1. Introduction
Unless otherwise stated, X stands for an infinite Tychonoff space. We denote by the linear space of real-valued continuous functions on X equipped with the pointwise topology . The topological dual of is denoted by , or by when provided with the weak* topology. We represent by the space equipped with the compact-open topology . A family of subsets of a set X is called a resolution for X if it covers X and verifies that for . Let us recall that a Hausdorff topological space X is called a -space if the closure of each functionally bounded set in X is compact. Every realcompact space is a -space, but the converse is not true. If E is a locally convex space, the bidual of E is the topological dual of the strong dual of E. By the strong dual of E, we mean the dual of E equipped with the strong topology of the uniform convergence on the bounded sets in E. We denote by and the weak locally convex topology of and the bounded-open topology of , respectively. If A is a nonempty set in a real linear space L, we represent by the (real) absolutely convex hull of A. The linear subspace of consisting of those bounded functions is denoted by .
A base
of (absolutely convex) neighborhoods of the origin in a locally convex space
E such that
if
is called a
-base. Let us mention that the notion of a
-base, originally introduced in the realm of locally convex spaces [
1], has been extended to topological groups and general topological spaces, sometimes under the name of
-base, by some authors (see [
2,
3]). However, in this paper we keep the original name. Trivially, if
E is a metrizable locally convex space with a decreasing base
of locally convex neighborhoods of the origin, the family
, where
for all
is a
-base for
E. The converse is not true, i.e., a locally convex space with a
-base need not be metrizable (see p. 107, [
4]).
A locally convex space
E with a
-base has metrizable compact sets, since, in this case, the weak dual
is quasi-Suslin [
5], hence trans-separable [
6] (properties not defined in this paper can be found in [
7,
8,
9,
10]). Therefore, if there is a locally convex topology
on
with a
-base such that
, every completely regular topology
on
such that
is angelic and has metrizable compact sets. Research on
-bases and their generalizations remain active since [
1]. For recent results on this topic see [
3,
11,
12,
13,
14]. Research on the existence of locally convex topologies on
stronger than the pointwise topology has been studied in [
15,
16]. In this paper, we enlarge the classification of topological properties on
X provided in (Theorem 3.1, [
15]).
3. A More Complete Classification
To enlarge the previous classification, we need the following result.
Lemma 1. Let X be completely regular. If Q is a metrizable and compact subspace of X, there exists a continuous linear extender map , i.e., such that for every .
Proof. Since
Q is (homeomorphic to) a metrizable compact subspace of the Stone–Čech compactification
of
X, according to [
18] (see also (Proposition 4.1, [
19])) there exists a continuous linear map
such that
which embeds
in
(as a closed subspace). If
is the restriction map
, it turns out that
is a continuous linear map from
into
such that
for every
.
Hence, has a closed graph when both spaces are regarded as Banach spaces. This implies in particular that is continuous, since the supremum norm topology on is stronger than the relative compact-open topology of . Thus, is a continuous linear extender map, as stated. □
Theorem 2. If X is a Tychonoff space, the following properties hold
- 1.
The compact-open topology on is metrizable if and only if X is a hemicompact space (Arens’ theorem (Theorem 7, [20]), see also (Theorem 2.5, [21])). - 2.
The bounded-open topology on is metrizable if and only if there is an increasing sequence of functionally bounded subsets of X swallowing the functionally bounded sets of X.
- 3.
The weak topology on is metrizable if and only if X is countable and compact sets in X are finite.
- 4.
The pointwise topology on is metrizable if and only X is countable.
- 5.
The compact-open topology on has a -base if and only if X has a compact resolution that swallows the compact sets (Theorem 2, [22]). - 6.
The bounded-open topology on has a -base if and only if X has a functionally bounded resolution that swallows the functionally bounded sets.
- 7.
The weak topology on has a -base if and only if X is countable and compact sets in X are finite.
- 8.
The pointwise topology on has a -base if and only X is countable (Corollary 15.2, [7]).
Proof. The proof of statements
and
is similar (for
see (Theorem 12, [
23])).
Let us prove statement
. If
is weakly metrizable, the weakly bounded sets are metrizable, so
X is countable by (Theorem 2.3, [
21]). This fact also follows from statement
of Theorem 1. We claim that, in addition, the compact sets of
X are finite. Otherwise, there exists an infinite compact set
Q in
X. However, since
X is countable,
Q is metrizable. By Lemma 1 there is a continuous linear extender
from
into
, i.e., such that
for every
. If
in
, given
there is
with
for every
, so
for
. Thus,
in
, and hence
in
, i.e.,
, which means that
embeds the Banach space
in
as a closed subspace. Since the weak topology is inherited by linear subspaces, the space
is linearly homeomorphic to a subspace of
. In other words, the Banach space
is weakly metrizable. As the weak topology of a Banach space is metrizable if and only if it is finite-dimensional, it turns out that
Q must be finite, a contradiction. Conversely, if
X is both countable and has finite compact sets, the former statement guarantees that
is metrizable whereas the latter implies that
coincides with
. Hence, the weak topology
is metrizable.
The proof of statement
is similar to that of statement
. The only difference is that the weak topology of the Banach space
now carries a
-base. However, if a locally convex space in its weak topology has a
-base, it is countable-dimensional (Proposition 11.2, [
7]). Thus,
must be countable-dimensional, so finite-dimensional by the Baire category theorem. This ensures that the compact set
Q must be finite. □
Remark 1. In (Theorem 2.3, [21]), it is shown that if X is countable, the bounded sets of are weakly metrizable. According to statement of the previous theorem, if X contains an infinite compact set, then cannot be weakly metrizable. Thus, if X is countable and contains an infinite compact set, then is not metrizable but has metrizable bounded sets. Example 1. As is a countable space with infinite compact sets which is not hemicompact, neither nor are metrizable, the former statement by Arens’ theorem and the latter by the previous remark, but is metrizable. Of course, has no -base. In fact, since is not a Polish space, Christensen’s theorem (Theorem 94, [4]) prevents from having a compact resolution that swallows the compact sets. This also implies that has no -base, by statement of Theorem 2. Example 2. Let be equipped with the discrete topology and choose . Then, provided with the relative topology of is not discrete but has finite compact sets. Thus, the weak topology on is metrizable. In fact, clearly, , although .
Example 3. is an uncountable hemicompact space, hence is metrizable, but and are not.
Example 4. If K is a countable infinite compact set, obviously both and are metrizable, but is not.
Example 5. The Sorgenfrey line is a Lindelöf space which is not σ-compact, so and by statement of Theorem 1 there is no metrizable, locally convex topology on such that .
Example 6. The space has a -base by statement of Theorem 2, but there is no metrizable topology such that , since is not σ-compact.
4. The Interval
In this section, we deal with the interval . Before stating our main result for this case, we need to establish two auxiliary results concerning the weak* dual of the space . We regard (the canonical homeomorphic copy of) X in as a Hamel basis of , and denote by the image of in . If and , we write to represent the action of the linear functional u on f, in particular .
Lemma 2. Let E denote the dual of . If X is a μ-space, then E coincides with the bidual of .
Proof. If
X is a
-space, then
is the strong dual of
(see (Lemma 2.2, [
24])). Therefore,
and hence,
E coincides with the bidual
of
. □
Lemma 3. Let X be a μ-space. If A is a bounded set in , there are a compact set in X and a real number with such that .
Proof. Since
X is a
-space, if
A is a bounded set in
, then
is a neighborhood of the origin in
, so there is
and a compact set
in
X such that
which implies that
, where the bipolar of
A is taken in
as well as the polar of
. The fact that
yields
so that
. We claim that
.
Assume that with . Since X is a Hamel basis of , we have that , with for . First, we show that the support of u is contained in , so that for . Indeed if for some , there is such that and for every . Note that since if with and for , then . As , this contradicts the fact that . Thus, for .
Now, we show that , so that . Since the support of u is finite and is contained in , there are pairwise disjoint open sets in with , so there are compactly supported continuous functions such that for . This means that , has compact support, , and for . For , set , where denotes the sign of the real number , and define . Clearly , , and for every , because of the supports of the functions are pairwise disjoint.
Since
and
there exists
with
and
for
such that
. Consequently,
and due to the fact that
and
it follows that
as required. □
If
X is a
-space and
denotes the family of all bounded sets in
, as a consequence of Lemma 2, the space
E coincides with (see (23.2.(1), [
8]))
where the closure is in
E under the weak* topology
of
E. Hence, if
there is a bounded set
A in
with
. However according to Lemma 3, given
, there is a compact
in
X together with some
with
. Thus, if
designates the family of all functionally bounded sets in
X, then
where the closures are in
. Therefore, the following property holds.
Remark 2. If X is a μ-space and stands for the family of all compact sets in X, thenwhere E is the dual of . Proof. Because of the bipolar theorem, one has . □
This motivates the following definition.
Definition 1. Let E be the dual of . We say that a countable family of compact sets in X is a complete sequence if there exists a sequence of positive integers such that If
X is a metrizable hemicompact space and
is a family of compact sets that swallows the compact sets in
X, then
is a complete sequence. If
is a base of neighborhoods of the origin in
, there are by Lemma 3 a compact set
S in
X and
such that
. If
then
for every
. If we define
if
for some
and
otherwise, then
If is a dual pair, a locally convex topology for F is called a polar topology if is the topology of the uniform convergence on the sets of a family of -bounded sets in E. Obviously, both the weak topology and the strong topology are polar topologies for F. We are ready to establish our last classification result, which reads as follows.
Theorem 3. Let X be a Tychonoff space. There is a metrizable polar topology on such that if and only if X contains a complete sequence.
Proof. Let us denote by E the dual of the locally convex space . Assume that is a metrizable polar topology on with a decreasing base of neighborhoods of the origin enjoying that . By the first statement of Theorem 1, we know that X is a -compact space, so a -space.
As
is a polar topology, there exists a family
of bounded sets in
such that for each
, there are some
with
. According to Lemma 3, there is a compact set
in
X and some
such that
, so that
. Set
. Since
for each
and
, we get
Thus, the sequence is complete.
Assume conversely that there is a complete sequence
in
X. We may suppose that
as well as that
. If not, replace
by the sequence
with
and the sequence
by
with
. Since
for every
, if
is complete so is
. Then, for each
, define
. Clearly,
and
is a base of neighborhoods of the origin of a metrizable locally convex topology
for
. Observe that
is stronger than
, since
On the other hand, as , the family of all compact sets in X, we see that is weaker than . □