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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.