Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 11 (Topics in Banach Spaces)

The infinite Ramsey theorem used here says that every finite colouring of the $m$-element subsets of an infinite subset of $\mathbb{N}$ has an infinite homogeneous subset. Let $(x_n)$ be normalized with basis constant $K$.

For each positive integer $m$, rational scalar tuple $a=(a_1,\dots ,a_m)$ and precision $2^{-r}$, colour increasing $m$-tuples according to intervals of length $2^{-r}$ containing $\Vert {\sum }_i{a}_i{x}_{n_i}\Vert$. This is a finite colouring, because all these norms lie between zero and ${\sum }_i|a_i|$. Enumerate the countably many triples $(m,a,r)$ and repeatedly apply the Ramsey theorem, obtaining decreasing infinite sets on which each specified norm has oscillation at most $2^{-r}$. Choose a diagonal subsequence $(y_j)$ whose tail is contained in every previously selected infinite set.

Consequently, for every rational tuple $a$, the net of norms over increasing tuples from $(y_j)$ is Cauchy as its first index tends to infinity. Denote its limit by $N(a)$. This extends to arbitrary scalar tuples becauseuniformly in the indices. For complex scalars use rational real and imaginary parts. The same estimate proves convergence uniformly over all sufficiently late increasing tuples for each fixed scalar tuple.

Homogeneity and the triangle inequality pass to the limit. The coordinate functionals of a normalized basic sequence have norm at most $2K$, since the coordinate projection is the difference of two initial projections. Thusso $N$ is a genuine norm. Zero coefficients may be inserted anywhere without changing the value: removing those coordinates simply selects another increasing tuple for the same limiting combination. Also every initial-segment projection has norm at most $K$, by the corresponding inequality for $(y_j)$. Completing the finitely supported scalar sequences in $N$ therefore gives a Banach space with a normalized basic sequence $(e_i)$.

Its finite norms are invariant under increasing relabellings, by the zero-insertion observation, andThis is the required spreading model, generated by the diagonal subsequence.

Write $N(a_1,\dots ,a_m)$ for the limiting spreading model norm. We show that deleting any one coefficient does not increase it; iteration then gives every finite suppression.

Fix an index $r$, the coefficients, and a small $\eta >0$. Choose a generation threshold so that all the relevant full and shortened combinations beyond it approximate their model norms within $\eta$. Fix the indices before position $r$ beyond this threshold. Since $(x_n)$ is a weakly null sequence, the Mazur lemma gives a finite convex combinationwith every $j$ after the fixed prefix. Choose the remaining indices after $\max F$. For every $j\in F$, the full ordered combination with $x_j$ in slot $r$ has norm at most $N(a)+\eta$. Averaging these combinations givesThe shortened combination has norm at most $N(a)+(1+|a_r|)\eta$, and approximates its own model norm within $\eta$. Letting $\eta \downarrow 0$ provesfor every finite set $A$. Thus the model is a suppression-unconditional basic sequence with constant one. The convex-combination argument allows the omitted coordinate to sit between two retained coordinates; simply letting a single omitted index tend to infinity with all others fixed would not respect their order.

Choose a normalized basic sequence in the infinite-dimensional Banach space, and apply the Rosenthal l1 theorem. If a subsequence is equivalent to the unit vector basis of ${\ell }^1$, any spreading model generated from a further subsequence retains its two-sided ${\ell }^1$ estimates and is therefore unconditional.

Otherwise there is a weakly Cauchy sequence as a subsequence $(z_n)$, still basic. Its consecutive differences $d_n=z_{2n}-z_{2n-1}$ are weakly null and are a block sequence. The coordinate-functional bound for its normalized parent basis gives $1/(2K)\le \Vert d_n\Vert \le 2$. Hence $d_n/\Vert d_n\Vert$ is a normalized weakly null basic sequence. Part (a) gives a spreading model, and part (b) makes that model one-suppression-unconditional. Thus every infinite-dimensional Banach space has an unconditional spreading model, even though it need not contain an unconditional basic sequence.

An infinite-dimensional Banach space $X$ is a hereditarily indecomposable Banach space when no closed infinite-dimensional subspace of $X$ splits as a topological direct sum of two infinite-dimensional closed subspaces. Finite-dimensional summands are not forbidden.

A useful equivalent formulation is that for every pair of infinite-dimensional closed subspaces $U,V$ and every $\epsilon >0$, there are unit vectors $u\in U,v\in V$ with $\Vert u-v\Vert <\epsilon$. To see the equivalence, a topological direct sum has bounded coordinate projections and hence a positive separation between its two unit spheres. Conversely, if the two unit spheres have separation $\delta >0$, then $U\cap V=0$ and addition is bounded below on $U\oplus V$: after factoring out the larger of $\Vert u\Vert ,\Vert v\Vert$, the triangle inequality and the separation bound give a positive lower bound for $\Vert u+v\Vert /(\Vert u\Vert +\Vert v\Vert )$. Its range is therefore closed and supplies the forbidden decomposition.

For the construction, use the standard basic-sequence selection procedure. At stage $n$, choose finitely many functionals which almost norm the unit sphere of the finite-dimensional span of the vectors already chosen. Put the next vector in their common kernels, and in the analogous kernels retained from earlier stages. These impose only finite codimension. Choosing the norming tolerances with a finite product gives a uniformly bounded basis constant, which may be fixed in advance, for example below two. The hypothesis then supplies a unit vector in this finite-codimensional subspace with image smaller than the prescribed ${\epsilon }_n$. This yields the requested normalized basic sequence.

For strict singularity, fix $\eta >0$ and make the construction with a summable sequence satisfying $2K{\sum }_n{\epsilon }_n<\eta$, where $K$ is the controlled basis constant. For $z=\sum a_n{e}_n$, the coordinate bound $|a_n|\le 2K\Vert z\Vert$ givesBy density this holds on the infinite-dimensional closed span $E$.

If $T$ were bounded below by $c>0$ on some infinite-dimensional subspace $Y$, choose $\eta <c/2$. Hereditary indecomposability supplies unit vectors $y\in Y,e\in E$ arbitrarily close. Thena contradiction for sufficiently close vectors. Therefore $T$ is a strictly singular operator, equivalently every infinite-dimensional subspace contains a unit vector with arbitrarily small image. The case $T=0$ is immediate.

The Gowers dichotomy theorem states that every infinite-dimensional Banach space contains either an unconditional basic sequence or a hereditarily indecomposable subspace. Choose an initial basic sequence; it suffices to prove the result for its closed span $X$, with basis constant $K$.

We use the following precise finite-sequence form of the Gowers Ramsey theorem for Banach spaces. Let $\mathcal{A}$ be an analytic family of finite normalized block sequences, large in the sense that every infinite-dimensional block subspace contains a member. Given positive errors $\Delta =({\delta }_i)$, some infinite-dimensional block subspace $Y$ makes ${\mathcal{A}}_{\Delta }$ strategically large. Here ${\mathcal{A}}_{\Delta }$ consists of tuples within ${\delta }_i$ coordinatewise of a member of $\mathcal{A}$. In the block game the subspace player chooses an infinite-dimensional block subspace at each turn; the vector player chooses a successive unit vector in it and can force a finite initial segment into ${\mathcal{A}}_{\Delta }$. This is the Ramsey theorem permitted without proof, not the dichotomy conclusion.

Assume $X$ contains no unconditional basic sequence. For $\epsilon >0$, let ${\mathcal{A}}_{\epsilon }$ be the finite normalized block tuples $(x_1,\dots ,x_n)$ for which the odd and even spans contain unit vectors $u,v$ with $\Vert u-v\Vert <\epsilon$. This family is analytic: in each finite length the witness coefficients and vectors range over finite-dimensional scalar spaces, and the strict distance test is continuous; taking a countable union of finite-support choices gives the block family.

It is large. Indeed, every block subspace has a conditional block basis, with coordinate projections of unbounded norms. Given a finite combination $z$, choose a set of coordinates $A$ with $\Vert P_{A}z\Vert$ arbitrarily large relative to $\Vert z\Vert$. Put $u_0=P_{A}z$, $v_0=-P_{A^c}z$. Then $u_0-v_0=z$ and $|\Vert u_0\Vert -\Vert v_0\Vert |\le \Vert z\Vert$, soGroup consecutive nonzero terms belonging to the same side into a single block, and normalize each group. The groups alternate between the two sides, and their odd and even spans contain these two normalized vectors. Interchanging their names if necessary gives a member of ${\mathcal{A}}_{\epsilon }$.

Choose $\Delta$ with $8K{\sum }_i{\delta }_i<\epsilon$. A game tuple in ${\mathcal{A}}_{\epsilon ,\Delta }$ actually belongs to ${\mathcal{A}}_{2\epsilon }$. To check the perturbation explicitly, each coefficient of a unit vector in a normalized block tuple is bounded by $2K$. Replacing its coordinates by their nearby game vectors changes it by at most $2K\sum {\delta }_i$. Normalizing again changes the vector by at most twice that amount. Doing this for the two witnesses adds at most $8K\sum {\delta }_i$ to their distance.

The Ramsey theorem thus supplies a block subspace on which the vector player can force odd and even spans to contain unit vectors within $2\epsilon$. Given any two infinite-dimensional block subspaces $U,V$ of that space, let the subspace player choose them alternately, using their successive tails to satisfy support order. The strategy produces the odd vectors in $U$ and the even vectors in $V$, so these subspaces contain the required nearby unit vectors.

Repeat inside decreasing block subspaces $Y_n$ with ${\epsilon }_n\downarrow 0$, and choose a diagonal normalized block sequence $(z_n)$ with $z_n\in Y_n$. Its tail from $n$ onward lies in $Y_n$. In $Z=[z_n]$, any two infinite-dimensional block subspaces can be restricted to this tail, hence contain unit vectors at distance less than $2{\epsilon }_n$ for every $n$.

Finally this zero-angle statement extends to arbitrary infinite-dimensional closed subspaces $U,V$ of $Z$. Choose unit vectors in each subspace annihilating the first successively many basis coordinates, and approximate them by finite successive blocks with summably small errors. The coordinate bound $2K$ makes the resulting map from the block span to the original subspace arbitrarily close to the identity in operator norm. Apply the block-subspace zero-angle result and normalize the transported vectors; their added error can be made arbitrarily small. Thus the unit spheres of $U,V$ have distance zero.

By the equivalence proved in part (a), $Z$ is hereditarily indecomposable. This proves the dichotomy, including the passage from block subspaces to all subspaces.

The Rosenthal l1 theorem states that every bounded sequence in a real or complex Banach space has either a weakly Cauchy subsequence or a subsequence equivalent to the unit vector basis of ${\ell }^1$. Equivalence means that there are $0<c\le C<\infty$ such that $c\sum |a_i|\le \Vert \sum a_i{x}_{n_i}\Vert \le C\sum |a_i|$ for every finite scalar combination.

We use the infinite Ramsey result that every closed subset of $[\mathbb{N}{]}^{\omega }$ is a Ramsey set of infinite subsets. It follows by taking complements in the open-set theorem proved in Question 6(a), and is also among the infinite Ramsey results permitted here.

Let $E$ be the closed span of the sequence and $K=B_{E^{\ast }}$ with its weak-star topology. It is compact by the Banach-Alaoglu theorem. The functions $f_n(\varphi )=\varphi (x_n)$ are continuous and uniformly bounded. Pointwise convergence of a subsequence on $K$ is precisely the weakly Cauchy conclusion.

Choose a countable collection of pairs $(Q_0,Q_1)$ of disjoint closed rational boxes in the scalar plane, or intervals for real scalars. Write $u,v$ for their centres, $d=|u-v|>0$, and require their diameters to be at most some $\eta <d/(4\pi )$. Include arbitrarily small such boxes, so any two distinct scalar cluster points lie in the interiors of one such pair. For a fixed pair, say an infinite set $M$ oscillates if some $\varphi \in K$ has $f_n(\varphi )\in Q_0$ and $f_n(\varphi )\in Q_1$ each for infinitely many $n\in M$.

There are two possibilities. If at some pair and some infinite $M$ every infinite subset oscillates, retain them. Otherwise enumerate the box pairs and successively thin so that no point oscillates for the current pair. A diagonal subsequence is eventually in each refined set, so has no point oscillating for any pair. A bounded scalar sequence with two cluster points would oscillate for a pair of boxes enclosing those points. Hence every scalar evaluation on that subsequence converges, giving the weakly Cauchy alternative.

In the remaining case put $A_n=f_n^{-1}(Q_0)$ and $B_n=f_n^{-1}(Q_1)$, closed subsets of $K$. ConsiderThis is closed. If the infinite intersection is empty, compactness gives an empty finite intersection, and every infinite sequence with the same sufficiently long initial segment also has an empty intersection. The complement is therefore open. Apply closed-set Ramsey to obtain an infinite $L\subseteq M$ homogeneous for $\mathcal{C}$.

It cannot be homogeneous outside $\mathcal{C}$: the retained oscillation property provides a point visiting both level sets infinitely often on $L$, from which we can select an alternating sequence belonging to $\mathcal{C}$. Thus every infinite subset of $L$ is in $\mathcal{C}$.

Pass to the even-indexed members $L^{\prime }=(l_2,l_4,\dots )$. The pairs $(A_n,B_n)$ for $n\in L^{\prime }$ are independent: every assignment of either $A$ or $B$ to finitely many chosen indices has a common point. Indeed insert zero or one unused members of $L$ before and between the specified indices to arrange their required odd or even positions, then complete to an infinite sequence. The gaps between even-indexed members supply these spare indices. Membership of that infinite sequence in $\mathcal{C}$ supplies the point realizing the finite assignment.

Take any finite complex coefficients $a_i$. Some unit scalar $\zeta$ satisfiesbecause the average over the unit circle is exactly the right side. Put ${\epsilon }_i=\mathrm{sign}\mathrm{Re}(\zeta a_i)$, choosing either sign at zero. Independence supplies functionals $\varphi ,\psi \in K$ such that $f_{n_i}(\varphi )$ lies near $u$ and $f_{n_i}(\psi )$ near $v$ when ${\epsilon }_i=1$, with the assignments reversed when ${\epsilon }_i=-1$. ThusAlso $|\sum {\epsilon }_i{a}_i|\ge \sum |\mathrm{Re}(\zeta a_i)|$. Since $\Vert \varphi -\psi \Vert \le 2$, we concludeThe coefficient is positive by the box choice, and boundedness of the original sequence gives the upper estimate. This proves the ${\ell }^1$ alternative over complex scalars. For real scalars simply partition by the signs of the coefficients; the same calculation works and gives the stronger lower constant $d/2-\eta$. This completes the proof.

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.

The norm comparison in bounded distortion of a Banach space is understood up to a positive rescaling of the new norm. Thus, after replacing $D$ by $D^2$ if one's convention uses symmetric comparison constants, we may use the following form: for every equivalent norm $q$ on a block subspace, a further block subspace $Z$ and a number $a>0$ satisfyThe number $a$ may depend on the norm and the subspace; $D$ is fixed. Without this rescaling convention the premise would be impossible, since arbitrarily large scalar multiples of the original norm are equivalent norms. We prove the substantive, scale-invariant assertion for an infinite Schauder basis.

Let $K$ be the basis constant of the original basis. Every block sequence has basis constant at most $K$, so its initial, tail and finite interval projections have norms at most $K$, $1+K$ and $2K$, respectively. We shall construct increasingly complicated finite unconditional configurations while keeping their unconditional constant fixed at $C=2D$.

First construct the Maurey hierarchy of finite sets. Write ${\mathcal{M}}_0=\{\varnothing \}$, and definewhere prepending to the empty set is permitted. For each nonzero countable limit ordinal $\lambda$, fix positive ordinals ${\lambda }_j\uparrow \lambda$ and putInduction shows that every family is hereditary and spreading: subsets remain in the family, and increasing any of the coordinates preserves membership. Every nonzero stage contains all singletons. These families are also well-founded as extension trees. At a successor stage, an infinite branch would yield an infinite branch at the previous stage after deleting its first integer. At a limit stage, a hypothetical branch with first integer $m$ has every initial segment in the finite union ${\bigcup }_{j\le m}{\mathcal{M}}_{{\lambda }_j}$. One family then contains arbitrarily long initial segments, and heredity would give it an infinite branch.

For every infinite set $L$, the extension tree ${\mathcal{M}}_{\alpha }\cap [L{]}^{<\infty }$ has empty-root rank of a well-founded tree at least $\alpha$. Here is the induction proving this lower bound. At a successor stage fix $n\in L$; the tree above the node $\{n\}$ contains a copy of ${\mathcal{M}}_{\alpha }$ restricted to the infinite tail of $L$ after $n$, and hence has rank at least $\alpha$. The root therefore has rank at least $\alpha +1$. At a limit stage, for every $j$ the restriction of ${\mathcal{M}}_{{\lambda }_j}$ to a sufficiently late infinite tail of $L$ is contained in ${\mathcal{M}}_{\lambda }$. The root rank is consequently at least ${\sup }_j{\lambda }_j=\lambda$.

The crucial elementary estimate concerns signed interval projection norms. Given a hereditary family $\mathcal{F}$ containing all singletons and a basis $(w_i)$, define on finitely supported vectorsChoosing one interval containing the support gives $q_{\mathcal{F}}^{(w)}(x)\ge \Vert x\Vert$. Whenever this supremum is bounded by a multiple of $\Vert x\Vert$, it extends to a norm on the closed span giving equivalent norms with the original norm.

If $S={\sum }_i{\epsilon }_i{P}_{E_i}$ is one of the admitted signed interval projections, thenTo prove this, test $Sx$ with another admitted operator $R={\sum }_j{\delta }_j{P}_{F_j}$. The composition is the signed sum of projections onto the nonempty intervals $E_i\cap F_j$. Colour each intersection according to which of $E_i,F_j$ supplies its left endpoint, breaking ties in favour of $E_i$. In the first colour the list of minima is a subset of $(\min E_i)$, with no repetitions; in the second it is a subset of $(\min F_j)$. Both lists belong to $\mathcal{F}$ by heredity. Each colour therefore contributes an admitted signed interval projection, and the triangle inequality gives $\Vert RSx\Vert \le 2q_{\mathcal{F}}^{(w)}(x)$. Taking the supremum over $R$ proves the estimate.

Now suppose $q_{\mathcal{F}}^{(w)}$ is an equivalent norm on a block subspace $W$. Apply bounded distortion to obtain a further block basis $(z_i)$ spanning $Z$, with $a\Vert z\Vert \le q_{\mathcal{F}}^{(w)}(z)\le aD\Vert z\Vert$. A signed interval projection relative to $(z_i)$, with an admissible list of minima, is the restriction of an admitted signed interval projection relative to $(w_i)$. Indeed, replace each interval of block indices by the interval from the first to the last original coordinate of its blocks; its minimum only increases, so the spreading property preserves admissibility. The elementary estimate and the norm comparison therefore giveIn particular $q_{\mathcal{F}}^{(z)}\le C\Vert \cdot \Vert$, with the same $C=2D$ for every family to which we apply this procedure.

We prove by transfinite induction that, in every block subspace and for every $1\le \alpha <{\omega }_1$, there is a normalized block basis $(z_i)$ for whichFor $\alpha =1$ only one interval is permitted, so the initial signed interval projection norm is bounded by $2K\Vert x\Vert$. Apply the preceding renorming procedure.

For a successor stage, first choose $(w_i)$ satisfying the induction hypothesis for ${\mathcal{M}}_{\alpha }$. Delete the first interval from any ${\mathcal{M}}_{\alpha +1}$-admissible family. The remaining minima lie in ${\mathcal{M}}_{\alpha }$, soThis is an equivalent norm. The renorming procedure produces a further block basis with the constant reset to $C$.

At a limit stage $\lambda$, choose decreasing block subspaces $W_j$, each with a block basis satisfying the induction hypothesis for ${\mathcal{M}}_{{\lambda }_j}$. Choose successive normalized $z_j\in W_j$. Make each $z_j$ sufficiently far out that its first coordinate in the basis of every $W_k$, $k\le j$, is at least $j$; only finitely many restrictions are imposed at each choice. An admissible interval family for ${\mathcal{M}}_{\lambda }$ belongs to ${\mathcal{M}}_{{\lambda }_j}$ for some $j\le \min E_1$. All blocks on which it acts lie in the tail $[z_i:i\ge j]\subset W_j$. Transferring its intervals to the basis of $W_j$ increases their minima, so the induction hypothesis givesThus $q_{{\mathcal{M}}_{\lambda }}^{(z)}$ is again an equivalent norm, and one more application of bounded distortion resets its bound to $C$. This completes the transfinite induction. The tail-projection factor in this step is important; the diagonal tail need not have the same norm as the original vector.

Finally let $T_C(X)$ be the tree of finite normalized sequences $(x_1,\dots ,x_n)$ satisfyingfor all scalars and all signs. Each finite level is closed in the corresponding power of the unit sphere, since the norm inequalities are closed conditions. The unit sphere of the separable Banach space $X$ is a Polish space. An infinite branch is an unconditional basic sequence: sign changes give coordinate projections with norm at most $(1+C)/2$, imply linear independence, and give uniformly bounded initial projections, hence the basis property on the closed span. These projection bounds also give unconditional convergence, including over complex scalars.

For each countable $\alpha$, the block basis constructed above maps the extension tree ${\mathcal{M}}_{\alpha }$ into $T_C(X)$ by $\{i_1<\cdots <i_r\}\mapsto (z_{i_1},\dots ,z_{i_r})$. To check the sign inequality, use the admissible singleton intervals $\{i_1\},\dots ,\{i_r\}$ on the vector $\sum a_k{z}_{i_k}$. Thus the unconditional tree index is at least $\alpha$ for every countable ordinal. If $T_C(X)$ were well-founded, the closed-tree result would give it countable height, contradicting these lower bounds. It has an infinite branch. Therefore $X$ contains an unconditional basic sequence.

Identify $[\mathbb{N}{]}^{\omega }$ with increasing infinite sequences, with the product topology. A Ramsey set of infinite subsets $\mathcal{A}$ is one such that for every infinite $M$, some infinite $L\subseteq M$ satisfies $[L{]}^{\omega }\subseteq \mathcal{A}$ or $[L{]}^{\omega }\cap \mathcal{A}=\varnothing$. We prove this for open $\mathcal{A}$.

For a finite increasing stem $s$ and an infinite tail $B$ above it, say that $B$ accepts $s$ if every infinite $N\subseteq B$ has $s^{⌢}N\in \mathcal{A}$. Say that $B$ rejects $s$ if no infinite subset of $B$ accepts it. Either a tail has an accepting refinement, or it rejects. Both acceptance and rejection persist under passage to infinite subsets.

Starting with any infinite ground set $M$, use fusion to obtain an infinite $B$ deciding every finite subset of $B$: after selecting its first $n$ elements, refine the remaining tail successively to decide each of the finitely many subsets of that prefix. Then select the next element from the refined tail. The final tail beyond a decided stem is contained in the tail which decided it, so the decision survives. In particular $B$ decides the empty stem.

If $B$ accepts the empty stem, it is homogeneous inside $\mathcal{A}$. Otherwise it rejects. We thin it further so that every finite subset is rejected. The key observation is that if $B/s$ rejects $s$, there can be only finitely many $n\in B$ above $s$ for which $B/n$ accepts $s^{⌢}n$. If there were infinitely many, call their set $C$. For any infinite $N\subseteq C$, its first element $n$ and the accepting tail $B/n$ would imply $s^{⌢}N\in \mathcal{A}$. Thus $C$ would accept $s$, contradicting rejection.

Build $L$ inductively. Given a finite prefix whose subsets are all rejected, exclude the finitely many bad next elements for each of those subsets, and choose the next element outside their union. All new stems are decided by the first fusion and are not accepted, hence are rejected. This preserves the induction.

If an infinite $N\subseteq L$ belonged to the open set $\mathcal{A}$, some finite initial segment $s$ of $N$ would have every infinite extension above $\max s$ in $\mathcal{A}$. Then $B/s$ would accept $s$, contradicting the rejection just arranged. Therefore $[L{]}^{\omega }$ misses $\mathcal{A}$. We have provedTaking complements gives the same result for closed subsets. The proof also works with any fixed finite stem, yielding the stronger finite-stem version when desired.

LetThis scalar function is continuous. The first element is locally constant; deleting it is continuous on increasing sequences; and evaluation on a fixed $x_m$ is continuous in the weak-star topology. Thus $\mathcal{U}=\{M:|h(M)|<\delta \}$ is open.

By part (a), choose an infinite $L_0$ homogeneous for $\mathcal{U}$. If all its infinite subsets are in $\mathcal{U}$, the required conclusion follows. Suppose instead every one has $|h(M)|\ge \delta$. Choose finitely many unit scalars ${\zeta }_1,\dots ,{\zeta }_q$ such that every scalar $z$ with $|z|\ge \delta$ has $\mathrm{Re}({\zeta }_{j}z)>\delta /2$ for some $j$; for real scalars the two signs suffice. The open setscover $[L_0{]}^{\omega }$. Apply the open Ramsey set of infinite subsets result successively to them. After finitely many refinements, at least one, say ${\mathcal{U}}_j$, contains every infinite subset of a remaining infinite $L$; otherwise their cover would be violated.

For any finite $F\subseteq L$, choose an infinite $N\subseteq L$ after $\max F$. For each $m\in F$, apply this homogeneity to $M=\{m\}\cup N$. Its deleted tail is always the same $N$, so the same functional satisfiesBut $(x_m{)}_{m\in L}$ is a weakly null sequence. The Mazur lemma gives a finite convex combination $z={\sum }_{m\in F}{\lambda }_m{x}_m$ with $\Vert z\Vert <\delta /2$. The displayed inequalities give $\mathrm{Re}({\zeta }_j{x}_N^{\ast }(z))>\delta /2$, contradicting $\Vert x_N^{\ast }\Vert \le 1$. The bad homogeneous case is impossible. ThereforeContinuity is used to make the scalar tests open; weak nullity is used only for the finite convex-combination contradiction. The argument works over both real and complex scalars.