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Proof of the main result

In this section we show that a rather strong set existence axiom, $\Pi^1_1$ comprehension, is needed to prove a very elementary (indeed trivial-sounding) fact of separable Banach space theory. Namely, $\Pi^1_1$- $\mathsf{CA}_0$ is equivalent over $\mathsf{ACA}_0$ to the following statement (S):

\begin{displaymath}\vcenter{\hsize=4.5truein \baselineskip=14.4truept
\noindent ...
...
subspace $C$\space of $X^*$\space such that $Y\subseteq C$ .}
\end{displaymath}

This equivalence is the content of our main result, Theorem 5.6 below.

The forward direction is the assertion that (S) is provable in $\Pi^1_1$- $\mathsf{CA}_0$. This will be obtained as a special case of:

Lemma 5.1   The following is provable in $\Pi^1_1$- $\mathsf{CA}_0$. Let X be a separable Banach space. Given a countable set $Y\subset X^*$, there is a smallest weak-*-closed set $C\subseteq X^*$ such that $Y\subseteq C$.  

Proof. We reason in $\Pi^1_1$- $\mathsf{CA}_0$. For each $x_0^*\in X^*$ and each finite set $F\subset X$, there is a weak-*-open set

\begin{displaymath}U(x_0^*,F)=\{x^*\in
X^*\mid\forall x\in F\,(\vert x^*(x)-x_0^*(x)\vert<1)\}\,.\end{displaymath}

By $\Sigma^1_1$comprehension, let U be the weak-*-open set which is the union of the U(x0*,F) for all $x_0^*\in X^*$ and all finite $F\subset X$satisfying the arithmetical condition $Y\cap U(x_0^*,F)=\emptyset$. In terms of Definitions 4.2 and 4.13, the code of $U_n=U\cap B_n(X^*)$ for each $n\in{\mathbb{N} }$ is the union of the codes of $U(x_0^*,F)\cap B_n(X^*)$ for all $x_0^*\in X^*$ and all finite $F\subset X$ such that $Y\cap U(x_0^*,F)=\emptyset$. Clearly $C=X^*\setminus U$ is the smallest weak-*-closed set that includes Y. This completes the proof.

For the reversal, we must show that (S) implies $\Pi^1_1$- $\mathsf{CA}_0$. The standard method of proving that a mathematical statement implies $\Pi^1_1$- $\mathsf{CA}_0$ is to apply the following lemma [12,23]:

Lemma 5.2   It is provable in $\mathsf{RCA}_0$ that the following are equivalent:
1.
 $\Pi^1_1$- $\mathsf{CA}_0$;
2.
 For any sequence of trees $\langle{}T_n\mid n\in{\mathbb{N} }\rangle{}$, there exists a set $W\subseteq{\mathbb{N} }$ consisting of all $n\in{\mathbb{N} }$ such that Tn is well-founded.
 

Proof. The assertion that Tn is well-founded is $\Pi^1_1$ (using the sequence $\langle{}T_n\mid n\in{\mathbb{N} }\rangle{}$ as a parameter). The implication from (1) to (2) is therefore obvious. For the converse, reasoning in $\mathsf{RCA}_0$, first we show that (2) implies $\mathsf{ACA}_0$. It is well-known (see [23]) that $\mathsf{ACA}_0$ is equivalent over $\mathsf{RCA}_0$ to the statement that, for any one-to-one function $f\colon{\mathbb{N} }\to{\mathbb{N} }$, the range of f exists. Accordingly, let $f\colon{\mathbb{N} }\to{\mathbb{N} }$ be one-to-one. By recursive comprehension (using f as a parameter), we define a sequence  $\langle{}T_n\mid n\in{\mathbb{N} }\rangle{}$ by $s\in T_n$ if and only if $\forall
k<\mbox{\rm lh}(s)\,(n\ne f(k))$. Note that Tn is well-founded if and only if n is in the range of f. By (2), there is a set Wsuch that $n\in W$ if and only if Tn is well-founded, and so $n\in W$ if and only if n is in the range of f, i.e., the range of f exists, as desired. This proves $\mathsf{ACA}_0$.

Now, reasoning in $\mathsf{ACA}_0$, let $\varphi(n)$ be a $\Pi^1_1$ formula. By the Normal Form Theorem formalized within $\mathsf{ACA}_0$ (see [23]), we can write $\varphi(n)\equiv\forall f\,\exists m\,\theta(f[m],n)$, where $\theta$ is $\Sigma_0^0$. Use recursive comprehension to form a sequence of trees

\begin{displaymath}T_n=\{s\mid\forall
m\le\mbox{\rm lh}(s)\,\,\neg\theta(s[m],n)\}\,,\end{displaymath}

$n\in{\mathbb{N} }$. Note that, for all n, $\varphi(n)$ holds if and only if Tn is well-founded. By (2) there exists a set W consisting of all n such that Tn is well-founded. Thus for all $n\in{\mathbb{N} }$ we have $\varphi(n)$ if and only if $n\in W$. This proves $\Pi^1_1$comprehension. Thus we have the implication from (2) to (1). This completes the proof.

In showing that the implication from (S) to $\Pi^1_1$- $\mathsf{CA}_0$ is provable in $\mathsf{ACA}_0$, we shall want to know that some of our results from Section 3 are provable in $\mathsf{ACA}_0$. Our $\mathsf{ACA}_0$ version of part of Lemma 3.14 is:

Lemma 5.3   The following is provable in $\mathsf{ACA}_0$ (actually $\mathsf{WKL}_0$). Let $T\subseteq\mbox{\rm Seq}$ be a tree and let

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

be the upward closure of T under majorization. Then T is well-founded if and only if T* is well-founded.

Proof. We reason in $\mathsf{WKL}_0$. If T* is well-founded then obviously T is well-founded, since $T\subseteq T^*$. Suppose now that T* is not well-founded. Let f be a path through T*. By recursive comprehension form the tree $T_f=\{t\in T\mid t\,\underline\ll\,f[\mbox{\rm lh}(t)]\}$. Since f is a path through T*, we have that for each n there exists $t\in T_f$ such that $t\,\underline\ll\,f[n]$, and hence $\mbox{\rm lh}(t)=n$, so Tfis infinite. Thus Tf is a bounded infinite tree. By Bounded König's Lemma, Tf has a path. Here we are using the fact that Bounded König's Lemma is provable in $\mathsf{WKL}_0$ (see [12] and Lemma IV.1.4 of [23]). Since $T_f\subseteq T$, it follows that T has a path. This completes the proof.

Recall that a tree T is said to be smooth if T*=T. The previous lemma implies the following refinement of Lemma 5.2.

Lemma 5.4   It is provable in $\mathsf{ACA}_0$ (actually in $\mathsf{WKL}_0$) that the following are pairwise equivalent:
1.
 $\Pi^1_1$- $\mathsf{CA}_0$;
2.
 For any sequence of trees $\langle{}T_n\mid n\in{\mathbb{N} }\rangle{}$, there exists a set $W\subseteq{\mathbb{N} }$ consisting of all $n\in{\mathbb{N} }$ such that Tn is well-founded.
3.
 For any sequence of smooth trees $\langle{}T_n\mid n\in{\mathbb{N} }\rangle{}$, there exists a set $W\subseteq{\mathbb{N} }$ consisting of all $n\in{\mathbb{N} }$ such that Tn is well-founded.
 

Proof. The equivalence of (1) and (2) is Lemma 5.2. Given a sequence of trees $\langle{}T_n\mid n\in{\mathbb{N} }\rangle{}$, we can use recursive comprehension to form a sequence of smooth trees $\langle{}T_n^*\mid n\in{\mathbb{N} }\rangle{}$. By the previous lemma we have in $\mathsf{WKL}_0$ that for all n, Tn is well-founded if and only if Tn* is well-founded. The equivalence of (2) and (3) follows.

Our $\mathsf{ACA}_0$ version of Corollary 3.12 is:

Lemma 5.5   The following is provable in $\mathsf{ACA}_0$. 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. Then ZS is a weak-*-closed subspace of $\ell_1(\mbox{\rm Seq})=c_0^*(\mbox{\rm Seq})$.

Proof. To simplify notation, let us write $X=c_0(\mbox{\rm Seq})$, so that $X^*=\ell_1(\mbox{\rm Seq})$. Clearly ZS is a subspace of X*. In order to show that ZS is weak-*-closed, we shall first show that $Z_S\cap B_1(X^*)$ is weak-*-closed.

By arithmetical comprehension, there exists a sequence $\langle{}M_s\mid
s\in S\rangle{}$ where Ms is the least M such that $s^\smallfrown{}\langle m\rangle\in S$ for all $m\ge M$. Using $\langle{}M_s\mid
s\in S\rangle{}$ as a parameter, let $\varphi(z)$ be a $\Sigma^0_1$ formula asserting for $z\in X^*$the existence of $s\in S$ and $M\ge M_s$ such that

 \begin{displaymath}\left\vert{\frac1{s'}}z(s)-{\frac1{s}}\sum_{m<M}z(s^\smallfro...
...{})\right\vert>{\frac1{s^\smallfrown{}\langle{}M\rangle{}}}\,.
\end{displaymath} (1)

Since $\varphi(z)$ is $\Sigma^0_1$, it follows by standard $\mathsf{RCA}_0$ techniques (see Lemma II.5.7 of [23]) that there exists a relatively weak-*-open set $U\subseteq B_1(X^*)$ consisting of all $z\in B_1(X^*)$ such that $\varphi(z)$ holds.

We claim that $U=B_1(X^*)\setminus Z_S$. To see this, let $z\in B_1(X^*)$ be given. If $z\notin Z_S$, then for some $s\in S$ we have

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

whence (1) holds for all sufficiently large $M\ge M_s$, whence $z\in U$. Conversely, if $z\in U\cap Z_S$, then by (1) we have

\begin{eqnarray*}{\frac1{s^\smallfrown{}\langle{}M\rangle{}}}
&<& \left\vert{\fr...
...le& {\frac{\Vert z\Vert}{s^\smallfrown{}\langle{}M\rangle{}}}\,,
\end{eqnarray*}


whence $\Vert z\Vert>1$, i.e., $z\notin B_1(X^*)$, a contradiction. This proves the claim.

The claim implies that $Z_S\cap B_1(X^*)$ is weak-*-closed. By Corollary 4.15, it follows in $\mathsf{ACA}_0$ that ZS is weak-*-closed. This completes the proof.

We are now ready to prove our main result:

Theorem 5.6 (tex2html_wrap_inline$RCA_0$)   The following are pairwise equivalent:
1.
  $\Pi^1_1$- $\mathsf{CA}_0$.
2.
  For every separable Banach space X and countable set $Y\subseteq X^*$, there exists a smallest weak-*-closed set in X* containing Y.
3.
  For every separable Banach space X and countable set $Y\subseteq X^*$, there exists a smallest weak-*-closed convex set in X* containing Y.
4.
  For every separable Banach space X and countable set $Y\subseteq X^*$, there exists a smallest weak-*-closed subspace of X* containing Y;
5.
  For every countable set $Y\subseteq\ell_1=c_0^*$, there exists a smallest weak-*-closed set in $\ell _1$ containing Y.
6.
  For every countable set $Y\subseteq\ell_1=c_0^*$, there exists a smallest weak-*-closed convex set in $\ell _1$ containing Y.
7.
  For every countable set $Y\subseteq\ell_1=c_0^*$, there exists a smallest weak-*-closed subspace of $\ell _1$ containing Y.

Proof. The implication from (1) to (2) is just Lemma 5.1. The implications from (2) to (3) and (5) to (6) are straightforward by considering the countable set

\begin{displaymath}\left\{\sum_{i=0}^n q_i\,y^*_i\Bigm\vert
\sum_{i=0}^nq_i=1,\ ...
...mathbb{Q} },\ q_i>0,\ y^*_i\in Y,\ n\in{\mathbb{N} }\right\}\,.\end{displaymath}

The implications from (3) to (4) and (6) to (7) are straightforward by considering the countable set

\begin{displaymath}\left\{\sum_{i=0}^n q_i\,y^*_i\Bigm\vert
q_i\in{\mathbb{Q} },\ y^*_i\in Y,\ n\in{\mathbb{N} }\right\}\,.\end{displaymath}

The implications from (2) to (5) and (3) to (6) and (4) to (7) are trivial.

It remains to prove in $\mathsf{RCA}_0$ that (7) implies $\Pi^1_1$- $\mathsf{CA}_0$. First we show in $\mathsf{RCA}_0$ that (7) implies $\mathsf{ACA}_0$. Toward that end, let $f\colon{\mathbb{N} }\to{\mathbb{N} }$ be one-to-one; we want to show that the range of f exists. Using f as a parameter, let $Y=\{y_{\langle{}n,m\rangle{}}\mid\exists k\le
m\,(f(k)=n)\}\subset\ell_1$. By (7), let C be the smallest weak-*-closed subspace of $\ell _1$ containing Y. Claim: for all $n\in{\mathbb{N} }$, $y_{\langle{}n\rangle{}}\in C$ if and only if n is in the range of f. To see this, first suppose n=f(k) for some k. Then $y_{\langle{}n,m\rangle{}}\in Y$ for all $m\ge k$, and, since $y_{\langle{}
n,m\rangle{}}\to y_{\langle{}n\rangle{}}$ weak-* as $m\to\infty$, we have $y_{\langle{}n\rangle{}}\in C$. Conversely, suppose n is not in the range of f. Then $y_{\langle{}n,m\rangle{}}\notin Y$ for all $m\in{\mathbb{N} }$, and hence, for all $y_{\langle{}
k,m\rangle{}}\in Y$, $y_{\langle{}k,m\rangle{}}(\langle{}n\rangle{})=0$. Let C' be the set of all $y\in\ell_1(\mbox{\rm Seq})$ such that $y(\langle{}n\rangle{})=0$. Then C' is a weak-*-closed subspace of $\ell _1$ which contains Y (and hence contains C), but $y_{\langle{}n\rangle{}}\notin C'\supseteq C$. This proves the claim. Now, ` $y_{\langle{}n\rangle{}}\in C$' is $\Pi^0_1$ (as C is a code for a weak-*-closed set), whereas `n is in the range of f' is  $\Sigma^0_1$, so by recursive comprehension, the range of f exists, as desired. Thus (7) implies  $\mathsf{ACA}_0$.

So, reasoning within $\mathsf{ACA}_0$, assume (7). Instead of proving $\Pi^1_1$- $\mathsf{CA}_0$ directly, we shall prove the equivalent statement given by Lemma 5.4.

Let $\langle{}T_n\mid n\in{\mathbb{N} }\rangle{}$ be a sequence of smooth trees. By recursive comprehension, form the set

\begin{displaymath}T=\{\langle{}n\rangle{}^\smallfrown{}s\mid
n\in{\mathbb{N} },\ s\in T_n\}\,.\end{displaymath}

Using the notation of Definition 3.6, consider the countable set $Y=\{y_t\mid
t\notin T\}$. Note that $Y\subseteq Z_T$. By (7) there is a smallest weak-*-closed subspace C of $\ell_1(\mbox{\rm Seq})$ such that $C\supseteq Y$. Since the predicate $y\in C$ is arithmetical (in fact $\Pi^0_1$, using a code for C as a parameter), we can use arithmetical comprehension to form the set

\begin{displaymath}W=\{n\mid y_{\langle{}n\rangle{}}\in
C\}\,.\end{displaymath}

We claim that, for all $n\in{\mathbb{N} }$, Tn is well-founded if and only if $n\in W$.

To prove the claim, let n be such that Tn is well-founded. We shall argue by arithmetical transfinite induction on the well-founded tree Tn that $y_{\langle{}n\rangle{}^\smallfrown{}s}\in C$ for all $s\in\mbox{\rm Seq}$. (Note that arithmetical transfinite induction is available in $\mathsf{ACA}_0$; see Lemma V.2.1 of [23].) The base step consists of observing that $s\notin T_n$ implies $\langle{}n\rangle{}^\smallfrown{}s\notin T$ which implies $y_{\langle{}n\rangle{}^\smallfrown{}s}\in Y\subseteq C$. For the inductive step, let $s\in T_n$ be given such that $y_{\langle{}n\rangle{}^\smallfrown{}s^\smallfrown{}\langle{}m\rangle{}}\in
C$ for all $m\in{\mathbb{N} }$. Clearly $y_{\langle{}n\rangle{}^\smallfrown{}s^\smallfrown{}\langle{}m\rangle{}}$converges weak-* to $y_{\langle{}n\rangle{}^\smallfrown{}s}$ as $m\to\infty$. Since C is weak-*-closed, it follows that $y_{\langle{}n\rangle{}^\smallfrown{}s}\in C$. This gives the inductive step. Thus $y_{\langle{}n\rangle{}^\smallfrown{}s}\in C$ for all $s\in\mbox{\rm Seq}$. In particular $y_{\langle{}n\rangle{}}\in C$, i.e., $n\in W$. This proves half of the claim.

For the other half, let n be such that Tn is not well-founded. We shall show that $n\notin W$. Let f be a path through Tn. By recursive comprehension form the set

\begin{displaymath}S=\{\langle{}n\rangle{}^\smallfrown{}s\mid
f[\mbox{\rm lh}(s)]\,\underline\ll\,s\}\,.\end{displaymath}

Note that $\langle{}n\rangle{}\in S$. Since Tn is smooth, we have $S\subseteq T$, and hence $y_{\langle{}n\rangle{}}\notin Z_S\supseteq
Z_T\supseteq Y$. Moreover, for any $\langle{}n\rangle{}^\smallfrown{}s\in S$, we have $\langle{}n\rangle{}^\smallfrown{}s^\smallfrown{}\langle{}m\rangle{}\in S$ for all $m\ge f(\mbox{\rm lh}(s))$, whence by Lemma 5.5 ZS is a weak-*-closed subspace of X*. It follows that $Z_S\supseteq C$, and hence $y_{\langle{}n\rangle{}}\notin
C$, i.e., $n\notin W$. This completes the proof of the claim.

From our assumption (7) we have shown that for all sequences of smooth trees $\langle{}T_n\mid n\in{\mathbb{N} }\rangle{}$, there exists a set W consisting of all n such that Tn is well-founded. Hence by Lemma 5.4 we see that (7) implies $\Pi^1_1$comprehension, i.e., (1). This completes the proof of Theorem 5.6.

Theorem 5.7 (tex2html_wrap_inline$RCA_0$)   The following are pairwise equivalent:
1.
  $\Pi^1_1$- $\mathsf{CA}_0$.
2.
  For every separable Banach space X such that X* is also separable, and for every norm closed subspace $Z\subseteq X^*$, there exists a smallest weak-*-closed subspace of X* containing Z.
3.
  For every separable Banach space X such that X* is also separable, and for every weakly closed subspace $Z\subseteq X^*$, there exists a smallest weak-*-closed subspace of X* containing Z.
4.
  For every norm closed subspace $Z\subseteq\ell_1=c_0^*$, there exists a smallest weak-*-closed subspace of $\ell _1$ containing Z.
5.
  For every weakly closed subspace $Z\subseteq\ell_1=c_0^*$, there exists a smallest weak-*-closed subspace of $\ell _1$ containing Z.

Of course, in light of Theorem 3.21, with enough comprehension the equivalences (2) $\Leftrightarrow$(3) and (4) $\Leftrightarrow$(5) are trivial; however, we do not know the status of Theorem 3.21 in $\mathsf{RCA}_0$.

Proof. Let X be a separable Banach space, with $Z\subseteq X^*$ a norm closed subspace. In $\Pi^1_1$- $\mathsf{CA}_0$, since Z is norm closed there is a countable set $S\subset X^*$ such that Z is the norm closure of S(see [4], [7]). By Theorem 5.6, there is a smallest weak-* closed subspace of X* containing S, which must also be the smallest weak-* closed subspace of X*containing Z. Thus (1) implies (2). The implications (2) $\Rightarrow
$(4) and (3) $\Rightarrow
$(5) are trivial, and the implications (2) $\Rightarrow
$(3) and (4) $\Rightarrow
$(5) follow from the norm topology being stronger than the weak topology.

It remains only to prove in $\mathsf{RCA}_0$ that (5) $\Rightarrow
$(1). As in the proof of Theorem 5.6, we first show that (5) implies $\mathsf{ACA}_0$. So assume (5) and let $f\colon{\mathbb{N} }\to{\mathbb{N} }$be one-to-one; we want to show in $\mathsf{RCA}_0$ that the range of fexists. By recursive comprehension (using f as a parameter), define a tree T by

\begin{displaymath}T=\{\langle{}\rangle{}\}\cup\{\langle{}n\rangle{}^\smallfrown{}s\mid \forall k\le\mbox{\rm lh}(s)\,(f(k)\ne
n)\}.\end{displaymath}

Then ZT is a weakly closed subspace of $\ell_1=c_0^*$. By (5), let C be the smallest weak-* closed subspace of $\ell _1$ containing ZT. We claim that for all $n\in{\mathbb{N} }$, n is in the range of f if and only if $y_{\langle{}n\rangle{}}\in C$. To see this, first suppose n is in the range of f, say f(m)=n. We'll show by $\Pi^0_1$-induction (which is available in $\mathsf{RCA}_0$; see Corollary 3.10 in [23]) that for all $s\in\mbox{\rm Seq}$, $y_{\langle{}n\rangle{}^\smallfrown{}s}\in C$. Let $\varphi(i)$ be the $\Pi^0_1$ formula $\forall s\in\mbox{\rm Seq}\;(\mbox{\rm lh}(s)=m-i\to y_{\langle{}n\rangle{}^\smallfrown{}s}\in C)$. By the definition of T, if $\mbox{\rm lh}(s)\ge m$ then $\langle{}n\rangle{}\notin T$, whence $y_{\langle{}n\rangle{}^\smallfrown{}s}\in Z_T\subseteq C$, and so $\varphi(0)$holds. Now suppose $\mbox{\rm lh}(s)=m-i$, and $y_{\langle{}n\rangle{}^\smallfrown{}t}\in C$ for all $t\in\mbox{\rm Seq}$ with $\mbox{\rm lh}(t)=m-i+1$. Then $y_{\langle{}n\rangle{}^\smallfrown{}
s^\smallfrown{}\langle{}k\rangle{}}\in C$ for all $k\in{\mathbb{N} }$, and $y_{\langle{}n\rangle{}^\smallfrown{}
s^\smallfrown{}\langle{}k\rangle{}}\to y_{\langle{}n\rangle{}^\smallfrown{}s}$ weak-* as $k\to\infty$, so $y_{\langle{}n\rangle{}^\smallfrown{}s}\in C$ since C is weak-* closed. Thus $\varphi(i-1)$ implies $\varphi(i)$, so it follows by $\Pi^0_1$-induction that $\varphi(i)$ holds for all $i\in{\mathbb{N} }$. In particular, $\varphi(m)$holds, i.e., $y_{\langle{}n\rangle{}}\in C$. This proves one half of the claim.

Conversely, suppose n is not in the range of f; we want to show that $y_{\langle{}n\rangle{}}\notin
C$. Let $S=\{\langle{}
n\rangle{}^\smallfrown{}s\mid s\in\mbox{\rm Seq}\}$. Then $S\subseteq T$, whence $Z_S\supseteq Z_T$. It suffices therefore to show that ZS is weak-* closed, because then $Z_S\supseteq C$, and $y_{\langle{}
n\rangle{}}\notin Z_S$ since $\langle{}n\rangle{}\in S$. In fact, we'll show that $Z_S=\{z\in\ell_1(\mbox{\rm Seq})\mid\forall t\in S\,(z(t)=0)\}$, which is clearly weak-* closed. Obviously, if z(t)=0 for all $t\in S$, it follows from the definition of ZS that $z\in Z_S$. On the other hand, suppose $z\in Z_S$ and let $t\in S$ be given. Since $z\in
Z_S\subset Z_{t^\smallfrown{}s}$ for all $s\in\mbox{\rm Seq}$, it follows that $(1/{t'})z(t)
=(1/t)\sum_{m_1\in{\mathbb{N} }}z(t^\smallfrown{}\langle{}m_1\rangle{})$, which in turn equals $\sum_{m_1\in{\mathbb{N} }}(1/(t^\smallfrown{}\langle{}
m_1\rangle{}))\sum_{m_2\in{\mathbb{N} }}z(t^\smallfrown{}\langle{}m_1,m_2\rangle{})$, which equals

\begin{displaymath}\sum_{m_1,m_2\in{\mathbb{N} }}\frac1{t^\smallfrown{}\langle{}...
...n{\mathbb{N} }}z(t^\smallfrown{}\langle{}m_1,m_2,m_3\rangle{}),\end{displaymath}

and so on. Thus for all $k\in{\mathbb{N} }$ we have

\begin{displaymath}\frac1{t'}z(t)=
\sum_{s\in{\mathbb{N} }^k}\frac1{t^\smallfrow...
...thbb{N} }}z(t^\smallfrown{}s^\smallfrown{}\langle{}m\rangle{}).\end{displaymath}

From this it follows easily that

\begin{displaymath}\frac1{t'}\vert z(t)\vert\le\frac1{t^\smallfrown{}\langle{}0\rangle{}^k}\Vert z\Vert _1 \quad\mbox{for
all }k\in{\mathbb{N} },\end{displaymath}

where $\langle{}0\rangle{}^k$ is the sequence of length k, all of whose terms are 0. The right hand side of this inequality approaches 0 as $k\to\infty$. Thus z(t)=0. We have shown that $z\in Z_S$ if and only if z(t)=0 for all $t\in S$. This completes the proof of the claim. Since `n is in the range of f' is $\Sigma^0_1$ and ` $y_{\langle{}n\rangle{}}\in C$' is $\Pi^0_1$, it follows by recursive comprehension that the range of f exists. Thus (5) implies $\mathsf{ACA}_0$.

To prove that (5) implies $\Pi^1_1$- $\mathsf{CA}_0$, we observe that the subspace $Z_T\subseteq\ell_1$ in the proof of Theorem 5.6 is weakly closed, so we need only repeat the $\mathsf{ACA}_0$ part of that argument. This completes the proof of Theorem 5.7.


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