Mathematics

D. R. ISBELL

ON THE INDUCTIVE DIMENSION OF PROXIMITY SPACES

(Presented by Academician P. S. Aleksandrov, 11 VI 1960)

1. Yu. M. Smirnov introduced the notion of the \(\delta\)-dimension \(\delta dP\) of a proximity space \(P\), defined by means of coverings. He also raised the question of the possibility of defining the inductive dimension \(\delta \operatorname{Ind} P\) \((^5)\), p. 289). In this note I propose a natural definition of inductive dimension such that \(\delta \operatorname{Ind} P \geqslant \delta dP\) for every space \(P\). The author knows of no example for which one would have \(\delta \operatorname{Ind} P > \delta dP\); equality holds for every subspace of Euclidean space and for every metric space \(M\) of finite dimension \(\Delta dM\) (in the sense of \((^4)\)).

The inequality of N. Vedenisov \((^1)\), \(\operatorname{Ind} X \geqslant \dim X\), proved by him for normal spaces \(X\), follows easily from the inequality \(\delta \operatorname{Ind} P \geqslant \delta dP\).

1. Let \(P\) be a proximity space. A set \(C\), by definition, \(\delta\)-separates the sets \(A, B\) if there exists a decomposition \(P \setminus C = A' \cup B'\), \(A'\) is far from \(B'\), \(A' \supset A\), and \(B' \supset B\). A set \(H\) frees \(A, B\) if \(H\) is far from \(A \cup B\) and every \(\delta\)-neighborhood of the set \(H\) which does not meet the set \(A \cup B\), \(\delta\)-separates \(A\) and \(B\).

Lemma 1. If \(H\) frees \(A\) and \(B\), then the closure of the set \(H\) frees the closures of \(A\) and \(B\), and conversely.

A chain between \(A\) and \(B\) is such a finite sequence of sets \(\Gamma_1,\ldots,\Gamma_n\) that \(A \cap \Gamma_1 \ne \Lambda\), \(B \cap \Gamma_n \ne \Lambda\), and \(\Gamma_i \cap \Gamma_{i+1} \ne \Lambda\) for \(i=1,\ldots,n-1\).

Remark. In order that the set \(H\), assumed far from \(A \cup B\), free \(A\) and \(B\), it is necessary and sufficient that into every \(\delta\)-covering \(\gamma\) one can inscribe such a \(\delta\)-covering \(\gamma'\) that every chain of elements of \(\gamma'\) between \(A\) and \(B\) contains an element intersecting \(H\).

Let the equality \(\delta \operatorname{Ind} P = -1\) mean that \(P\) is the empty space. We shall say that \(\delta \operatorname{Ind} P \leqslant n\) if for every pair of far sets \(A, B\) there exists a set \(H\) which frees \(A\) and \(B\) and \(\delta \operatorname{Ind} H \leqslant n-1\). If there is no such number \(n\), then \(\delta \operatorname{Ind} P = \infty\).

Lemma 2. Let the space \(P\) be normal, and let the proximity be the largest of those that generate the topology of \(P\). If \(H\) separates \(A\) and \(B\) in the topological sense, then \(H\) frees \(A\) and \(B\).

The converse assertion is false. This is easily seen from the example of a sequence of intervals converging to an interval.

By easy induction from the lemma we obtain the following theorem.

Theorem 1. Let the space \(P\) be normal, and let the proximity be the largest of those that generate the topology of \(P\). Then \(\operatorname{Ind} P \geqslant \delta \operatorname{Ind} P\).

Let now \(uP\) be the bicompact \(\delta\)-extension of the \(\delta\)-space \(P\). It is known \((^5)\) that \(\delta duP = \delta dP\).

Theorem 2. If the space \(P\) is dense in the space \(\widetilde{P}\), then \(\delta \operatorname{Ind} P \geqslant \delta \operatorname{Ind} \widetilde{P}\); in particular, \(\delta \operatorname{Ind} P \geqslant \delta \operatorname{Ind} uP\). Moreover, \(\delta \operatorname{Ind} uP \geqslant \delta dP\).

The first inequality follows almost directly from Lemma 1. To prove the last inequality one must apply induction and the sum theorem (⁵) for δ-dimension. For every δ-covering \(\gamma\) there exists a set \(H\) such that \(H\) consists of a finite sum of sets \(H_i\) having \(\delta\operatorname{Ind} H_i \leqslant n-1\), and \(P \setminus H\) is a sum of pairwise disjoint sets, lying singly in some \(\Gamma_i \in \gamma\).

Corollary (Vedenisov (¹)). For every normal space \(X\),
\[ \operatorname{Ind} X \geqslant \dim X . \]

3. We shall consider a metric space \(R\) as a uniform space. It is known (²) that a δ-homeomorphism of metric spaces is simply a uniform homeomorphism. A subset \(M\) of a uniform space \(R\) is called uniformly discrete if every covering of the set \(M\) is a uniform covering; a system \(\gamma\) of sets \(\Gamma_\alpha\) is called uniformly discrete if the sets \(\Gamma_\alpha\) are pairwise disjoint and \(\gamma\) is a uniform covering of the sum \(\bigcup \Gamma_\alpha\). The large dimension \(\Delta d R\) is defined as the least of all such integers \(n \geqslant 0\) that in every uniform covering of the space \(R\) one can inscribe a uniform covering of multiplicity \(\leqslant n+1\).

Lemma 3. Let \(R\) be a uniform space, \(H\) a nonempty subset of the space \(R\), and \(Q\) such a subset of the space \(R\) that every set \(M\) lying in \(Q\) and far from \(H\) is uniformly discrete. Then
\[ \delta\operatorname{Ind} Q \leqslant \delta\operatorname{Ind} H . \]

Lemma 4. Let \(R\) be a metric space, \(\widetilde R\) the completion of the space \(R\), and \(H\) any subset of the completion \(\widetilde R\). There exists such a set \(Q\) of the space \(R\) that its closure contains \(H\), and that every set \(M\) from \(Q\) which is far from \(H\) is uniformly discrete.

Indeed, if \(H\) separates two sets and they are far from \(Q\), then \(Q\) separates them. By applying induction from Theorem 2 one obtains:

Theorem 3. Let \(R\) be a metric space, and \(\widetilde R\) its completion. Then
\[ \delta\operatorname{Ind} R = \delta\operatorname{Ind} \widetilde R . \]

For the next theorem we need two lemmas. If \(\Delta dR \leqslant n\), then in every uniform covering of the space \(R\) one can inscribe a uniform covering \(\gamma\), consisting of the sum of \(n+1\) uniformly discrete systems \(\gamma_0,\ldots,\gamma_n\) (⁴). Secondly, in the proof we shall use the notion of a stable centered system. In 1954 S. Ginsburg and the author studied such systems (unpublished). A centered system \(\varphi\) of sets in a uniform space is called stable if for every uniform covering \(\gamma\) there exists a set \(A \in \varphi\) such that, for every set \(B \in \varphi\), the sum of all elements of the covering \(\gamma\) that meet \(B\) contains \(A\).

Lemma 5. Let \(R\) be a complete metric space, \(\varphi\) a stable centered system in the space \(R\), and \(H\) the set of all points of proximity to every element of the system \(\varphi\). Then every δ-neighborhood of the set \(H\) contains an element of \(\varphi\).

Theorem 4. For every metric space \(R\) we have:
\[ \delta\operatorname{Ind} R = \Delta d R . \]

Sketch of proof. Recalling Theorem 3 and the corresponding theorem from (³), we may assume that \(R\) is complete. It suffices to prove that if \(\Delta dR = n\), then for any two far sets \(A, B\) there exists a set \(H\) separating them such that \(\Delta dH \leqslant n-1\). Then there exists a sequence of coverings \(\gamma^i\), where \(\gamma^{i+1}\) is inscribed in \(\gamma^i\) and each \(\gamma^i\) consists of the \((n+1)\)-st uniformly discrete system \(\gamma_0^i,\ldots,\gamma_n^i\) (⁴). In order to construct a stable centered system, we remove, first, the δ-neighborhoods of the sets \(A\) and \(B\), and then such a subset of the sum of the system \(\gamma_0^i\) that \(\gamma_1^i \cup \cdots \cup \gamma_n^i\) is a uniform covering of the remaining set \(D_i\), which δ-separates \(A\) and \(B\).

Then the desired set \(H\) is the set of all points close to every element of this system.

We note that, by the method described above, releasing points of a line, one can obtain a perfect Cantor set!

It is known \((^4)\) that for any uniform space \(R\) of finite dimension \(\Delta dR\) the equality \(\delta dR=\Delta dR\) follows.

Corollary. If the space \(R\) is metric and \(\Delta dR<\infty\), then
\(\delta dR=\Delta dR=\delta \operatorname{Ind} R\).

1. Apart from the results set forth above, little is known about the dimensions \(\Delta d\) and \(\delta \operatorname{Ind}\). In \((^4)\) an example is given of a space \(R\) for which \(\Delta dR=\infty\), while \(\delta dR=0\). As Yu. M. Smirnov noted \((^5)\), \(\delta dR=0\) if and only if every two distant sets are \(\delta\)-separable by the empty set, i.e. when \(\delta \operatorname{Ind} R=0\).

It is not difficult to prove:

Theorem 5. For a metric space \(R\), from the equality \(sdR=0\) it follows that \(\Delta dR=0\).

It is clear from this that if, for a metric space \(R\), the dimensions \(\delta dR\), \(\Delta dR\), \(\delta \operatorname{Ind} R\) do not coincide, then
\(\delta \operatorname{Ind} R \ge \delta dR \ge 1\), \(\Delta dR=\infty\).

Purdue University
Lafayette, USA

Received
11 VI 1960

References

1. N. Vedenisov, Izv. Akad. Nauk SSSR, Ser. Mat., 5, 211 (1941).
2. V. A. Efremovich, Mat. Sb., 31, 189 (1952).
3. J. Isbell, Tohoku Math. J., 7, 1 (1955).
4. J. Isbell, Pacific J. Math., 9, 107 (1959).
5. Yu. M. Smirnov, Mat. Sb., 38, 283 (1956).