Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 11 / 4

Use a compatible complete metric on the Polish space $P$. For a tree on a Polish space $T\subseteq {\bigcup }_n{P}^n$, letbe its rank of a well-founded tree, with terminal rank zero. We bound this rank using a countable approximation tree, rather than assuming that the original tree has countably many nodes.

Choose a countable base of open balls. A node of the approximation tree $S$ at depth $n$ records, for every coordinate $i\le j\le n$, a basic ball $U_{j,i}$. Require its diameter to be less than $2^{-j}$, require $\overline{U_{j+1,i}}\subseteq U_{j,i}$, and requireInitial truncation removes the last row of balls. There are only countably many such finite records, so $S$ is countable.

An infinite branch in $S$ would give, by completeness and shrinking nested balls, a limit $p_i\in P$ for each coordinate. For fixed $j$, witnesses from later levels of $T$ have their first $j$ coordinates converging to $(p_1,\dots ,p_j)$. Initial-segment closure puts those witnesses in $T_j$, and closedness of $T_j$ puts the limit there too. Hence $(p_i)$ would be an infinite branch of $T$. Thus $S$ is well-founded.

Every tuple $t=(p_1,\dots ,p_n)\in T_n$ can be fitted inside such a record with each $p_i\in U_{n,i}$. Every extension $t^{⌢}p$ can be fitted in a child record: shrink the old balls around their actual points and choose a small ball around the new point. A well-founded induction therefore gives $r_T(t)\le r_S(s)$ whenever the record $s$ fits $t$. At the empty record this bounds the whole tree's rank by $r_S(\varnothing )$.

The rank of a countable well-founded tree is countable: at each node it is a supremum of countably many successors of countable ordinals, by well-founded recursion. Consequently a closed well-founded tree on a Polish space has countable height. The completeness and closedness assumptions are precisely what turn a branch of approximation records into an actual branch.

For $K\ge 1$, define the normalized ${\ell }^1$ treeThe upper bound follows from normalization. This is a closed tree: it suffices to impose the lower inequalities for a countable dense set of scalar tuples. The Bourgain l1-index iswhere $o$ is its well-founded tree order, and the index is ${\omega }_1$ if a tree is ill-founded. An infinite branch is exactly a sequence equivalent to the unit vector basis of ${\ell }^1$.

If $X$ is separable and contains no ${\ell }^1$, its unit sphere is Polish and every $T_K(X)$ is closed and well-founded. Part (a) makes each order countable, and their countable supremum is countable. In particular every separable reflexive Banach space has countable index, since its closed subspaces are reflexive whereas ${\ell }^1$ is not.

We now construct separable reflexive spaces with arbitrarily high index. Take any countable well-founded rooted tree $T$. Recursively, at each node $t$, defineAt a terminal node this is just $\mathbb{R}$. The recursion is legitimate by well-foundedness. Each $X_t$ is separable and reflexive: a countable ${\ell }^2$ sum of separable reflexive spaces is separable and reflexive, and adjoining one dimension in a finite direct sum preserves reflexivity. Let $X_T$ be the root space, and let $e_t$ be the unit vector at each node coordinate.

Along every finite branch $t_1\prec \cdots \prec t_n$, the recursive norm is exactlyIndeed, at every step only one child component is occupied, so its ${\ell }^2$ norm equals that component's norm, while the node coordinate adds by the ${\ell }^1$ sum. Thus $T$ embeds by extension into $T_1(X_T)$, and its rank is a lower bound for the index, up to the harmless empty-root convention. Countable well-founded trees have arbitrarily high countable ranks: add a root for a successor stage and attach countably many trees of cofinal ranks at a limit stage.

Suppose now a separable reflexive space $U$ were universal. Put $\beta =I_1(U)<{\omega }_1$ and choose $T$ of height larger than $\beta$. An isomorphic embedding $J:X_T\to U$ has bounds $c\Vert x\Vert \le \Vert Jx\Vert \le C\Vert x\Vert$. The normalized images $J{e}_t/\Vert J{e}_t\Vert$ along each branch satisfy the lower ${\ell }^1$ estimate $c/C$, so the tree embeds into $T_K(U)$ for some integer $K\ge C/c$. It would have order greater than $\beta$, contradicting its definition. Therefore no separable reflexive Banach space is universal for all separable reflexive Banach spaces. The embedding constant need not be uniform across spaces; the supremum over all $K$ handles this.