UDC 513.83

MATHEMATICS

N. HADZHIIVANOV (BULGARIA)

AN (n)-DIMENSIONAL CUBE CANNOT BE DECOMPOSED INTO A COUNTABLE UNION OF PROPER CLOSED SUBSETS WHOSE PAIRWISE INTERSECTIONS ARE AT MOST ((n-2))-DIMENSIONAL

(Presented by Academician P. S. Aleksandrov, 14 IV 1970)

The assertion announced in the title, for two summands, was proved by P. S. Urysohn in his famous paper “On Cantorian manifolds” (see (¹), p. 435).

Let an (n)-dimensional parallelepiped (P) be contained in a countable union of closed subsets (\Phi_k) of the space (E^n), (P \not\subset \Phi_k), and (\dim(\Phi_i \cap \Phi_j) \leq n - 2) for any distinct (i) and (j).

Lemma 1. For every (i), at least one of the summands (\Phi_k) contains a continuum joining the opposite faces (P_{-i}) and (P_{+i}) of the parallelepiped (P).

By symmetry, it suffices to restrict ourselves to the case (i = 1). Put
[
M = \bigcup_{i \ne j}(\Phi_i \cap \Phi_j).
]
The set (M) has dimension (\dim M \leq n - 2), while ((P_{-i}, P_{+i})), (i = 2, 3, \ldots, n), are ((n-1)) pairs of closed sets for which (P_{-i} \cap P_{+i} = \varnothing). Then in (P) there exist closed partitions (C_i) between (P_{-i}) and (P_{+i}) such that the set
[
C = \bigcap_{i=2}^{n} C_i
]
does not intersect the set (M).

To prove the lemma, it is enough to find at least one component of connectedness (K) of the compactum (C) joining (P_{-1}) with (P_{+1}). Indeed, since (K \cap M = \varnothing), the continuum (K) is decomposed into a union of pairwise disjoint closed sets (K \cap \Phi_k). Hence, by the well-known theorem of Sierpiński (²), (K = K \cap \Phi_k) for some (k), i.e. (K \subset \Phi_k), as required by the lemma. Let us show that such a component (K) can be found. Indeed, if there were no such components, then, by a known assertion that every component of connectedness of a compactum is its quasicomponent, we could, for every component of connectedness of the compactum (C), take a clopen neighborhood (O) in (C) not intersecting both faces (P_{-1}) and (P_{+1}) at the same time. These neighborhoods (O) cover the compactum (C), and from them one can choose a finite subcover (O_1, O_2, \ldots, O_p). Denote by (C_-) the union of the face (P_{-1}) with all those neighborhoods (O_i) which do not intersect the face (P_{+1}), and by (C_+) the union of the face (P_{+1}) with the complement (C \setminus C_-), which is clopen in (C). It is clear that the closed sets (C_-) and (C_+) do not intersect. Hence they can be separated in the parallelepiped (P) by a closed partition (C_1). Thus the opposite faces of the parallelepiped (P) would be separated by partitions (C_1, C_2, \ldots, C_n) with empty intersection, which is impossible. Consequently, contrary to the supposition, there exists a component of connectedness (K) of the compactum (C) joining the face (P_{-1}) with the face (P_{+1}), which was to be proved.

Lemma 2. No summand (\Phi_k) contains any parallelepiped (N) two faces of which are parallel to the faces (P_{-i}) and (P_{+i}), while the remaining ones lie on the faces of the parallelepiped (P).

Of course, we may again assume that (i = 1). Suppose that, for example, the summand (\Phi_1) contains entirely some parallelepiped (N) of the indicated...

of the prescribed form. Then the union of the closed differences (\Phi_k \setminus \operatorname{Int} N), (k \ne 1), together with the first summand (\Phi_1), covers the parallelepiped (P); none of them contains (P), and all their pairwise intersections are at most ((n-2))-dimensional. Therefore one may assume that the given decomposition is such that a certain parallelepiped (N) of the indicated form does not meet any of the summands (\Phi_k), (k \ne 1). Since the closed set (\Phi_1) does not contain (P), there is in (P) a cube (Q), with faces parallel to the faces of the parallelepiped (P), which does not meet the set (\Phi_1). Extending its faces, parallel to the faces (P_i), (i=2,3,\ldots,n), up to their intersection with the faces (P_{-1}) and (P_{+1}), we obtain a parallelepiped (R) which contains (Q). The parallelepiped (R) does not lie in any of the summands (\Phi_k), (k \ne 1), since otherwise (N) would meet one of them. Nor does (R) lie in (\Phi_1), since the cube (Q) does not lie in (\Phi_1). Hence, by Lemma 1, there exists a continuum (K) lying in some (R \cap \Phi_k) and joining the faces of the parallelepiped (R) that lie on the faces (P_{-1}) and (P_{+1}). This continuum necessarily intersects the parallelepiped (N), and consequently (\Phi_k \cap N \ne \varnothing), which means that (k=1). Thus (K \subset \Phi_1). But the continuum (K) also intersects the cube (Q), so that (Q \cap \Phi_1 \ne \varnothing), and this contradicts the choice of the cube (Q). The lemma is proved.

Theorem. An (n)-dimensional cube cannot be decomposed into a countable union of closed subsets distinct from the whole cube, whose pairwise intersections are no more than ((n-2))-dimensional.

Let the (n)-dimensional cube (I^n) be decomposed into a countable union of closed subsets (\Phi_k), (\Phi_k \ne I^n), with (\dim(\Phi_i \cap \Phi_j) \le n-2) for any distinct (i) and (j). Since (I^n) is a complete metric space, all summands (\Phi_k) cannot be nowhere dense. Hence at least one of the summands contains entirely some cube (Q) with faces parallel to the faces of the cube (I^n). Let this be the summand (\Phi_1). There exists a sequence of parallelepipeds (P_i), (i=1,2,\ldots,n+1), with faces parallel to the faces of the cube (I^n), the first of which coincides with (I^n), and the last with (Q), and, moreover, all faces of the parallelepiped (P_{i+1}), except for one pair of opposite ones, lie on faces of the parallelepiped (P_i). We obtain the parallelepiped (P_n) as follows. Extend all faces of the cube (Q), except for one single pair of opposite faces, until they meet the faces of the cube (I^n). The extended faces divide the cube (I^n) into (3^{\,n-1}) parallelepipeds. (P_n) will be the one among them that contains (Q). We construct the parallelepiped (P_{n-1}) in the same way, but this time starting from (P_n) and extending all its faces except for one single pair of opposite faces, different from those that lie on the faces of the cube (I^n). It is clear that this process will stop at the ((n-1))-st step, at which we obtain the parallelepiped (P_2). All its faces, except for one single pair of opposite ones, will lie on faces of the cube (I^n).

By Lemma 2, the parallelepiped (P_2) is not contained entirely in any of the summands (\Phi_k). Then, by Lemma 2, (P_3) is not contained in any of the summands (\Phi_k). Continuing this reasoning, at the (n)-th step we obtain that (Q) is not contained in any of the summands (\Phi_k), which contradicts the choice of the cube (Q). The theorem is proved.

Corollary 1. If a separable topological vector space (L) has a basis consisting of at least (n) elements, then (L) cannot be decomposed into a countable union of closed subsets, distinct from the whole (L), whose pairwise intersections are no more than ((n-2))-dimensional.

This corollary is obtained quite easily from the theorem with the aid of the following lemma:

Lemma 3. Let ({X_\alpha}) be a family of such closed subsets of a space (X) that, for every (\alpha), (X_\alpha) cannot be decomposed into a countable union of closed subsets distinct from (X_\alpha), whose pairwise intersections are no more than ((n-1))-dimensional; then, if (\dim(X_{\alpha_0} \cap X_\alpha) \ge n) for every (\alpha \ne \alpha_0), then (X'=\bigcup_\alpha X_\alpha) also cannot be decomposed into a countable union of the indicated form.

Suppose there exists a decomposition (X'=\bigcup_{k=1}^{\infty}\Phi_k), where the summands (\Phi_k) are closed, (\Phi_k\ne X') for every (k), and (\dim(\Phi_i\cap\Phi_j)\le n-1) for distinct (i) and (j). Since (X_\alpha) is not decomposed in this way, for every (\alpha) there exists a (\Phi_{k_\alpha}) containing (X_\alpha). From the inclusion (X_{\alpha_0}\cap X_\alpha\subset \Phi_{k_{\alpha_0}}\cap\Phi_{k_\alpha}) it follows that (\dim(\Phi_{k_{\alpha_0}}\cap\Phi_{k_\alpha})\ge n), and this entails (k_\alpha=k_{\alpha_0}), i.e. (\Phi_{k_\alpha}=\Phi_{k_{\alpha_0}}). Then (X_\alpha\subset \Phi_{k_{\alpha_0}}) for every (\alpha) and, consequently, (X'=\Phi_{k_{\alpha_0}}), which is a contradiction. The lemma is proved.

Proof of Corollary 1. Let (l_1,l_2,\ldots,l_n) be distinct basis elements of the space (L), and let (a) be an arbitrary element of the space (L). Denote by (L_a) the convex hull spanned by the elements (a,l_1,l_2,\ldots,l_n). (L_a) is a closed subset of the space (L), homeomorphic to an (m)-dimensional simplex (T^m), where (m\ge n). By the theorem, the set (L_a) is not decomposed into a countable union of closed subsets different from (L_a), the pairwise intersections of which are at most ((n-2))-dimensional. Moreover, the intersection (L_{l_1}\cap L_a) contains (L_{l_1}), so that (\lim(L_{l_1}\cap L_a)\ge n-1). Finally, (L=\bigcup_\alpha L_\alpha). To complete the proof of Corollary 1, it remains only to apply Lemma 3.

Corollary 2. If (M) is a connected topological space, every point (x) of which has a neighborhood homeomorphic to the (m(x))-dimensional Euclidean space (E^{m(x)}), where (m(x)\ge n), then (M) is not decomposed into a countable union of closed subsets different from all of (M), the pairwise intersections of which are at most ((n-2))-dimensional.

Let (M=\bigcup_{k=1}^{\infty}\Phi_k), where (M\ne \Phi_k) for every (k), (\dim(\Phi_i\cap\Phi_j)\le n-2) whenever (i\ne j), and the sets (\Phi_k) are closed. Let (x\in M) and let (\overline{O_x}) be a closed neighborhood of the point (x), homeomorphic to the cube (I^{m(x)}), where (m(x)\ge n). From the theorem it follows that (\overline{O_x}) necessarily lies in one of the summands (\Phi_k), say in (\Phi_1). Hence (U=\operatorname{Int}\Phi_1\ne\varnothing). Since the space (M) is connected, (\operatorname{Fr}U\ne\varnothing). Now let (y\in\operatorname{Fr}U), and let (\overline{O_y}) be a closed neighborhood of the point (y), homeomorphic to the cube (I^{m(y)}), where (m(y)\ge n). There will be a summand (\Phi_k), (k\ne1), such that (\overline{O_y}\subset \Phi_k). Put (F_1=\Phi_1\cap\overline{O_y}) and (F_2=\overline{O_y}\setminus U). It is clear that (F_1\ne\overline{O_y}), (F_2\ne\overline{O_y}), (F_1\cup F_2=\overline{O_y}), and (\dim(F_1\cap F_2)\le \dim(\Phi_1\cap\Phi_k)\le n-1), which contradicts the theorem. Consequently, a decomposition of the space (M) of the indicated kind is impossible.

I consider it my duty to express my gratitude to Prof. Yu. M. Smirnov for the attention he has shown to this work.

Moscow State University
named after M. V. Lomonosov

Received
4 IV 1970

REFERENCES

1. P. S. Urysohn, Works on topology and other areas of mathematics, 1, Moscow–Leningrad, 1951.
2. W. Sierpinski, Tôhoku Math. J., 13, 300 (1918).