Banach space preliminaries

The purpose of this section is to review some well-known concepts and results from separable Banach space theory. Our focus is the weak-*topology on the dual of a separable Banach space. A reference for most of this material is Chapter V of Dunford and Schwartz [11].

Let *X* be a Banach space. The *weak topology* on *X* is the
weakest topology such that every bounded linear functional on *X* is
continuous. The *dual space* of *X* is the space *X*^{*} of all
bounded linear functionals on *X*. The *norm* of
is defined by

The

for some finite set .

A key theorem concerning the weak-* topology is:

is weak-*closed and weak-* compact. Furthermore, if

*Proof. *See [11], page 424.

In considering weak-* sequential convergence, an important fact to keep in mind is the following consequence of the Banach-Steinhaus theorem:

*Proof. *See [11], page 66.

A set
is said to be *weak-*sequentially closed* if it is closed under weak-* limits of
sequences, *i.e.*,
weak-* and
for all
imply .

For an arbitrary set
,
being weak-* sequentially
closed is not in general equivalent to being weak-* closed. In
particular, the weak-* topology is not in general metrizable, even
when *X* is separable. This is shown by the following theorem.

*Proof. *We first prove the following lemma.

*Proof. *For any finite set
,
is a subspace of *X* of codimension at most the
cardinality of *F*, and hence in particular it intersects
for each
.
By the Banach-Alaoglu theorem,
*B*_{n}(*X*^{*}) is weak-* compact, and so
*B*_{n}(*X*^{*})^{n} is weak-* compact
as well. Thus for each *n* we can find a finite set
such that, for each finite set
of
cardinality *n*,
intersects *G*_{n}. Letting
be an enumeration
without repetition of
,
we obtain the
desired sequence.

*Proof of Theorem 2.5.* Since *X* is
separable, it follows by Theorem 2.1 that *X*^{*} is weak-*separable, so let
be a countable weak-* dense
subset of *X*^{*}. By the preceding lemma, for each *n* we can find a
sequence
such that
for
all *k* and
and *x*^{*}_{n} belongs to the
weak, and hence weak-*, closure of
.
Thus
is a countable set which is weak-*dense in *X*^{*}. On the other hand, for each *n*,
is
finite, so by Corollary 2.4 *Z* is weak-* sequentially
closed.

The previous theorem shows that a set in *X*^{*} can be weak-*sequentially closed yet far from weak-* closed, even when *X* is
separable. Nevertheless, it turns out that for *convex* sets in
*X*^{*}, being weak-* sequentially closed is equivalent to being
weak-* closed, provided *X* is separable. We shall obtain this
result as a consequence of the following well-known theorem:

*Proof. *See [11], page 429.

The special case of the Krein-Smulian theorem for subspaces of
*X*^{*} is originally due to Banach ([2] page 124):

*Proof. *For a subspace *Z* of *X*^{*}, we have
.
Hence
is weak-* closed for all *r* if
and only if
is weak-* closed. The desired result
follows immediately from the Krein-Smulian theorem.

A set *Z* in *X*^{*} is said to be *bounded-weak-* closed* if
is weak-*-closed for all *r*>0. This defines yet
another topology on *X*^{*}, the *bounded-weak-* topology*. By
the Banach-Alaoglu theorem, weak-* closed sets are bounded-weak-*closed, but the converse does not hold in general. We can paraphrase
the Krein-Smulian theorem by saying that a convex set in *X*^{*} is
weak-* closed if and only if it is bounded-weak-* closed.

*Proof. *By definition, *Z* is bounded-weak-* closed if and only if
is weak-* closed for all *r*. Since *X* is separable, we
have by Theorem 2.1 that *B*_{r}(*X*^{*}) is weak-* compact and
weak-* metrizable. Hence, for all *r*,
is weak-*closed if and only if
is weak-* sequentially
closed. By Corollary 2.4 it now follows that
is weak-* closed for all *r* if and only if *Z* is
weak-* sequentially closed. This completes the proof.

We now obtain the desired result:

*Proof. *Immediate from Theorem 2.7 plus Lemma 2.9.

Again, the special case when *Z* is a subspace of *X*^{*} is due to
Banach ([2] page 124):

We now turn to a discussion of weak-* sequential closure ordinals.
For an arbitrary set
,
let *Z*' denote the set of
weak-* limits of sequences from *Z*. Define a transfinite sequence
of sets
,
an ordinal, by

Clearly implies . Define to be the least ordinal such that . Thus is the weak-*sequential closure of

In the remainder of this section and the next section, we shall prove
some results which completely answer the question of which ordinals
can arise as closure ordinals of subspaces of *X*^{*} where *X* is a
separable Banach space. This question was first answered completely
by McGehee [18] and Sarason
[19,20,21].

*Proof. *For *r*>0 and
an ordinal, let
be the weak-*closure of
.
For fixed *r*>0, the sets
form an increasing, transfinite sequence of compact
subsets of a compact metric space, namely *B*_{r}(*X*^{*}) with the weak-*topology (Theorem 2.1). This transfinite sequence must
therefore stabilize at some countable ordinal. Since
for
all ,
it follows that the transfinite sequence
also stabilizes at a countable ordinal,
call it .
Put
.
Then
is a countable ordinal. We claim that
is
weak-* sequentially closed. To see this, suppose that *x*^{*} is the
weak-* limit of a sequence
from
.
By Corollary 2.3, there exists
such that
for all
.
Hence
,
so in
particular
and our claim is proved. Thus
,
*i.e.*,
is countable.

Specializing to subspaces of *X*^{*}, we can say more:

*Proof. *By the previous lemma,
is
countable. Suppose that
,
for some countable limit ordinal
.
As
is norm closed, this implies that
,
where
denotes
the norm closure of
.
Thus
is a closed
subset of a complete metric space, written as a countable union of
closed sets. By the Baire category theorem, there must be an ordinal
such that
has non-void interior as a
subset of
,
*i.e.*,
contains the
intersection of an open (in norm) ball with
.
But
is a subspace of
,
so it must be all of
.
Since
,
it
follows that
,
whence
.
We
have now shown that
is not a limit ordinal. Thus
must be either 0 or a successor ordinal. If
then *Z* is
weak-* sequentially closed, and hence is weak-* closed by
Corollary 2.11. This completes the proof. See also
Kechris and Louveau [14], page 157.

It is known that the converse of the previous theorem also holds: For
every countable successor ordinal ,
we can find a subspace
*Z* of the dual *X*^{*} of a separable Banach space *X* such that
.
In the next section we present a
proof of this result, with
and *Z* weak-*dense in .

The study of closure ordinals of subspaces of *X*^{*} has an interesting
history. Von Neumann ([24] page 380) exhibits a set
such that 0 is in the weak closure of *S* yet no
sequence from *S* converges weakly to 0. The first example of a
subspace *Z* of *X*^{*} such that
is due to Mazurkiewicz
[17]. Banach ([2] pages 209-213) proves
that for every
there is a subspace *Z* of
such
that
and states the analogous result for all countable
ordinals. For the proof Banach refers to a ``forthcoming'' paper
which seems never to have appeared, and this reference is omitted from
the English translation [1]. Later, McGehee
[18] proves the stronger result that for each countable
ordinal
there exists a weak-* dense subspace of
whose closure ordinal is exactly
(but note
that McGehee's notation differs from ours). In the next section we
reprove this result of McGehee. Sarason
[19,20,21] proves a similar result for the
spaces
and
.
In view of the theorem above,
these results of McGehee and Sarason are in a sense best possible.
While McGehee's proof uses sets of synthesis and uniqueness, our proof
in the next section is much more elementary.