MATHEMATICS

A. I. VEKSLER

ON THE TOPOLOGICAL AND STRUCTURAL COMPLETENESS OF NORMED AND LINEAR TOPOLOGICAL STRUCTURES

(Presented by Academician V. I. Smirnov, 9 XI 1961)

The connection between topological completeness and structural completeness (i.e. completeness with respect to ordering) is investigated in certain classes of normed and linear topological structures.

We shall use, for the most part, the terminology of (1). By a \(KN\)-lineal we shall mean a normed structure, i.e. a \(K\)-lineal which is a normed space with a monotone norm; by a \(KB\)-lineal—a Banach structure, or a \((b)\)-complete \(KN\)-lineal; by a \(KN\)-space—a structurally or conditionally complete (i.e. one which is a \(K\)-space) \(KN\)-lineal. For a \((b)\)-complete \(KN\)-space we shall use no special term. By a \(K_\sigma\)-space we shall mean a conditionally \(\sigma\)-complete \(K\)-lineal.

Let us recall some definitions. A component of an arbitrary \(K\)-lineal \(X\) is any set \(X'\subset X\) that is the disjunctive complement of some set \(E\subset X\). For an Archimedean \(K\)-lineal \(X\) this definition is equivalent to the usual definition of a component in a \(K_\sigma\)-space ((1), p. 97). The component generated by a given set \(H\) is the smallest component containing \(H\); it coincides with the disjunctive complement of the disjunctive complement of \(H\).

If \(X'\) is a component of \(X\), \(x\in X\), and \(x\geqslant 0\), then \(\sup x'\) over all \(x'\in X'\), \(x'\leqslant x\) (if it exists), is called the projection of \(x\) onto \(X'\). For arbitrary \(x\in X\), its projection onto \(X'\) is called the difference of the projections onto \(X'\) of its positive and negative parts.

\(X\) is called a \(K\)-lineal with projections onto the component generated by an element if the projection of every element exists onto every component generated by a one-element set. If in \(X\) the projection of every element exists onto any component, then \(X\) is called a \(K\)-lineal with projections.

Theorem 1. A \(KB\)-lineal with projections is always a \(K\)-space.

Theorem 2. A \(KB\)-lineal with projections onto the component generated by an element is always a \(K_\sigma\)-space.

In the proof of Theorem 2 the following auxiliary result is used.

Lemma 1. Let \(X\) be an Archimedean \(K\)-lineal with unit \(1\), in which the projection of every element exists onto any component generated by an arbitrary countable set. Then the base of all unit elements* of \(X\) is a \(\sigma\)-complete Boolean algebra.

Are Theorems 1 and 2 reversible? For Theorem 1, for example, the question is formulated as follows. Will not every \(KN\)-space be \((b)\)-complete (it is well known that every \(K\)-space is a \(K\)-lineal with projections; see, for example, (1), p. 100).

* An element \(x\in X\) is called unitary if \(x\wedge(1-x)=0\).

The following example shows that such assumptions do not hold in the general case.

Example 1. \(X\) is the \(K\)-space of all terminating sequences

\[ x=\{\xi_1,\xi_2,\ldots,\xi_{n(x)},0,0,\ldots\}. \]

It is clear that \(X\) can be made into a \(KN\)-space, for example with respect to the norm

\[ \|x\|=\max_k|\xi_k|. \]

However, it is obviously not \((b)\)-complete. Moreover, \(X\) cannot be made complete for any choice of norm. This follows at least from the fact that \(X\) has a countable algebraic basis, whereas the cardinality of any algebraic basis in a Banach space is not less than \(c\) (see ([2])).

We shall single out a certain class of \(K\)-lineals for which Theorems 1 and 2 are reversible.

Let \(X\) be an Archimedean \(K\)-lineal of bounded elements with unit \(1\). In it one introduces the norm

\[ \|x\|=\inf\{\lambda: |x|\leq \lambda 1\}. \]

It is known that this norm is monotone. With this norm \(X\) is called a \(KN\)-lineal of bounded elements. A \(KN\)-lineal of bounded elements that is a \(K_\sigma\)-space is always \((b)\)-complete (([1], p. 200). Hence we obtain the following theorems.

Theorem 3. In order that a \(KN\)-lineal of bounded elements be a \(K\)-space, it is necessary and sufficient that it be a \(KB\)-lineal with projections.

Theorem 4. In order that a \(KN\)-lineal of bounded elements be a \(K_\sigma\)-space, it is necessary and sufficient that it be a \(KB\)-lineal with projections onto the component generated by an element.

Theorems 1 and 2 admit generalizations. These generalizations can be made simultaneously in two directions. First, one may consider not only normed structures but also linear topological structures \(X\); in doing so it is enough to require completeness of \(X\) only with respect to sequences. Secondly, the very condition of completeness of \(X\) with respect to sequences can be weakened by considering not all convergent-to-itself sequences.

We shall call a locally convex linear topological structure a \(KT\)-lineal, i.e. a \(K\)-lineal that is a locally convex space in which every seminorm from the family of seminorms defining the topology is monotone. A \(KT\)-lineal \(X\) will be called boundedly semicomplete if in it every structurally bounded sequence that converges to itself converges to some element. A boundedly semicomplete \(KN\)-lineal will simply be called boundedly complete*.

Theorem 5. A boundedly semicomplete \(KT\)-lineal with projections is a \(K\)-space.

Theorem 6. A boundedly semicomplete \(KT\)-lineal with projections onto the component generated by an element is a \(K_\sigma\)-space.

Let us clarify the question of the reversibility of these theorems. Let us consider, for example, Theorem 5 and restrict ourselves to the case of a \(KN\)-lineal. If the converse assertion were true here, this would mean that every \(KN\)-space is boundedly complete. This assertion would be weaker than the assertion converse to Theorem 1, but it too is false. Here is the corresponding example.

* A set \(P\subset X\) is called structurally bounded if for some \(y\in X\) and every \(x\in P\) the inequality \(|x|\leq y\) holds.

Example 2. \(X\) is the \(K\)-space of bounded elements \(m\), but with a different norm. Let \(f(x)\) be a Banach limit for the bounded numerical sequence \(x=\{\xi_k\}\). Put

\[ \|x\|=\sup_k \frac{|\xi_k|}{k}+f(|x|). \]

In \(X\) take the sequence \(\{e_n\}\), setting

\[ \xi_k(e_n)= \begin{cases} 0 & \text{for } k<n,\\ 1 & \text{for } k\ge n \end{cases} \qquad (n=1,2,\ldots). \]

It is obvious that \(\{e_n\}\) is a structurally bounded sequence, and

\[ \|e_{n+p}-e_n\|=\frac{1}{n}, \]

i.e. \(\{e_n\}\) converges to itself. Suppose that \(e_n \xrightarrow{(b)} x\) in \(X\). Then, obviously,

\[ 0 \le x \le e_n \qquad (n=1,2,\ldots), \]

whence

\[ 0 \le x=\inf_n e_n=0, \]

i.e. \(x=0\); but

\[ \|e_n-0\|=\|e_n\|=1+\frac{1}{n}, \]

i.e. \(e_n \not\xrightarrow{b} 0\).

Consequently, the \(KN\)-space \(X\) is not boundedly complete.

Various topologies compatible with the order relation in the given \(K\)-linear space are possible (i.e. topologies that turn \(X\) into a \(KT\)-linear space). At the same time, the introduction of a Banach topology is possible only in a unique way up to equivalence \({}^{(3)}\). There may, however, generally speaking, be many simply normed compatible topologies. The structural completeness of \(X\) does not change the situation. If, for example, in the space of Example 1 all such topologies are boundedly complete, then already in the space of Example 2 this is not so, despite the fact that in this space there even exists a Banach topology compatible with the order relation.

Leningrad Textile Institute
named after S. M. Kirov

Received
18 X 1961

REFERENCES

1. B. Z. Vulikh, Introduction to the Theory of Partially Ordered Spaces, 1961.
2. G. Mackey, Trans. Am. Math. Soc., 57, No. 1, 155 (1945).
3. B. M. Makarov, DAN, 107, No. 1, 17 (1956).