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Next: The weak-* topology in Z Up: Separable Banach space theory Previous: Banach space preliminaries

   
Trees and subspaces of $\ell _1$

In this section we prove the following result: For each countable ordinal $\alpha$, there exists a weak-*dense subspace Z of $\ell_1=c_0^*$ such that $\mbox{\rm ord}(Z)=\alpha+1$. Our proof uses some simple concepts and results concerning trees. We give a self-contained treatment of these auxiliary results.

Definition 3.1   Let $\mbox{\rm Seq}$ denote the set of finite sequences of natural numbers, i.e.,

\begin{displaymath}\mbox{\rm Seq}=\bigcup_{k=0}^\infty{\mathbb{N} }^k=\left\{\la...
...mathbb{N} },\ n_i\in{\mathbb{N} }\hbox{ for all
}i<k\right\}\,.\end{displaymath}

For $s=\langle{}n_0,\ldots,n_{k-1}\rangle{}\in\mbox{\rm Seq}$, we write

\begin{displaymath}s=\langle{}s(0),s(1),\ldots,s(\mbox{\rm lh}(s)-1)\rangle{}\,,\end{displaymath}

where $\mbox{\rm lh}(s)=k$ denotes the length of s, and s(i)=ni for all $i<\mbox{\rm lh}(s)$. In particular $\langle{}\rangle{}$ is the empty sequence, the unique sequence of length 0. For $s,t\in\mbox{\rm Seq}$ we denote by $s^\smallfrown{}t$ the concatenation of s and t, i.e., the sequence of length $\mbox{\rm lh}(s)+\mbox{\rm lh}(t)$ given by

\begin{displaymath}s^\smallfrown{}t=\langle{}
s(0),\ldots,s(\mbox{\rm lh}(s)-1),t(0),\ldots,t(\mbox{\rm lh}(t)-1)\rangle{}\,.\end{displaymath}

For $s,t\in\mbox{\rm Seq}$, we write $s\subseteq t$ to mean that s is an initial segment of t, i.e., $\mbox{\rm lh}(s)\le\mbox{\rm lh}(t)$ and, for all $i<\mbox{\rm lh}(s)$, s(i)=t(i). Given $s\in\mbox{\rm Seq}$, if $s\ne\langle\rangle$let

\begin{displaymath}s'=\langle s(0),s(1),\ldots,s(\mbox{\rm lh}(s)-2)\rangle\,,\end{displaymath}

i.e., s' is the initial segment of s of length $\mbox{\rm lh}(s)-1$. For $s=\langle{}\rangle{}$ we put $s'=\langle\rangle'=\langle\rangle$.

Definition 3.2   We define a tree to be a nonempty set $T\subseteq\mbox{\rm Seq}$ which is closed under taking initial segments, i.e., for all $s,t\in\mbox{\rm Seq}$, if $t\in T$ and $s\subseteq t$ then $s\in T$. If T is a tree and $s\in T$, we say that s is a node of T. If s is a node of Tsuch that $s^\smallfrown{}\langle n\rangle\notin T$ for all $n\in{\mathbb{N} }$, then s is called an end node; otherwise s is called an interior node of T. Given a tree T, a function $f:{\mathbb{N} }\to{\mathbb{N} }$is called a path through T if for all $n\in{\mathbb{N} }$ we have $f[n]\in T$, where $f[n]=\langle f(0),f(1),\ldots,f(n-1)\rangle$. A tree T is said to be well-founded if it has no path.

Definition 3.3   If T is a tree, let

\begin{displaymath}T'=\{t'\mid t\in T\} =\{\mbox{interior nodes of
}T\}\cup\{\langle{}\rangle{}\}\,.\end{displaymath}

Note that T' is a subtree of T. We define a transfinite sequence of subtrees $T^{(\alpha)}$ of T by

\begin{eqnarray*}T^{(0)} & = & T\,,\\
T^{(\alpha+1)} & = & (T^{(\alpha)})'\,,\\...
...elta}T^{(\alpha)}
\quad\mbox{for }\delta\mbox{ a limit ordinal.}
\end{eqnarray*}


Note that T is well-founded if and only if $T^{(\alpha)}=\{\langle{}\rangle{}\}$for some countable ordinal $\alpha$. The least such $\alpha$ is called the height of T, denoted h(T). Given a well-founded tree T, we define a function $h_T:\mbox{\rm Seq}\to\mbox{\rm Ord}\cup\{-1\}$ (where $\mbox{\rm Ord}$ denotes the set of countable ordinals) by hT(s)=-1 for $s\notin T$ and, for $s\in T$, hT(s)=the least $\alpha$ such that s is an end node of $T^{(\alpha)}$. In particular $h_T(\langle{}\rangle{})=h(T)$. Note that, for all $s\in T$, $h_T(s)=\sup\{h_T(s^\smallfrown{}\langle{}n\rangle{})+1\mid n\in{\mathbb{N} }\}$.

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.

Theorem 3.4   For any countable ordinal $\alpha$, we can construct a well-founded tree T such that $h(T)=\alpha$.  

Proof. We prove this by transfinite induction on $\alpha$. For $\alpha=0$ we have $h(\{\langle\rangle\})=0$. For successor ordinals, note that if $h(T)=\alpha$, then $h(T^+)=\alpha+1$ where $T^+=\{\langle\rangle\}\cup\{\langle 0\rangle^\smallfrown{}s\mid s\in T\}$. Suppose now that $\delta$ is a limit ordinal, say $\delta=\sup\{\alpha_n\mid n\in{\mathbb{N} }\}$, where $\alpha_n<\delta$ for all $n\in{\mathbb{N} }$. For each n let Tn be a tree of height $\alpha_n$, and put $T=\{\langle\rangle\}\cup\{\langle n\rangle^\smallfrown{}s\mid n\in{\mathbb{N} },\
s\in T_n\}$. Then T is well-founded and, for each $n\in{\mathbb{N} }$, $h_T(\langle n\rangle)=\alpha_n$. Thus $h(T)=\sup\{\alpha_n+1\mid
n\in{\mathbb{N} }\}=\delta$.

Definition 3.5   Fix an injection $\char93 :\mbox{\rm Seq}\to{\mathbb{N} }$ with the following properties:
1.
$\char93 (\langle{}\rangle{})=1$;
2.
  $s\subseteq t$ implies $\char93 (s)\le\char93 (t)$;
3.
  m<n implies $\char93 (s^\smallfrown{}\langle{}
m\rangle{})<\char93 (s^\smallfrown{}\langle{}n\rangle{})$.
Given $s\in\mbox{\rm Seq}$ we refer to $\char93 (s)$ as the Gödel number of s. To simplify notation in what follows, we shall often identify sequences with their Gödel numbers, i.e., we write s instead of $\char93 (s)$.

Let $X=c_0(\mbox{\rm Seq})$, the space of sequences of real numbers converging to 0 indexed by $\mbox{\rm Seq}$. Then we may identify X* with $\ell_1(\mbox{\rm Seq})$, the space of absolutely summable sequences of real numbers indexed by $\mbox{\rm Seq}$. In the rest of this section we shall mainly be interested in the weak-* topology on $\ell_1(\mbox{\rm Seq})=c_0^*(\mbox{\rm Seq})$.

Definition 3.6   For each $s\in\mbox{\rm Seq}$ we define a distinguished point $y_s\in\ell_1(\mbox{\rm Seq})$ by

\begin{displaymath}y_s(t)=\left\{
\begin{array}{ll}
\char93 (t')&\mbox{if }t\subseteq s,\\
0&\mbox{otherwise.}
\end{array}\right.\end{displaymath}

Using the convention that sequences are to be identified with their Gödel numbers, we can write ys(t)=t' if $t\subseteq s$, 0otherwise.  Note that, for all $s\in\mbox{\rm Seq}$, the sequence $\langle{}y_{s^\smallfrown{}\langle{}n\rangle{}}\colon
n\in{\mathbb{N} }\rangle{}$, converges weak-* to ys in $\ell_1(\mbox{\rm Seq})$.

Definition 3.7   Given $s\in\mbox{\rm Seq}$ we set

\begin{displaymath}Z_s=\Bigl\{z\in\ell_1(\mbox{\rm Seq})\Bigm\vert\frac1{s'}z(s)...
...{m\in{\mathbb{N} }}z(s^\smallfrown{}\langle m\rangle)\Bigr\}\,.\end{displaymath}

If $S\subseteq\mbox{\rm Seq}$, let $Z_S=\bigcap_{s\in S}Z_s$. Note that ZS is a norm closed subspace of $\ell_1(\mbox{\rm Seq})$. Note also that $y_s\in Z_S$ if and only if $s\notin S$.

The next two lemmas imply that, for any well-founded tree T, ZTis weak-* dense in $\ell_1(\mbox{\rm Seq})$.

Lemma 3.8   If T is a well-founded tree, then $y_s\in
Z_T^{(h_T(s)+1)}$ for all $s\in\mbox{\rm Seq}$.

Proof. We proceed by induction on hT(s). If hT(s)=-1 then $s\notin T$ and $y_s\in Z_T=Z_T^{(0)}$. Suppose now that $h_T(s)=\alpha\ge0$, and that the theorem holds for all t such that $h_T(t)<\alpha$. Then for each $n\in{\mathbb{N} }$, $h_T(s^\smallfrown{}\langle{}n\rangle{})<\alpha$, so $y_{s^\smallfrown{}\langle{}n\rangle{}}\in
Z_T^{(\alpha)}$ for all $n\in{\mathbb{N} }$. Since ys is the weak-* limit of the sequence $\langle{}y_{s^\smallfrown{}\langle{}n\rangle{}}\colon
n\in{\mathbb{N} }\rangle{}$, it follows that $y_s\in Z_T^{(\alpha+1)}$, as desired.

Lemma 3.9   If T is a well-founded tree, then ZT is weak-* dense in $\ell_1(\mbox{\rm Seq})$; in fact, $Z_T^{(h(T)+1)}=\ell_1(\mbox{\rm Seq})$.  

Proof. Let $z\in\ell_1(\mbox{\rm Seq})$ be given. We can write $z=\sum_{s\in\mbox{\rm Seq}}z(s)\chi_s$, where $\chi_s\in\ell_1(\mbox{\rm Seq})$ is the characteristic function of $\{s\}$, i.e., $\chi_s(t)=1$ if t=s, 0otherwise. Note that if $s\ne\langle\rangle$ then $y_s-y_{s'}=s'\chi_s$, whereas $y_{\langle\rangle}=\chi_{\langle\rangle}$. Also, by the previous lemma, if $s\ne\langle\rangle$ then $y_s\in Z_T^{(h(T))}$, so, if $\mbox{\rm lh}(s)>1$ then $\chi_s\in Z_T^{(h(T))}$. Since z is an absolutely summable series, we have

\begin{eqnarray*}\sum_{s\in\mbox{\rm Seq}}z(s)\chi_s
&=&z(\langle\rangle)y_{\lan...
...ngle)y_{\langle m\rangle}+\sum_{\mbox{\rm lh}(s)>1}z(s)\chi_s\,.
\end{eqnarray*}


Now, $z(\langle\rangle)-\sum_{m\in{\mathbb{N} }}z(\langle m\rangle)$ is just a real number since z is absolutely summable, and $y_{\langle\rangle}$ is the weak-* limit of the sequence $\langle{}y_{\langle{}m\rangle{}}\colon m\in{\mathbb{N} }\rangle{}$, so the first term in this last sum is in ZT(h(T)+1). Likewise $\sum_{m\in{\mathbb{N} }}z(\langle m\rangle)y_{\langle
m\rangle}+\sum_{\mbox{\rm lh}(s)>1}z(s)\chi_s$, viewed as a series in $\ell_1(\mbox{\rm Seq})$, converges in norm, and hence converges weak-*; since the functionals $y_{\langle m\rangle}$ and $\chi_s$ are in ZT(h(T)) for all $m\in{\mathbb{N} }$ and all $s\in\mbox{\rm Seq}$ with $\mbox{\rm lh}(s)>1$, it, too, is the weak-* limit of a sequence from ZT(h(T)). Hence z is the sum of two weak-* limits of sequences from ZT(h(T)), whence z is in ZT(h(T)+1) as desired.

Corollary 3.10   If T is a well-founded tree, then $\mbox{\rm ord}(Z_T)\le h(T)+1$.

Thus we have an upper bound on the closure ordinal of ZT 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 $Z_T^{(\alpha)}$.

Lemma 3.11   Suppose that $z_k\to z$ weak-* in $\ell_1(\mbox{\rm Seq})=c_0^*(\mbox{\rm Seq})$. Let $s\in\mbox{\rm Seq}$ be given and suppose that $z_k\in Z_s$ for all $k\in{\mathbb{N} }$. A sufficient condition for $z\in Z_s$is the existence of $M\in{\mathbb{N} }$ such that $z_k\in Z_{s^\smallfrown{}\langle
m\rangle}$ for all $k\in{\mathbb{N} }$ and all $m\ge M$.  

Proof.

Assume that the stated condition holds. Suppose for a contradiction that $z\notin Z_s$, say

\begin{displaymath}\left\vert{\frac1{s'}}z(s)-{\frac1{s}}\sum_{m=0}^\infty z(s^\smallfrown{}\langle{}
m\rangle{})\right\vert>\varepsilon>0\,.\end{displaymath}

Then for all sufficiently large M we have

\begin{displaymath}\left\vert{\frac1{s'}}z(s)-{\frac1{s}}\sum_{m<M}z(s^\smallfrown{}\langle{}
m\rangle{})\right\vert>\varepsilon\,.\end{displaymath}

Fix such an M, with $z_k\in Z_{s^\smallfrown{}\langle
m\rangle}$ for all $k\in{\mathbb{N} }$ and all $m\ge M$ as well. Then for all sufficiently large k we have

\begin{displaymath}\left\vert{\frac1{s'}}z_k(s)-{\frac1{s}}\sum_{m<M}z_k(s^\smallfrown{}\langle{}
m\rangle{})\right\vert>\varepsilon\,,\end{displaymath}

and hence

\begin{eqnarray*}\varepsilon
& <&\left\vert{\frac1{s}}\sum_{m\ge M}z_k(s^\smallf...
...\frac{\Vert z_k\Vert}{s^\smallfrown{}\langle{}M\rangle{}}}\,,\\
\end{eqnarray*}


and hence $\Vert z_k\Vert>(s^\smallfrown{}\langle{}M\rangle{})\varepsilon$. Thus $\{\Vert z_k\Vert\mid
k\in{\mathbb{N} }\}$is unbounded, contradicting Corollary 2.3.

In particular, we have the following result.

Corollary 3.12   Let $S\subseteq\mbox{\rm Seq}$ be such that, for each $s\in S$, $s^\smallfrown{}\langle m\rangle\in S$ for all but finitely many $m\in{\mathbb{N} }$. Then ZS is weak-* closed.  

Proof. By Lemma 3.11 ZS is weak-* sequentially closed. Hence by Corollary 2.11 ZS is weak-* closed.

In order to make use of this lemma, we consider a special class of trees known as smooth trees:

Definition 3.13   For $s,t\in\mbox{\rm Seq}$ we say s is majorized by t, written $s\,\underline\ll\,t$, if $\mbox{\rm lh}(s)=\mbox{\rm lh}(t)$and $s(i)\le t(i)$ for all $i<\mbox{\rm lh}(s)$. For any tree T, we define T* to be the upward closure of T under majorization, i.e.,

\begin{displaymath}T^*=\{t\in\mbox{\rm Seq}\mid\exists s\in T(s\,\underline\ll\,t)\}\,.\end{displaymath}

A tree T is said to be smooth if it is upward closed under majorization, i.e., T*=T.

Lemma 3.14 (Marcone [15,16])   Let T be a tree. Then T is well-founded if and only if T* is well-founded, in which case h(T)=h(T*).  

Proof. Note first that $T\subseteq T^*$, so if T* is well-founded then so is T. Conversely, suppose T* has a path f; let $T_f=\{s\in T\mid s\,\underline\ll\,f[\mbox{\rm lh}(s)]\}$. Then Tf is a finitely-branching subtree of T, and, since f is a path through T*, Tf must be infinite. Hence by König's Lemma Tf has a path, whence T has a path.

Assuming T and T* are well-founded, we obviously have $h(T)\le
h(T^*)$. For the opposite inequality, we claim that for all s, $h_{T^*}(s)=\max\{h_T(t)\mid t\,\underline\ll\,s\}$. (Note that $\{t\mid t\,\underline\ll\,s\}$ is a finite set, so we may take max rather than sup.) We prove the claim by induction on hT*(s). If hT*(s)=-1 then $s\notin T^*$, so for any t with $t\,\underline\ll\,s$we have $t\notin T$, whence hT(t)=-1 for all such t. Otherwise $s\in T^*$ and we have

\begin{eqnarray*}h_{T^*}(s)
&=&\sup\{h_{T^*}(s^\smallfrown{}\langle n\rangle)+1\...
...nderline\ll\,s\}\\
&=&\max\{h_T(t)\mid t\,\underline\ll\,s\}\,.
\end{eqnarray*}


This proves our claim. In particular $h(T^*)=h_{T^*}(\langle{}\rangle{})=h_T(\langle{}\rangle{})=h(T)$and the proof of the lemma is complete.

Corollary 3.15   For any countable ordinal $\alpha$, there exists a smooth well-founded tree T such that $h(T)=\alpha$.  

Proof. This follows immediately from Theorem 3.4 and the previous lemma.

Lemma 3.16   If T is a smooth tree, then $T^{(\alpha)}$ is smooth for all $\alpha$.

Proof. We proceed by induction on $\alpha$. For $\alpha=0$ there's nothing to prove. Assume $T^{(\alpha)}$ is smooth, and let $s\in T^{(\alpha+1)}$ be given. Suppose $s\,\underline\ll\,t$; since $s\in T^{(\alpha+1)}\subseteq T^{(\alpha)}$ which is smooth, tmust be in $T^{(\alpha)}$; furthermore, since s is an interior node of $T^{(\alpha)}$, there is an $m\in{\mathbb{N} }$ such that $s^\smallfrown{}\langle
m\rangle\in T^{(\alpha)}$. But $s^\smallfrown{}\langle m\rangle\,\underline\ll\,
t^\smallfrown{}\langle m\rangle$, so $t^\smallfrown{}\langle m\rangle\in
T^{(\alpha)}$, whence $t\in T^{(\alpha+1)}$. Finally, smoothness is clearly preserved under intersections, so the induction goes through at limit stages.

Lemma 3.17   Let T be a smooth well-founded tree. Then for all $\alpha\le h(T)$, $Z_T^{(\alpha)}\subseteq Z_{T^{(\alpha)}}$.

Proof. We proceed by induction on $\alpha$. If $\alpha=0$there's nothing to prove. Assume $Z_T^{(\alpha)}\subseteq Z_{T^{(\alpha)}}$ and $\alpha<h(T)$ and let $z\in Z_T^{(\alpha+1)}$ be given. Then z is the weak-* limit of some sequence $\langle{}z_k\mid
k\in{\mathbb{N} }\rangle{}$ from $Z_T^{(\alpha)}\subseteq Z_{T^{(\alpha)}}$. Since $\alpha<h(T)$, we have that any $s\in T^{(\alpha+1)}$ is an interior node of $T^{(\alpha)}$, i.e., $s\in T^{(\alpha)}$ and $s^\smallfrown{}\langle
M\rangle\in T^{(\alpha)}$ for some M. Since T is smooth, so is $T^{(\alpha)}$, and hence $s^\smallfrown{}\langle
m\rangle\in T^{(\alpha)}$for all $m\ge M$. Hence, for each $k\in{\mathbb{N} }$ we have $z_k\in Z_s$ and $z_k\in Z_{s^\smallfrown{}\langle
m\rangle}$ for all $m\ge M$. By Lemma 3.11 it follows that $z\in Z_s$. Since s is an arbitrary node in $T^{(\alpha+1)}$, we have $z\in Z_T^{(\alpha+1)}$. This shows that $Z_T^{(\alpha+1)}\subseteq Z_{T^{(\alpha+1)}}$. Finally, if $\delta$ is a limit ordinal $\le h(T)$ and $Z_T^{(\alpha)}\subseteq Z_{T^{(\alpha)}}$ for all $\alpha<\delta$, it follows easily that $Z_{T^{(\alpha)}}\subseteq Z_{T^{(\delta)}}$ for all $\alpha<\delta$, whence $Z_T^{(\delta)}\subseteq
Z_{T^{(\delta)}}$. This completes the proof.

Corollary 3.18   If T is a smooth well-founded tree, then $\mbox{\rm ord}(Z_T)=h(T)+1$.  

Proof. By Lemma 3.9 we have $Z_T^{(\mbox{\rm ord}(Z_T))}=\ell_1(\mbox{\rm Seq})$ and $\mbox{\rm ord}(Z_T)\le h(T)+1$. On the other hand, $T^{(h(T))}=\{\langle\rangle\}$ so $y_{\langle\rangle}\notin Z_{T^{(h(T))}}$; hence, by the previous lemma, $y_{\langle\rangle}\notin Z_{T^{(h(T))}}$ so in particular $Z_T^{(h(T))}\ne\ell_1(\mbox{\rm Seq})$, and hence $h(T)<\mbox{\rm ord}(Z_T)$. This completes the proof.

We now obtain the main result of this section, originally due to McGehee [18]:

Theorem 3.19   For any countable ordinal $\alpha$, there exists a weak-* dense subspace Z of $\ell_1=c_0^*$such that $\mbox{\rm ord}(Z)=\alpha+1$.  

Proof. Since $\mbox{\rm Seq}$ is a countably infinite set, we may identify $\ell_1=c_0^*$ with $\ell_1(\mbox{\rm Seq})=c_0^*(\mbox{\rm Seq})$. By Corollary 3.15 let T be a smooth well-founded tree of height $\alpha$. By Lemma 3.9 ZT is weak-* dense in $\ell _1$ and by Corollary 3.18 we have $\mbox{\rm ord}(Z_T)=h(T)+1=\alpha+1$.

Corollary 3.20   The ordinals which can occur as closure ordinals of subspaces of the dual of a separable Banach space are precisely the countable non-limit ordinals.

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 $\ell_1=c_0^*$, but what about $\ell_2$? It turns out that the answer is much simpler (see Corollary 3.23 below).

Theorem 3.21   Let X be a Banach space, and let $C\subseteq X$ be convex. Then for any $x\in X$, the following conditions are equivalent:
1.
  x is in the weak closure of C.
2.
  x is in the norm closure of C.
3.
  x is the norm limit of a sequence of points in C.
4.
  x is the weak limit of a sequence of points in C.
 

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) $\Rightarrow
$(3) $\Rightarrow
$(4) $\Rightarrow
$(1) are all trivial.

Corollary 3.22   If X is a reflexive Banach space and $C\subseteq X^*$ is convex, then the weak-* sequential closure ordinal of C is 0 if C is weak-* closed, and is 1 otherwise.

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.

Corollary 3.23   If $C\subseteq\ell_2$ is convex, then the weak-* (i.e., weak) sequential closure ordinal of C is 0 if C is weak-*(i.e., weakly) closed, 1 otherwise.  

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.


next up previous
Next: The weak-* topology in Z Up: Separable Banach space theory Previous: Banach space preliminaries
Stephen G Simpson
1998-10-25