L. A. Skornyakov

THE SPACE OF CONVERGENT SEQUENCES

(Presented by Academician A. I. Mal’cev on 21 XI 1961)

It is known that the topological product of groups coincides with the closure of the subspace of their discrete direct product. However, the discrete direct product of groups, considered as a subspace of some topological group, cannot always be embedded homeomorphically in the corresponding topological product. Therefore the question of describing the closure of a discrete direct product of groups is a topical one. In the present note a partial solution of this question is derived from results proved for uniform spaces.

Uniform spaces will be called similar if the partially ordered sets of their entourages are isomorphic ((²), p. 154). Let (R_i) ((i \in I)) be similar uniform spaces, in each (R_i) let an element (O_i) be distinguished, and let (\Omega) be the partially ordered set of their entourages. If (a) and (b) are close of order (\alpha) ((\alpha \in \Omega)), then we shall write (a \equiv b\;(\alpha)). Further, let (R) be the direct product of the sets (R_i). Denote by (\pi_i) the projection of (R) onto (R_i). An element ((a_i) \in R) will be called convergent if, whatever (\alpha \in \Omega) may be, (a_i \equiv O_i\;(\alpha)) for almost all (i). Consider the set of convergent elements of (R), and put ((a_i) \equiv (b_i)\;(\alpha)) ((\alpha \in \Omega)) if (a_i \equiv b_i\;(\alpha)) for all (i). The resulting uniform space will be called the space of convergent sequences of the space (R) and will be denoted by (\Pi^c R_i). An element ((a_i) \in R) will be called finite if (a_i = O_i) for almost all (i).

Putting (S = \Pi^c R_i), it is easy to verify the following facts:

Proposition 1. The natural embedding of (R_i) in (S) is a uniform homeomorphism*; moreover, if all (R_i) are separated, then (R_i) is closed in (S).

Proposition 2. The set of finite elements is everywhere dense in (S).

Proposition 3. If all (R_i) are separated, then (S) is also separated.

Theorem 1. The space (S) is bicompact if and only if: 1) all (R_i) are bicompact; 2) whatever (\alpha \in \Omega) may be, almost all (R_i) are small of order (\alpha).

Proof. If (S) is bicompact, then the validity of 1) follows from Proposition 1. If 2) does not hold, then there exists an infinite sequence (x^1, x^2, \ldots) of elements from (R_1, R_2, \ldots), such that (x^k \not\equiv O_k\;(\alpha)). Since (R_k \subset S), (x^k \to x). Suppose that (\pi_i(x) \ne O_i). Then some (\beta)-neighborhood of the element (\pi_i(x)) ((\beta \in \Omega)) does not contain (O_i). But then the (\beta)-neighborhood of the element (\pi_i(x)) contains (x^i). In view of the arbitrariness of the smallness of (\beta), we have (\pi_i(x) = x^i). Since (x \in S), this is possible only for a finite set of values of (i). Consequently, (\pi_i(x) = O_i) for almost all (i). But then the infinite set of elements (x^k) does not belong to the (\alpha)-neighborhood of the element (x)—a contradiction. If, conversely, conditions 1) and 2) are fulfilled, then (S = R), and the topology on (S) coincides with the Tychonoff topology. It remains to take into account Tychonoff’s theorem ((¹), p. 394; (²), p. 114).

* In (²) (p. 160) the term “isomorphism” is used.

Theorem 2. The space (S) is bicompact at the point ((O_i)) (i.e., has a neighborhood of the point ((O_i)) with bicompact closure) if and only if there exists a neighborhood (\alpha \in \Omega) such that: 1) the closure of the (\alpha)-neighborhood of (O_i) in each (R_i) is bicompact; 2) for any (\beta \in \Omega), for almost all (i) the closure of the (\alpha)-neighborhood of (O_i) in (R_i) is of order less than (\beta).

We shall precede the proof by two lemmas.

Lemma 1. If (U) is an (\alpha)-neighborhood of the point ((O_i)) in (S), then (\overline U = \prod^c \pi_i(\overline U))*.

Indeed, the inclusion (U \subset A), where (A=\prod^c \pi_i(\overline U)), is obvious. If (a \in A) and (\beta \in \Omega), then there is a finite set of indices (K) such that (\pi_i(a) \equiv O_i(\beta)) for (i \notin K). Since (\pi_i(a) \in \pi_i(\overline U)), there exist (u^k \in U) ((k \in K)) such that (\pi_k(u^k) \equiv \pi_k(a)\,(\beta)). Let (u \in R) be such that (\pi_i(u)=\pi_i(u^i)) for (i \in K) and (\pi_i(u)=O_i) for (i \notin K). Then (u \in U) and (a \equiv u\,(\beta)). Consequently, (a \in \overline U), and hence (A \subset \overline U).

Lemma 2. If (U) is an (\alpha)-neighborhood of the point ((O_i)) in (S), and (V) is the (\alpha)-neighborhood of the point (O_i) in (R_i), then (\pi_i(\overline U)=\overline V).

Indeed, if (x \in \pi_i(\overline U)), then (x=\pi_i(y)), where (y \in \overline U). If (\beta \in \Omega), then (y \equiv u(\beta)), where (u \in U). Then (x \equiv \pi_i(u)(\beta)). But (\pi_i(u) \equiv O_i(\alpha)), i.e. (\pi_i(u) \in V). Consequently, (x \in \overline V), i.e. (\pi_i(\overline U)\subset \overline V). Suppose now that (x \in \overline V) and (\beta \in \Omega). Then (x \equiv y(\beta)), where (y \equiv O_i(\alpha)). In view of Proposition 1, (x,y \in S). But then (y \in U). Consequently, (x \in \overline U), and hence (x \in \pi_i(\overline U)). Thus, (\overline V \subset \pi_i(\overline U)).

Now let (S) be bicompact at the point ((O_i)). Choose an (\alpha)-neighborhood (U) of the point ((O_i)) with bicompact closure. In view of Lemmas 1 and 2, (\overline U=\prod^c \overline V_i), where (V_i) is the (\alpha)-neighborhood of (O_i) in (R_i), so that the validity of 1) and 2) follows from Theorem 1. The sufficiency of conditions 1) and 2) also follows from Theorem 1 with the help of Lemmas 1 and 2.

Theorem 3. If all (R_i) are separated and complete, then (S) is the completion of the set of finite elements.

First we prove two lemmas.

Lemma 3. Let (\mathfrak F) be a base of a Cauchy filter (\left({}^{2}\right)), p. 168, in the set of finite elements, (\alpha \in \Omega), and (\beta \overset{2}{<} \alpha). Then there exists a finite set of indices (K) such that (\pi_i(A)\equiv O_i(\alpha)) for all (i \notin K), if (A \in \mathfrak F) and is of order less than (\beta).

Indeed, suppose that there is an infinite set of indices (s) such that (A^s \in \mathfrak F), (A^s) is of order less than (\beta), (a^s \in A^s), (\pi_s(a^s)\not\equiv O_s(\alpha)). Renumbering, if necessary, we shall have (\pi_2(a^1)=O_2). Consequently, (a^1 \not\equiv a^2(\alpha)). Since (A^1) and (A^2) are of order less than (\beta) and the intersection (A^1 \cap A^2) is nonempty, this is impossible.

Lemma 4. If (R_i) are complete, then (S) is complete.

Indeed, let (\mathfrak F) be a base of a Cauchy filter in the set of finite elements. Put (\mathfrak F_i=\pi_i(\mathfrak F)). It is clear that (\mathfrak F_i) is a base of a Cauchy filter in (R_i). Let (x_i) be the limit of this base (\left({}^{2}\right)), p. 62. Denote by (x) the element of (R) such that (\pi_i(x)=x_i) for all (i). If (x \notin S), then there exist (\alpha \in \Omega) and an infinite sequence of indices (\Lambda) such that (x_i\not\equiv O_i(\alpha)) for all (i \in \Lambda). Let (\beta \overset{3}{<} \alpha), (A \in \mathfrak F), and let (A) be of order less than (\beta). In view of Lemma 3, for some (k \in \Lambda) we shall have (\pi_k(A)\equiv O_k(\beta)). But then (\pi_k(A)\not\equiv x_k(\beta)), which is incompatible with (\pi_k(A)\in \mathfrak F_k) (see (\left({}^{2}\right)), p. 62, Proposition 1). Thus, (x \in S). Let again (\alpha \in \Omega) and (\beta \overset{6}{<} \alpha). Taking into account Lemma 3 and the relation (x \in S), we find a finite set of indices (K) such that, whatever the set (A) from (\mathfrak F) of order less than (\beta) may be, for all (i \notin K) one has

* By (\overline M) is denoted the closure of the set (M).

(x_i = O_i(\beta)) and (\pi_i(A) \equiv O_i(\beta)). Since (\mathfrak F_i \to x_i), there is a (B \in \mathfrak F) such that (\pi_i(B) \equiv x_i(\beta)) for all (i \in K). Since (B) is of order (\beta), (B \equiv x(\alpha)). Hence (\mathfrak F \to x). It remains to take into account Proposition 2 and ((^2)), p. 170, Proposition 7.

In view of Propositions 2 and 3, the validity of Theorem 3 follows from Lemma 2 and ((^2)), p. 175, Proposition 9.

Let (G) be a topological group. In a natural way it may be regarded as a uniform space (((^2)), p. 224). A system (\mathfrak H = {H_i, i \in I}) of closed subgroups of the group (G) will be called independent if each finite subsystem of it generates in (G) a subgroup that is the topological direct product of the corresponding subgroups from (\mathfrak H). Turn (H_i) into the uniform space (\hat H_i), taking as a system of neighborhoods of the identity the system of sets of the form (p_i(U)), where (p_i) is the projection (H = \sum H_i) onto (H_i), and (U) is a neighborhood of the identity in (H). The uniform space (S = \prod^{c} \hat H_i) will be called the accompanying space of the system (\mathfrak H). The mapping (h \to (p_i(h))) of the space (\bar H) onto the set of finite elements of (S) will be called natural. We shall say that the projections (p_i) are uniformly equicontinuous* if for every neighborhood (U) of the identity in (\bar H) there is a neighborhood (V) of the identity in (H) such that (p_i(V) \subset H_i \cap U) for all (i \in I). It is easy to prove:

Proposition 4. The natural mapping is one-to-one and uniformly continuous; if the group (H) admits a complete system of neighborhoods of the identity consisting of subgroups, and the projections are uniformly equicontinuous, then the natural mapping is a uniform homeomorphism.

Theorem 4. Let (G) be a complete separated topological group, (\mathfrak H = {H_i}) an independent system of subgroups of (G), and (H = \sum H_i). Then the natural mapping of the group (H) into the accompanying space (S) of the system (\mathfrak H) can be extended to a uniformly continuous mapping of the group (\bar H) onto (S); if the group (H) admits a system of neighborhoods of the identity consisting of subgroups, and the projections (H) onto (H_i) are uniformly equicontinuous, then the above-mentioned mapping of (\bar H) onto (S) is a uniform homeomorphism.

Proof. The space (\bar H) is complete (((^2)), p. 170, Proposition 6). Therefore the validity of the theorem follows from Proposition 4, Theorem 3, ((^2)), p. 171, Theorem 1, and ((^2)), p. 175, Proposition 9.

Theorem 5. If the topological groups (H_i) are complete uniform spaces and (H = \prod^{c} H_i) has a group neighborhood of zero (U) with bicompact closure, then (H) is a local direct product of the groups (H_i) with respect to the subgroups (\pi_i(U)).

For the proof it suffices to take into account Theorem 2 and the definition of a local direct product of groups (((^3)), p. 9).

Moscow State University
named after M. V. Lomonosov

Received
12 XI 1961

REFERENCES

1. P. S. Aleksandrov, Introduction to the General Theory of Sets and Functions, Moscow–Leningrad, 1948.
2. N. Bourbaki, General Topology. Basic Structures, Moscow, 1958.
3. J. Braconnier, J. Math. pure et appl., 87, 1 (1948).

* What is meant is the subgroup generated in (G) by the union (\bigcup H_i); obviously, it coincides with the discrete direct product of the groups (H_i).