Proposition 1 (Talagrand). Let be a regular space which admits a stronger topology ϑ such that is a Lindelöf Σ-space. Then , where and denote the density and the weight of X, respectively.
4.1. Proof of Theorem 3
Let us prove Theorem 3.
It is obvious that if Y is analytic then Y is separable and K-analytic, so Theorem 3 holds.
To prove the converse of the statement of this theorem it is enough to show that
Y admits a weaker metrizable topology because then, by [
1] (Theorem 15), the space
Y is analytic.
Firstly we are going to check that to prove this converse we may suppose the additional condition that X is a strongly web-bounding space.
In fact, let X be a web-bounding space and suppose that there is a web-bounding representation of X whose union is dense in X. Then the restriction map defined by is an injective continuous linear map. Let be a subset with a compact resolution contained in a separable subset . Then for the assumptions are satisfied, so is analytic in the induced topology from and consequently admits a weaker metrizable topology . Then is a weaker metrizable topology on Y. Therefore we may assume that X is strongly web-bounding.
Hence, to finish the proof of Theorem 3 it is enough to prove the following Proposition.
Proposition 2. Let X be a strongly web-bounding space and let Y be a non-empty subset of such that Y has a compact resolution and is contained in a separable subset of . Then Y admits a weaker metrizable topology (hence, as was said before, Y is analytic).
Proof. Let
be the real-compactification of
X. Since
X is strongly web-bounding, we apply [
3] (Theorem 9.15) to deduce that
is Lindelöf
-space.
As a help to the reader we split the proof in two parts.
Step 1. Assume that Y is a subset of , Y has a compact resolution and it is contained in a separable subset . Now we prove that L (and also Y) admits a weaker metrizable topology. Let D be a countable dense subset of L. Let and be the weakest topologies on that make continuous the functions of D and L, respectively. By density for each implies for each , hence the topological quotients and of and respect to the relations if for all f of D and if for all f of L, respectively, are algebraically identical and we denote by is the quotient map.
If we define the map by , , then clearly F is continuous and if and only . is homeomorphic to a subspace of and consequently is metrizable and separable. On the other hand is a Lindelöf -space, since it is a continuous image of the Lindelöf -space . It follows from Proposition 1 that the space is separable.
Let be a countable subset of such that the set is dense in . For each let be the element of such that . Let be such that . Then, from if follows that and the density condition implies that . Therefore . Consequently, if f and g are two different elements of L there exists such that . This means that the weaker topology on L defined by the topology of the pointwise convergence on S is metrizable.
Step 2. Let
be equipped with a compact resolution and let
L be a separable set in
containing
Let
be defined by
where
is the unique continuous extension of
f to the whole
. Since
is continuous on each countable set, see [
13] (Theorem 4.6(3)),
has a resolution of countably compact sets. On the other hand, the space
is angelic, see [
5] (Theorem 3), so every countably compact set in
is compact. Hence,
has a compact resolution.
Let
be a dense subset of
L. Take any
, any
and let
be an arbitrary finite subset of
. Then there is
with
and by [
13] (Theorem 4.6(1)) for each
there exists
such that
and
for each
. Choose
such that
for each
. Hence,
for each
. This shows that
is a dense subset of
, so that
is separable. By Step 1 we derive that
is analytic in
. The continuity of the surjection
implies that
is also analytic. □
For a completely regular topological space
X, Tkachuk proved in [
14] that
is K-analytic if and only if it has a compact resolution. If
X is a separable metric space, then
is analytic if and only if it admits a resolution consisting of bounded sets, see [
15] (Corollary 2.5) and [
16] (Proposition 1).
From the proof of Proposition 2 follows immediately the following claim that enables to get in Corollary 2 the following variant for analyticity of for arbitrary X.
Claim 1. Let X be a topological space such that its real compactification is Lindelöf Σ-space and let Y be a non-empty subset of such that Y has a compact resolution and is contained in a separable subset of . Then, Y admits a weaker metrizable topology (hence, as was said before, Y is analytic).
Corollary 2. Let ξ be a topology on which is stronger or equal than the pointwise topology of . Then is analytic if and only if is separable and has a ξ-compact resolution.
Proof. It is enough to prove this Corollary when
, because a submetrizable topological space is analytic if and only if it admits a compact resolution (see [
1] (Theorem 15)). Assume that
is separable and has a compact resolution. Then by [
17] (Corollary 23) the space
is a Lindelöf
-space. Now, Claim 1 for
implies that
is analytic. The converse is clear. □
Hence, a separable space admits a compact resolution if and only if it is analytic, or, equivalently, there is an upper semi-continuous compact-valued map from covering if and only if is a continuous image of .
The following example shows that Corollary 2 does not work in general for the weak*-dual of .
Example 3. Corollary 2 fails for the weak*-dual of .
Proof. It is well known that the space
endowed with the product topology is K-analytic separable but not analytic. Consequently
is K-analytic and separable by [
6] (Proposition 0.5.14).
is not analytic, since
is a closed subspace of
and each closed subspace of an analytic space is analytic. □
4.2. Proofs of Theorems 1 and 2
We are ready to prove Theorem 1.
Proof. Note that is strongly web-bounding. Indeed, let be a -representation of . Then if and , , there exists for each a such that and . From these equalities for it follows that there exists with . Hence, for all , yielding equicontinuity of , so is functionally bounded. Finally, as is contained in the proof follows applying Theorem 3. □
We complete the paper with a short and elementary proof of Theorem 2. It is enough to make the proof for a compact subset K of E, because the completion of a locally convex space E in class belongs to class and the closure in the completion of a precompact subset of E is a compact subset.
Proof. Let be a -representation of . By we denote the topology of E and let K be a compact of We say that a subset M of is -separated if , for each . By Zorn’s lemma there exists a maximal -separated subset of and the maximal condition implies that .
Note that
is countable. Indeed, otherwise, since
and
whenever
, for
,
in
, we determine a sequence
such that each
,
, contains and uncountable subset of
and then by a very easy standard argument we obtain countable infinite subset
P of
and
such that
, see [
3,
10,
18].
Since
E belongs to
P is equicontinuous, so, by Grothendieck theorem of polar topologies ([
19] (Chapter IV,
$21.7))
P is precompact in the topology of uniform convergence on the
-precompact subsets of
E. Therefore there exists a finite set
such that
. Clearly there exists
such that the set
is infinite, contradicting the hypothesis that
(
) is
-separated.
Let
be a maximal subset of
that it is
-separated, for each
. The set
is countable. Let
be the weakest topology on
K that makes continuous the functions of
. If
are two points of
K then there exist
and
such that
. Since
, there exists
such that
. Hence,
Therefore
is metrizable, so
K is metrizable. □