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We prove a weighted version of the Szemerédi regularity lemma, which is sufficient for the grid application. For disjoint nonempty vertex sets $U,W$ in a graph, write $d(U,W)=e(U,W)/(|U||W|)$ for their edge density of a bipartite graph. Call the pair $\epsilon$-regular if every $U^{\prime }\subseteq U$, $W^{\prime }\subseteq W$ with $|U^{\prime }|\ge \epsilon |U|$ and $|W^{\prime }|\ge \epsilon |W|$ satisfies $|d(U^{\prime },W^{\prime })-d(U,W)|\le \epsilon$.

The weighted Szemerédi regularity lemma states that, for every $0<\epsilon <1$, there are integers $M,v_0$ such that every graph on $v\ge v_0$ vertices has a set partition into at most $M$ nonempty cells $V_1,\dots ,V_k$, each of size at most $\epsilon v$, for whichThe sum is ordered. The cells need not have equal sizes; their weights replace the usual unweighted count of bad pairs.

To prove it, start with $k_0=\lceil 2/\epsilon \rceil$ nearly equal cells, taking $v\ge k_0$. Each has size at most $v/k_0+1\le \epsilon v$. Define the regularity energyHere even diagonal rectangles use the mean adjacency indicator, counting ordered pairs of vertices. Thus $0\le q\le 1$. Under a refinement, the old rectangle density is the weighted mean of the new ones, and the refinement variance identity for regularity energy givesIn particular refinement cannot decrease energy.

If the irregular-pair weight exceeds $\epsilon v^2$, choose witness subsets $U^{\prime }\subseteq U$, $W^{\prime }\subseteq W$ for every irregular unordered pair, and split each cell simultaneously by all its witnesses. There are at most $k-1$ such subsets in any one cell, so the new number of cells is at most $k{2}^k$. In an irregular rectangle, its witness has relative area at least ${\epsilon }^2$ and density discrepancy greater than $\epsilon$. Since the witness rectangle is now a union of refined rectangles, the Cauchy-Schwarz inequality applied to their weighted density discrepancies bounds the variance contribution from below bySumming over both orientations of all irregular pairs raises $q$ by more than ${\epsilon }^5$. Since $q\le 1$, after at most $\lceil {\epsilon }^{-5}\rceil$ rounds the required weight bound must hold. The finite recurrence $k\mapsto k{2}^k$, starting at $k_0$, supplies $M(\epsilon )$; all cells remain small because we only refine. This proves the stated weighted Szemerédi regularity lemma.

We next derive the triangle removal lemma from this version. First, three disjoint classes $U,W,Z$ whose pairs are $\epsilon$-regular pairs of density at least $d$ have many triangles if $\epsilon \le d/2$ and $\epsilon \le 1/4$. Fewer than $\epsilon |U|$ vertices of $U$ have fewer than $(d-\epsilon )|W|$ neighbours in $W$: otherwise that set of vertices, paired with all of $W$, would violate regularity. The same holds for $Z$. For each of the at least $(1-2\epsilon )|U|$ remaining vertices, both neighbour sets have relative size at least $d-\epsilon \ge \epsilon$. Regularity of $(W,Z)$ then supplies at least $(d-\epsilon )$ times the product of those two neighbour-set sizes edges. Thus the regular triangle counting lemma, also for unequal class sizes, givesFor a desired deletion tolerance $0<\eta <1$, take $\epsilon =\eta /100$, $d=\eta /4$, and a weighted regularity partition with at most $M=M(\epsilon )$ cells. Put $\tau =\epsilon /M$. Delete edges inside cells, between irregular pairs, between regular pairs of density less than $d$, and incident to cells smaller than $\tau v$. The first two deletions each cost at most $\epsilon v^2/2$. The sparse-pair deletion costs at most $d{v}^2/2$. For the last deletion, small-cell deletion in weighted regularity removes at most $M\tau v=\epsilon v$ vertices and at most $\epsilon v^2$ incident edges. The total is bounded byA triangle surviving these deletions would lie in three distinct cells, all of size at least $\tau v$, with all their pairs regular and of density at least $d$. The original graph would then have at least $d^3{\tau }^3{v}^3/16$ triangles. Therefore, witha graph on sufficiently many vertices with at most $\kappa v^3$ triangles can be made triangle-free by deleting fewer than $\eta v^2$ edges. This establishes the triangle removal lemma with every parameter depending only on $\eta$.

Now apply a tripartite graph encoding of a grid. Take separate vertex classes $X=[N]$, $Y=[N]$, $Z=\{2,\dots ,2N\}$. For each grid point $(x,y)$ in the given set, put in the three edgeswhere their endpoints belong respectively to $X\times Y$, $X\times Z$, and $Y\times Z$. Each point supplies a triangle. These $|A|$ triangles are edge-disjoint triangles, since any one of the three edge types uniquely determines the point that supplied it.

Conversely, any triangle with vertices $x\in X$, $y\in Y$, $z\in Z$ forces the three pointsto belong to $A$. Put $d=z-x-y$. For $d\ne 0$ these form a corner in an integer grid; for $d=0$ they all arise from the same grid point. Thus if there is no corner, the graph has exactly $|A|\le N^2$ triangles, yet destroying all triangles requires at least $|A|\ge \delta N^2$ edge deletions.

We may suppose $0<\delta \le 1$, since larger densities make the request vacuous. Use the triangle removal lemma with $\eta =\delta /32$. Our graph has $v=4N-1$ vertices, so, for sufficiently large $N$, its bound $N^2$ on the number of triangles is at most $\kappa v^3$. The lemma would delete fewer thanedges to destroy all triangles. This contradicts the lower bound $\delta N^2$ from the edge-disjoint family. Hence every sufficiently large positive-density square grid contains a nondegenerate corner, proving the requested corners theorem and in particular the existence of a suitable $N$.

The uniform modular Frankl-Wilson theorem states the following. Let $p$ be prime, let $L\subseteq {\mathbb{F}}_p$ have size $s$, and let $\mathcal{F}$ be a rank-$k$ family on $[N]$, with $0\le s\le \min \{k,N-k\}$. Assume $k\mathrm{mod}p\notin L$ and $|A\cap B|\mathrm{mod}p\in L$ for every distinct $A,B\in \mathcal{F}$. ThenThe restrictions concern distinct pairs; the diagonal size residue is explicitly excluded from $L$.

For each member $A$, over ${\mathbb{F}}_p$ form the intersection polynomialAt the characteristic vector $1_B$, its value vanishes when $B\ne A$ and is nonzero when $B=A$. Thus these functions are linearly independent on the whole rank-$k$ layer: evaluate any linear relation at each family member in turn. Replacing every positive power of $z_i$ by $z_i$ preserves all Boolean evaluations, so the functions are restrictions of multilinear polynomials of degree at most $s$.

We need the sharp low-degree evaluation rank on a uniform layer, not the larger whole-cube dimension ${\sum }_{j\le s}\left(\genfrac{}{}{0ex}{}{N}{j}\right)$. Let $M$ be the integer incidence matrix with rows indexed by subsets $S$ of size at most $s$, columns by rank-$k$ sets $B$, and entries $1_{S\subseteq B}$. Over the rationals, if $|S|=j\le s$, thenThis coefficient counts the possible $T$ inside each containing column set and is positive because $s\le k$. Consequently the degree-$s$ rows span all rows over $\mathbb{Q}$, giving rank at most $\left(\genfrac{}{}{0ex}{}{N}{s}\right)$. Every larger square minor is therefore the integer zero. Reducing those minors modulo $p$ shows the rank over ${\mathbb{F}}_p$ is also at most $\left(\genfrac{}{}{0ex}{}{N}{s}\right)$. This transfer avoids dividing by a binomial coefficient that might vanish modulo $p$. The independent functions $f_A$ lie in this row space, so their number is at most its rank. This proves the theorem in full.

For the requested forbidden-intersection application, split $\mathcal{A}$ into sets containing coordinate one and sets avoiding it. In the first part, delete one. Distinct original sets have intersection size between one and $2p-1$, excluding $p$. Their reduced intersections therefore lie between zero and $2p-2$, excluding $p-1$, so their residues belong to $L=\{0,\dots ,p-2\}$. The reduced set size is $2p-1$, whose residue is $p-1\notin L$. Frankl-Wilson on the ground set of size $4p-1$ bounds this part by $\left(\genfrac{}{}{0ex}{}{4p-1}{p-1}\right)$.

For the avoiding part, first complement every set. The complements contain one, retain size $2p$, and satisfyDelete one and apply the same argument. The complement splitting for forbidden midpoint intersections yields the stronger boundThe split is essential: a family may contain complementary pairs with intersection zero, preventing direct use of $L=\{1,\dots ,p-1\}$ on the original layer.

The Borsuk conjecture asserted that a bounded subset of ${\mathbb{R}}^d$ of positive diameter can be partitioned into $d+1$ subsets of strictly smaller diameter. Here is the Kahn-Kalai counterexample to the Borsuk conjecture, implemented by balanced cut vectors. For each $X\in [4p{]}^{(2p)}$, define $v_X\in {\mathbb{R}}^{\left(\genfrac{}{}{0ex}{}{4p}{2}\right)}$ whose coordinate indexed by an unordered pair $\{i,j\}$ is one when exactly one endpoint lies in $X$, and zero otherwise. Complementary sets give the same cut. Choose the representative with $1\in X$, obtaining exactlydistinct points: a cut determines its two parts, and fixing the side containing one removes its only ambiguity. Every cut has $(2p{)}^2=4p^2$ occupied coordinates, so all points lie in the affine hyperplane with coordinate sum $4p^2$. Its dimension isWe can identify this hyperplane isometrically with ${\mathbb{R}}^d$; no claim of full affine span is needed.

For two side sets with $t=|X\cap Y|$, their symmetric difference has $4p-2t$ points. The two cuts disagree exactly on pairs with one endpoint in that symmetric difference, giving the balanced cut-vector distance formulaThe diameter is $2p$, attained exactly when $t=p$; such pairs exist even among representatives containing one. Thus a smaller-diameter part contains no pair with side intersection $p$. Since all representative sets already contain one, deleting that common coordinate and applying the preceding Frankl-Wilson argument bounds each part by $\left(\genfrac{}{}{0ex}{}{4p-1}{p-1}\right)$. There is no extra factor two at this step. This is the cut construction described in the original Kahn-Kalai paper; the distance and part-size calculations above give the required justification directly.

An explicit choice is $p=13$. The point set lies in dimensionThe exact counts areTheir ratio requires at least $\lceil M/B\rceil =1562$ smaller-diameter parts, exceeding $d+1=1326$. In particular,Rescaling the point set to diameter one changes none of these partition counts. Therefore the Borsuk counterexample in dimension 1325 is fully explicit. This is a concrete dimension, not a claim that it is the smallest possible counterexample dimension.

Use binary encoding $b(S)={\sum }_{i\in S}{2}^{i-1}$ for vertices of the hypercube graph, and write $I_m=\{S:b(S)<m\}$. Let $e(\mathcal{A})$ count edges with both endpoints in a family. The exact edge-isoperimetric theorem for binary initial segments, due to Harper, Lindsey, Bernstein and Hart, iswhere $s_2(j)$ is the number of ones in the binary expansion. Equivalently,The equality for $I_m$ follows by assigning each internal edge to its larger endpoint: each one-bit of that endpoint can be cleared, giving an earlier endpoint still in $I_m$. This is the exact extremal statement, stronger than the logarithmic entropy bound.

We prove it by induction on $n$. The case $n=1$ follows by checking sizes zero, one and two. Fix a coordinate $i$ and delete it from the two sections ${\mathcal{A}}_0,{\mathcal{A}}_1$. The edge count splits asReplace each section by a binary initial segment of the same size. By induction its internal edges cannot decrease. The two replacement segments are nested, so their intersection size is the smaller section size, at least the previous intersection size. Thus this section compression in binary order cannot decrease $e(\mathcal{A})$.

Repeatedly perform any nontrivial section compression. The sum ${\sum }_{S\in \mathcal{A}}b(S)$ strictly decreases each time: in a fixed section, the original numerical order is exactly the numerical order after deleting its fixed coordinate. This nonnegative integer potential ensures termination at a family compressed in every coordinate section.

Classify such a terminal family. If an absent vertex $S$ precedes a present vertex $T$, they cannot agree in any coordinate, because their common section is compressed and its earlier vertex would then have to be present. Hence $T=[n]\setminus S$. There can be no vertex strictly between them: if present, it would also have to be the complement of $S$; if absent, it would have to be the complement of $T$. Thus they are consecutive complementary binary integers, forcingNo other inversion is possible for the same reason. Therefore a terminal family is either a binary initial segment or the sole exceptional formThis exception has size $2^{n-1}$. For $n\ge 2$, compared with the half-cube $I_{2^{n-1}}$, it removes a vertex of internal degree $n-1$ and adds a vertex joined only to the retained empty set. Its edge count is thus $e(I_{2^{n-1}})-(n-1)+1\le e(I_{2^{n-1}})$. The $n=1$ case was already handled. Since compression never decreased edges, every original family has at most the initial segment's edge count. This proves the theorem, including the exceptional terminal configuration rather than assuming all compressed families are initial segments.

For the remaining requests it is useful to obtain the entropy proof of cube edge-isoperimetry, including its equality cases. Inductively a family of size $m$ satisfiesIndeed let the two sections have sizes $a\ge b\ge 0$, $m=a+b>0$. Their internal edges are bounded inductively and the crossing edges are at most $b$, soHere $0{\log }_{2}0=0$ and $H_2(t)=-t{\log }_{2}t-(1-t){\log }_2(1-t)$ is the binary entropy function. Its concavity and its endpoint values at zero and one-half give $H_2(t)\ge 2t$ for $0\le t\le 1/2$. This proves the bound, starting from the zero-dimensional cube.

Define the isoperimetric number of a graph bySince every cube vertex has degree $n$, the internal-edge bound giveswhen $|A|\le 2^{n-1}$. A coordinate half-cube has one crossing edge per vertex, attaining ratio one. Hence, for $n\ge 1$,

Now establish the equality cases of entropy cube edge-isoperimetry. Strict concavity of $H_2$ makes $H_2(t)>2t$ for $0<t<1/2$. Equality in the internal-edge bound therefore requires either one section to be empty, or two equal-size sections. In the equal-size case equality in the crossing bound forces the sections to coincide; each section must also attain its inductive internal-edge bound. Induction gives exactly coordinate subcubes: the empty-section case fixes the coordinate, and the coincident-section case frees it. Conversely every coordinate subcube attains that bound.

At size $m=2^{n-1}$, an edge boundary of size $m$ means $e(A)=\frac{1}{2}m(n-1)=\frac{1}{2}m{\log }_{2}m$, so these equality cases apply. The subcube has dimension $n-1$ and fixes precisely one coordinate. ThereforeThese $2n$ coordinate half-cubes are the only extremizers of the requested size.

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.

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.