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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.