 Solution

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.