant \sqrt{2n}\,d(x,y)\). By virtue of 2), the sets
\[ \Gamma_\lambda=\mathcal E\{f(x): f_\lambda(x)>0\} \]
form a uniform cover \(\gamma\) of the image \(Y=f(X)\), and the preimage \(f^{-1}(\gamma)\) is inscribed in \(\omega\).

Let us indicate the derivation of Theorem 1. Take a base \(\{\omega_\alpha\}\) consisting of finite-multiplicity covers, whose cardinality is equal to the uniform weight. We may assume that the cardinality of each cover \(\omega_\alpha\) is equal to the topological weight \(\tau\) of the space \(X\). By Lemma 3, for each \(\alpha\) there exists a mapping \(f_\alpha\) of the space \(X\) into the ball \(S^\tau\) and a cover \(\gamma_\alpha\) of the image \(f_\alpha(X)\) such that
\[ f_\alpha^{-1}(\gamma_\alpha) \]
is inscribed in \(\omega_\alpha\). The system of covers \(\{f_\alpha^{-1}(\gamma_\alpha)\}\) turns out to be a base. By Lemmas 1 and 2, the space \(X\) is uniformly homeomorphic to the image \(f(X)\) lying in \(\prod S^\tau\), where \(f=\{f_\alpha\}\).

Let now a countable number of uniformly metrizable spaces \(Y_i\) be given. It is known that the product \(\prod Y_i\) can be metrized in the following way: if \(\rho_i\) is a metric of the space \(Y_i\) (one may assume \((^5)\) that the diameter of the space \(Y_i\) in the metric \(\rho_i\) is not greater than 1), then for any points \(y'=\{y_i'\}\) and \(y''=\{y_i''\}\) of the product \(\prod Y_i\) we put
\[ \rho(y',y'')=\left[\sum_i 2^{-i}\rho_i^2(y_i',y_i'')\right]^{1/2}. \]
We shall agree to speak of the metric product of the spaces \(Y_i\), meaning precisely this metric \(\rho\).

Lemma 4. For a countable number of uniformly metrizable spaces \(Y_i\), their uniform and metric products are uniformly homeomorphic.

Indeed, let \(\gamma\) be a cover of the uniform product. Inscribe in it a cover \(\pi_k\) of the form
\[ \{\Gamma_1\times\cdots\times\Gamma_k\times\prod_{i>k}Y_i\}, \]
where \(\Gamma_i\in\gamma_i\), and \(\gamma_i\) is a cover of the space \(Y_i\) \((i\leqslant k)\). There exists \(\varepsilon>0\) such that the cover \(\omega_\varepsilon^i\) of the space \(Y_i\) (consisting of the \(\varepsilon\)-neighborhoods of all its points) is inscribed in \(\gamma_i\). For the number
\[ \delta=\varepsilon/\sqrt{2^k}, \]
the cover \(\omega_\delta=\{O_\delta y\}\) of the metric product is inscribed in the cover \(\pi_k\), and hence also in \(\gamma\). Conversely, for an arbitrary number \(\delta>0\) and a natural number \(N\), chosen according to \(\delta\) so that
\[ \frac12^N<\delta^2/2, \]
the cover \(\omega_{\delta,N}\) of the form
\[ \{O_r y_1\times\cdots\times O_r y_N\times\prod_{i>N}Y_i\}, \]
where
\[ r=\delta/\sqrt N, \]
and \(O_r y_i\) is the neighborhood of the point \(y_i\) in \(Y_i\) (\(y_i\) runs through the entire set of points of the space \(Y_i\)), is inscribed in the cover \(\omega_\delta=\{O_\delta y\}\) of the metric product.

Theorem 2. Every metric space of topological weight \(\tau\) satisfying condition K can be uniformly embedded in the unit ball of a generalized Hilbert space of weight \(\tau\).

Let us indicate the proof. Take a base \(\{\omega_i\}\) of the space \(Y\), consisting of a countable number of finite-multiplicity covers of cardinality \(\leqslant \tau\). By Theorem 1, the space \(Y\) is uniformly embedded in the uniform product
\[ \prod_{i=1}^{\infty} S_i^\tau, \]
and this, by Lemma 4, is uniformly embedded in the ball \(S^\tau\).

Corollary. Every metric space of finite uniform dimension can be uniformly embedded in the unit ball of a generalized Hilbert space of the same weight.

Example. Consider an abstract set \(R\), consisting of a countable number of points \(x^i_{nk}\), where \(n=1,2,\ldots;\ k=1,2,\ldots;\ i=1,2,\ldots,n\). Introduce in it a distance \(\rho\): let \(\rho(x^i_{nk},x^j_{mr})=1\), if \(n\ne m\) or \(k\ne r\); \(\rho(x^i_{nk},x^j_{nk})=1/k\), if \(i\ne j\), and, finally, \(\rho(x^i_{nk},x^i_{nk})=0\). Since in every ball of radius \(1/2\) there is only a finite number of points, \(R\) is uniformly locally bicompact. Let \(\varepsilon>0\), and let \(\delta=1/r<\varepsilon\). The system of \(\delta\)-neighborhoods of all points \(x^i_{nk}\) for which \(k\le r-1\), and of \(\varepsilon\)-neighborhoods of all points \(x^1_{nk}\) for which \(k\ge r\), is a uniform discrete cover (since \(O_\varepsilon x^i_{nk}=O_\varepsilon x^1_{nk}\) for \(k\ge r\)) inscribed in \(\omega_\varepsilon\). Thus, \(R\) is uniformly zero-dimensional.

Call a metric space \(\varepsilon\)-discrete if all positive distances in it are \(\ge \varepsilon\). If the “coordinates” \(n\) and \(k\) of the points \(x^i_{nk}\in R\) are fixed, then we obtain a \(1/k\)-discrete set \(D_{nk}\). Let \(\delta(\varepsilon)\) be the “modulus of continuity” of a one-to-one mapping \(f'\). Under this assumption, the complete preimage of an \(\varepsilon\)-discrete set will be \(\delta(\varepsilon)\)-discrete. Suppose, finally, that there exists a uniformly homeomorphic mapping \(f\) of the space \(R\) into a Euclidean space \(E^N\). Let \(d\) be the usual metric in \(E^N\). By uniform continuity there exists a number \(K\) such that, whenever \(\rho(x,x')\le 1/K\), necessarily \(d(f(x),f(x'))<1\). Hence,
\(d(f(x^i_{nK}), f(x^1_{nK}))<1\) for all \(n\) and \(i\), where \(i\le n\). Let \(\delta(\varepsilon)\) be the modulus of continuity of the inverse mapping \(f^{-1}\), and let \(\eta=\delta(1/K)\). The image of the set \(D_{nK}\) (for any \(n\)) is \(\eta\)-discrete and lies in the unit ball \(S^N\) of the Euclidean space \(E^N\), which is impossible, since the number of points of any \(\eta\)-discrete set of the ball \(S^N\) does not exceed the number

\[ \left(\frac{2+\eta}{\eta}\right)^N, \]

whereas the number of points in \(D_{nK}\) grows with \(n\).

Remark. No bounded set of the space \(R\) other than a singleton (being completely bounded) is uniformly separated \((^5)\). Therefore E. Gorin’s arguments \((^5)\) do not apply here.

I express my heartfelt gratitude to Yu. M. Smirnov, under whose supervision this work was done.

Sofia University
Sofia, Bulgaria

Received
3 VII 1960

REFERENCES

\(^1\) P. S. Urysohn, Math. Ann., 94, 309 (1925).
\(^2\) K. Menger, Proc. Acad. Wetensch. Amst., 29, 476 (1926).
\(^3\) Nöbeling, Math. Ann., 104, 71 (1930).
\(^4\) W. Hurewicz, Proc. Acad. Wetensch. Amst., 30, 425 (1927).
\(^5\) E. Gorin, UMN, 14, No. 5, 129 (1959).
\(^6\) Yu. M. Smirnov, Matem. sborn., 38, No. 3, 283 (1956).
\(^7\) J. Isbell, Pacif. J. Math., 9, 107 (1959).
\(^8\) Yu. M. Smirnov, Matem. sborn., 31, No. 3, 543 (1952).
\(^9\) J. Kelley, General Topology, N. Y., 1955.
\(^10\) Yu. M. Smirnov, UMN, 6, No. 6, 100 (1951).