In this section we prove the following result: For each countable ordinal , there exists a weak-*dense subspace Z of such that . Our proof uses some simple concepts and results concerning trees. We give a self-contained treatment of these auxiliary results.
From the definition above, the height of a well-founded tree is a countable ordinal. The following standard theorem shows that the converse holds as well.
Proof. We prove this by transfinite induction on . For we have . For successor ordinals, note that if , then where . Suppose now that is a limit ordinal, say , where for all . For each n let T_{n} be a tree of height , and put . Then T is well-founded and, for each , . Thus .
Let , the space of sequences of real numbers converging to 0 indexed by . Then we may identify X^{*} with , the space of absolutely summable sequences of real numbers indexed by . In the rest of this section we shall mainly be interested in the weak-* topology on .
The next two lemmas imply that, for any well-founded tree T, Z_{T}is weak-* dense in .
Proof. We proceed by induction on h_{T}(s). If h_{T}(s)=-1 then and . Suppose now that , and that the theorem holds for all t such that . Then for each , , so for all . Since y_{s} is the weak-* limit of the sequence , it follows that , as desired.
Proof. Let
be given. We can write
,
where
is the
characteristic function of ,
i.e.,
if t=s, 0otherwise. Note that if
then
,
whereas
.
Also, by the previous
lemma, if
then
,
so, if
then
.
Since z is an
absolutely summable series, we have
Thus we have an upper bound on the closure ordinal of Z_{T} in terms of the height of T. To get a lower bound, we use the following technical lemma, which gives us a handle on the growth of the spaces .
Proof.
Assume that the stated condition holds. Suppose for a
contradiction that
,
say
In particular, we have the following result.
Proof. By Lemma 3.11 Z_{S} is weak-* sequentially closed. Hence by Corollary 2.11 Z_{S} is weak-* closed.
In order to make use of this lemma, we consider a special class of trees known as smooth trees:
Proof. Note first that , so if T^{*} is well-founded then so is T. Conversely, suppose T^{*} has a path f; let . Then T_{f} is a finitely-branching subtree of T, and, since f is a path through T^{*}, T_{f} must be infinite. Hence by König's Lemma T_{f} has a path, whence T has a path.
Assuming T and T^{*} are well-founded, we obviously have
.
For the opposite inequality, we claim that for all s,
.
(Note that
is a finite set, so we may take max rather
than sup.) We prove the claim by induction on
h_{T*}(s). If
h_{T*}(s)=-1 then
,
so for any t with
we have ,
whence h_{T}(t)=-1 for all such t. Otherwise
and we have
Proof. This follows immediately from Theorem 3.4 and the previous lemma.
Proof. We proceed by induction on . For there's nothing to prove. Assume is smooth, and let be given. Suppose ; since which is smooth, tmust be in ; furthermore, since s is an interior node of , there is an such that . But , so , whence . Finally, smoothness is clearly preserved under intersections, so the induction goes through at limit stages.
Proof. We proceed by induction on . If there's nothing to prove. Assume and and let be given. Then z is the weak-* limit of some sequence from . Since , we have that any is an interior node of , i.e., and for some M. Since T is smooth, so is , and hence for all . Hence, for each we have and for all . By Lemma 3.11 it follows that . Since s is an arbitrary node in , we have . This shows that . Finally, if is a limit ordinal and for all , it follows easily that for all , whence . This completes the proof.
Proof. By Lemma 3.9 we have and . On the other hand, so ; hence, by the previous lemma, so in particular , and hence . This completes the proof.
We now obtain the main result of this section, originally due to McGehee [18]:
Proof. Since is a countably infinite set, we may identify with . By Corollary 3.15 let T be a smooth well-founded tree of height . By Lemma 3.9 Z_{T} is weak-* dense in and by Corollary 3.18 we have .
Proof. This is immediate from Theorems 2.13 and 3.19. The result is originally due to Sarason [19,20,21] and McGehee [18].
This settles the question of closure ordinals of subspaces of , but what about ? It turns out that the answer is much simpler (see Corollary 3.23 below).
Proof. The equivalence of (1) and (2) follows from the well-known fact (Theorem V.3.13 in [11]) that a convex set is weakly closed if and only if it is norm closed. The implications (2) (3) (4) (1) are all trivial.
Proof. If X is reflexive then so is X^{*} (see Corollary II.3.24 in [11]), and hence the weak and weak-* topologies on X^{*}coincide. The result now follows immediately from Theorem 3.21.
We would like to thank Howard Becker for pointing this out to us.
We do not know whether the closure ordinal of a convex set in the dual of a separable Banach space can be a limit ordinal.