Reports of the Academy of Sciences of the USSR
1962. Volume 143, No. 2

MATHEMATICS

Yu. OREVKOV

ON TOTALLY BOUNDED METRIC SPACES

(Presented by Academician P. S. Aleksandrov, 23 X 1961)

Let \(P\) be a metric space; \(uP\) its bicompact extension corresponding to the metric proximity \(\left(^{1}\right)\). Yu. M. Smirnov \(\left(^{2}\right)\) characterized completeness of a space in a given metric through the properties of the remainder of the bicompact extension \(uP \setminus P\). The property of a metric space \(P\) of being totally bounded, which, like completeness, is preserved under uniformly continuous mappings, can also be described with the aid of the remainder \(uP \setminus P\). This is done in the present note.

Theorem 1. In order that a metric space \(P\) be totally bounded, it is necessary and sufficient that \(uP \setminus P\) be hereditarily normal.

Necessity. Since in the case when \(P\) is totally bounded the extension \(uP\) is a compactum, \(uP \setminus P\) is evidently hereditarily normal.

Sufficiency. Let \(P\) be a metric space that is not totally bounded. Then there exists a countable set \(N=\{x_i\}\subset P\) such that for any two distinct points \(x_i\) and \(x_j\) one always has \(\rho(x_i,x_j)\ge \eta>0\). Consequently, the set \(N\) is uniformly equivalent to the set of natural numbers of the real line. The bicompact extension corresponding to the proximity space of the natural sequence is its Čech extension, and, since \(N\subset P\), \(P[N]=N\), it follows that \(\beta N=uN=uP[N]\). Therefore, in order to show that the difference \(uP\setminus P\) is not hereditarily normal, it suffices to prove the following assertion:

Lemma. \(\beta N\setminus N\) is not hereditarily normal.

Proof. Let \(I\) be the half-interval \([0,1)\), with the neighborhoods of any point \(x\in I\) being half-intervals of the form \([x,y)\). Consider the space \(R=I\times I\). As is known \(\left(^{3}\right)\), \(R\) is a completely regular, but not normal, space having a countable everywhere dense subset \(\{a_i\}\). Let \(\gamma R\) be some bicompact extension of the space \(R\). We construct a bicompact extension \(\tau N=\gamma R\cup N\) of the countable discrete space \(N\), where the topology is defined as follows: a neighborhood of a point from \(N\) is the point itself; a neighborhood of a point from \(\gamma R\) is an ordinary neighborhood in \(\gamma R\) with the addition of those natural numbers which serve as indices of the points from \(\{a_i\}\) falling into this neighborhood. The bicompact extension of the natural sequence thus obtained has a remainder that is not hereditarily normal, since \(R\) is its subspace. In view of the property* of the Čech extension, we have \(f(\beta N\setminus N)=\tau N\setminus N\). Hence \(f\) is a closed continuous mapping of the space \(f^{-1}(R)\) onto \(R\), and therefore \(f^{-1}(R)\) is a completely regular but not normal subspace of \(\beta N\setminus N\). The lemma and, consequently, Theorem 1 are proved.

The lemma makes it possible to prove the following assertion:

Theorem 2. Every completely regular non-pseudocompact space \(R\) has a Čech remainder which is not hereditarily normal.

* For otherwise, for some point \(y\in N\) (which is open in \(\tau N\)), we would have that \(f^{-1}y\) intersects \(\beta N\setminus N\) and, being an open set, contains infinitely many points from \(N\), which is impossible.

Proof. In \(R\) there exists a continuous unbounded function \(f\) and, consequently, a countable subset \(A=\{a_i\}\) such that \(|f(a_i)-f(a_j)|>1\) for \(i\ne j\). Let \(\varphi\) be any real function defined on \(A\). We shall prove that it can be extended continuously to all of \(R\). To this end consider neighborhoods \(Oa_i\) on which the oscillation of the function \(f\) does not exceed \(1/3\). Since \(R\) is completely regular, there exists a function \(\psi_i\), continuous on \(R\), such that \(\psi_i(a_i)=\varphi(a_i)\) and \(\psi_i(x)=0\) for \(x\in R\setminus Oa_i\), with \(0\le \psi_i(x)\le \varphi(a_i)\) at points \(x\in Oa_i\) when \(\varphi(a_i)\ge 0\), and \(0\ge \psi_i(x)\ge \varphi(a_i)\) at points \(x\in Oa_i\) when \(\varphi(a_i)\le 0\). It is clear that

\[ \text{the function }\psi(x)=\sum_{i=1}^{\infty}\psi_i(x) \]

is continuous on \(R\) and that \(\psi(a_i)=\varphi(a_i)\).

Consequently, \(\beta R[A]=\beta A\). Hence, by the lemma, the absence of hereditary normality in \(\beta R\setminus R\) follows. In Theorem 1 the condition of hereditary normality can be replaced by any condition from which hereditary normality follows and which holds for subsets of compacta. Therefore the following is true:

Theorem 3. Each of the following conditions is necessary and sufficient for the metric space \(P\) to be totally bounded:

a) \(uP\setminus P\) is perfectly normal (every open set is of type \(F_\sigma\));

b) \(uP\setminus P\) is hereditarily finally compact (Lindelöf);

c) \(uP\setminus P\) is hereditarily strongly paracompact;

d) \(uP\setminus P\) is hereditarily paracompact;

e) \(uP\setminus P\) has a countable base.

I express my deep gratitude to Yu. M. Smirnov for guidance and assistance in the work.

Mathematical Institute named after V. A. Steklov
Academy of Sciences of the USSR

Received
18 X 1961

REFERENCES

1. Yu. M. Smirnov, Matem. sborn., 31 (73), No. 3, 543 (1952).
2. Yu. M. Smirnov, Tr. Mosk. matem. obshch., 3, 271 (1954).
3. R. Sorgenfrey, Bull. Am. Math. Soc., 53, No. 6 (1947).