UDC 513.831

V. V. PROIZVOLOV

ON THE CARDINALITY OF SYSTEMS OF OPEN SUBSETS IN A DYADIC BICOMPACTUM

(Presented by Academician P. S. Aleksandrov on 11 III 1967)

In 1941, Shpilrain proved a theorem on the countability of an arbitrary point-countable system of open sets in a dyadic bicompactum ((^1)^*). Here it will be proved that a dyadic bicompactum has a stronger property of the same order. Namely, the following is true.

Theorem 1. The cardinality of an arbitrary system of open sets of a dyadic bicompactum does not exceed the cardinal number (\tau), if the pointwise cardinality of this system does not exceed (\tau^{}) (it is assumed that (\tau \geq \aleph_0)).

Proof. In a dyadic bicompactum (X) let there be given some system of open sets (W={W_\alpha}), whose pointwise cardinality does not exceed (\tau). The bicompactum (X) is a continuous image of (D^m): (f: D^m \to X). If we denote (U_\alpha=f^{-1}W_\alpha), then the system of open sets (U={U_\alpha}) in (D^m) has pointwise cardinality not exceeding (\tau). It is enough for us to prove that the cardinality of (U) does not exceed (\tau); then the cardinality of (W) also does not exceed (\tau).

Represent
[
D^m=\prod_\lambda D_\lambda,
]
where for every index (\lambda) the factor (D_\lambda) is the simple two-point space, and take in (D^m) the canonical base, each element of which has a finite number of distinguished indices; denote it by (B={B_\alpha}).

In each element of the system (U) take one element of the base (B). It may happen that for distinct (U_{\alpha_1}) and (U_{\alpha_2}) one and the same element (B_\beta\in B), belonging to their intersection, is chosen; in this case we shall count the element (B_\beta) only once. The totality of all such chosen elements forms a system (V={V_\alpha}), and if (\alpha\ne\beta), then (V_\alpha\ne V_\beta). The system (V) is inscribed in the system (U) in such a way that the pointwise cardinality of (V) does not exceed (\tau).

The systems (U) and (V) are equipotent. It is immediately clear that the cardinality of (U\geq) the cardinality of (V). Let us verify the reverse inequality. Suppose the contrary: (\tau_1>\tau_2), where (\tau_1) is the cardinality of the system (U), and (\tau_2) is the cardinality of the system (V), with (\tau_1>\tau) (otherwise everything would already have been proved). Since an element of the system (V) is inscribed in each element of the system (U), there will be such an element (V_\alpha\in V) that is inscribed simultaneously in (\tau_1) elements of the system (U), but this contradicts the fact that the pointwise cardinality of the system (U) does not exceed (\tau). Consequently, it remains only to prove that the cardinality of the system (V={V_\alpha}) does not exceed (\tau).

Suppose the contrary: the cardinality of the system (V) is (\tau_1), where (\tau_1>\tau). If (\tau_1>2^\tau), then from (V) one can select a subsystem of cardinality (2^\tau), so we shall assume that (\tau_1\leq 2^\tau).

Since each element (V_\alpha\in V) is an element of the base (B) in the bicompactum
[
D^m=\prod_\lambda D_\lambda,
]
(V_\alpha) has a finite number of distinguished indices (\lambda).

The union of all distinguished indices over all (\alpha) forms a set (A). The cardinality of (A) is equal to the cardinality of (V), and therefore the cardinality of (A\leq 2^\tau).

The bicompactum
[
R=\prod_{\lambda\in A}D_\lambda
]
may be regarded naturally as embedded in (D^m). In this case the system (V) induces in (R) the system (\widetilde V={\widetilde V_\alpha}), where

* A bicompactum is called dyadic if it is a continuous image of some (D^\tau) (the product of (\tau) “copies” of the two-point space).

** The pointwise cardinality of a system does not exceed (\tau) if every point of the bicompactum belongs to no more than (\tau) elements of the system.

(\tilde V_\alpha = V_\alpha \cap R). The pointwise cardinality of the system (\tilde V) does not exceed (\tau), while the cardinality of (\tilde V) coincides with the cardinality of (V).

Hewitt’s theorem ((^3)) asserts that the density of the space (D^n), where (n \le 2^\tau), does not exceed (\tau). Therefore the bicompactum (R) possesses a dense subset in (R) whose cardinality does not exceed (\tau). Hence the system (\tilde V), being a system of pointwise cardinality (\le \tau) in a space of density (\le \tau), has cardinality (\le \tau). This contradicts the assumption that the cardinality of (V > \tau). The theorem is proved(^*).

Corollary. A point-countable covering of a locally bicompact dyadic space is countable.

A theorem more general than Theorem 7 is true.

Theorem (1'). For a bicompactum coabsolute with a dyadic one, Theorem 1 is valid.

Proof. Thus, the bicompactum (X) is coabsolute with the dyadic bicompactum (Y). But for a pair of coabsolute bicompacta there is always a third bicompactum (Z) which maps irreducibly both onto (X) and onto (Y) (for a proof of this fact see, for example, ((^2))). We have irreducible mappings (f : Z \to X) and (g : Z \to Y).

Let the bicompactum (X) contain a system of open sets (U = {U_\alpha}) whose pointwise cardinality does not exceed (\tau). It is necessary to prove that the integral cardinality of the system (U) does not exceed (\tau). In the space (Z) consider the system of open sets (V = {V_\alpha}), where (V_\alpha = f^{-1}U_\alpha). Obviously, the cardinality of the system (V) is equal to the cardinality of the system (U), and the pointwise cardinality of the system (V) does not exceed (\tau). Consequently, it suffices to prove that the integral cardinality of the system (V) does not exceed (\tau).

In each element (V_\alpha) of the system (V) choose one open set (W_\alpha) such that ([W_\alpha] \subseteq V_\alpha). The resulting system (W = {W_\alpha}) has pointwise cardinality not exceeding (\tau), and is equipotent to the system (W). Obviously, the cardinality of (V \ge) the cardinality of (W). Let us verify the reverse inequality. Suppose the contrary: (\tau_1 > \tau_2), where (\tau_1) is the cardinality of the system (V), and (\tau_2) is the cardinality of the system (W), with (\tau_1 > \tau) (otherwise everything would already have been proved). Since an element of the system (W) is inscribed in each element of the system (V), there exists an element (W_\alpha) which is inscribed in (\tau_1) elements of the system (U), but this contradicts the fact that the pointwise cardinality of the system (U) does not exceed (\tau). Thus, (V) and (W) are equipotent. It now remains to prove that the cardinality of (W) does not exceed (\tau).

Since the mapping (g : Z \to Y) is irreducible, the set (\langle gW_\alpha\rangle) is nonempty for every (W_\alpha \in W), where (\langle R\rangle) denotes the set of all interior points of (R). Put (M_\alpha = \langle gW_\alpha\rangle); the system of open sets (M = {M_\alpha}) in the space (Y) has the same cardinality as (W). We show that the pointwise cardinality of (M) does not exceed (\tau). Suppose the contrary: there is a point (y \in Y) which belongs to more than (\tau) elements of the system (M). But since (g^{-1}M_\alpha \subseteq [W_\alpha]), in view of the irreducibility of (g), the set (g^{-1}y) is contained in more than (\tau) sets of the system (\bar W = {[W_\alpha]}). But the latter conclusion contradicts the fact that the pointwise cardinality of the system (\bar W) does not exceed (\tau).

Thus, the pointwise cardinality of (M) does not exceed (\tau), and since (Y) is a dyadic bicompactum, by Theorem 1 the integral cardinality of (M) does not exceed (\tau). Consequently, the integral cardinality of (W) does not exceed (\tau), which proves everything. From the proof it is clear that Theorem 1 is also valid for bicompacta that map irreducibly onto dyadic ones.

Moscow State University
named after M. V. Lomonosov

Received
22 II 1967

CITED LITERATURE

(^1) E. Shpilyrain, DAN, 31, 525 (1941).
(^2) S. Iliadis, S. Fomin, UMN, 21, no. 4 (130), 47 (1966).
(^3) E. Hewitt, Bull. Am. Math. Soc., 52, 641 (1946).

(^*) As B. Efimov observed, Theorem 1 can also be proved simply by using the well-known theorem of Shanin on calibers.