... are all isomorphic to $l^1$ on some other index set. At least, that much I "know" from 2nd-hand sources, since the original proof is apparently in a paper of Köthe from the ~~1930s~~ 1960s (in German) that I ~~can't get hold of~~ have had trouble digesting. Since there are some Banach space specialists reading MO, I wondered if someone could sketch how the proof differs from the countable case, or point to a more recent text, preferably in English or French, that gives the proof? I hope this is a well-defined question for MO, since I'm not baiting with something where I know the answer.

(Some background for other readers: the analogous result when $X$ is countably infinite follows from combining two steps: one first uses a block basis argument to show that a closed, complemented subspace $V$ inside $l^1(\bf N)$ must either be finite-dimensional, or contain an infinite-dimensional, closed complemented subspace $W$ that is isomorphic to $\ell^1({\bf N})$. In the former case, $V$ is then obviously isomorphic to some finite-dimensional $\ell^1$. In the latter case, one applies Pelczynski decomposition. My impression is that it's the *first step* which might prove problematic if attempted for $\ell^1(X)$ when $X$ is uncountably infinite, but I could well be wrong and would welcome corrections.)