TITLE: Closed, complemented subspaces of $l^1(X)$ when $X$ is uncountable
QUESTION [6 upvotes]: ... 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.)

REPLY [7 votes]: Nice answer, Phil.
The case  of  $Z=\ell_p(X)$, 1 < p < infinity, and $Z=c_0(X)$ with $X$ uncountable are a bit easier because if $T$ is a bounded linear operator from $Y$ into $Z$ and the density character of $Y$ is smaller than $|X|$, then $T$ cannot be one to one.  So if $Y$ is a subspace of $Z$ with density character $|X|$, then a maximal set of disjointly supported unit vectors in $Y$ has cardinality $|X|$ (use  the obvious fact that a subspace $W$ of $Z$ is contained in $\ell_p(Q)$
[or $c_0(Q)$] with $|Q|$ equal to the density character of $W$).  So every subspace of $Z$ with density character $|X|$ has a norm one complemented subspace that is isometric to $Z$.  This and a second use of the ``obvious fact" gives the desired result.