Predual of $l^1(\Gamma)$ Let $\Gamma$ be an uncountable index set. For example $\Gamma=\mathbb R$. Let 
$l^1(\Gamma)$ be the set of functions with countable support and
finite sum:
$$
\sum_{a\in\Gamma}|f(a)|<\infty.
$$
The space $l^1(\Gamma)$ is a Banach space with the norm $\|f\|:=\sum_{a\in\Gamma}|f(a)|$. It is not separable.
My question is: does $l^1(\Gamma)$ have a separable pre-dual space?
I have seen statements that $l^1(\Gamma)$ is isometric to the dual space of $c_0(\Gamma)$,
however I did not found a definition of $c_0(\Gamma)$ nor a statement about its separability.
The usual Krein-Milman based argument does not fail as in the $L^1$ case (the unit ball of $l^1(\Gamma)$ is the closed convex hull of its extreme points).
 A: We have 
$$ c_0(\Gamma) := \{x \in \mathbf K^\Gamma \mid \forall \epsilon > 0 \,\exists \Gamma' \subseteq \Gamma, |\Gamma'| < \infty \land \sup_{\gamma \in \Gamma - \Gamma'} |x(\gamma)| < \epsilon \} $$
which is a Banach space with respect to the norm 
$$ \def\norm#1{\left\|#1\right\|}\norm{x}_\infty := \sup_{\gamma \in \Gamma} \def\abs#1{\left|#1\right|}\abs{x(\gamma)} $$
$c_0(\Gamma)^*$ is isometric to $\ell^1(\Gamma)$ via 
$$ T \colon \ell^1(\Gamma) \to c_0(\Gamma)^*, \quad (Ty)(x) := \sum_{\gamma\in \Gamma} y(\gamma)x(\gamma), \qquad x \in c_0(\Gamma), y \in \ell^1(\Gamma) $$
Regarding seperability: $c_0(\Gamma)$ is not separable: For $\delta \in \Gamma$, consider $e_\delta \colon \Gamma \to \mathbf K$ given by $e_\delta(\gamma) = 0$ for $\gamma \ne \delta$ and $e_\delta(\delta) = 1$. Then $e_\delta \in c_0(\Gamma)$ and 
$$ \norm{e_\delta - e_{\delta'}} = 1, \qquad \delta \ne \delta' $$
So, if $\Gamma$ is uncountable, $c_0(\Gamma)$ is not separable.
A: When $\Gamma$ is uncountable, $c_0(\Gamma)$ is not separable.  It consists of all functions $g : \Gamma \to \mathbb R$ (with countable support) such that for every $\epsilon>0$, the set $\{\gamma \in \Gamma : |g(\gamma)|>\epsilon\}$ is finite.
But (you may ask) even if this particular pre-dual is nonseparable, might there not be some other pre-dual that is separable?  The answer to that is "no".  (But the proof I have in mind is a theorem of Stegall about "Asplund space" ... Israel J. Math. 29 (1978), 408–412.  Maybe there is a more conventional proof?)

Stegall:
Stegall, Charles, The duality between Asplund spaces and spaces with the Radon-Nikodym property, Isr. J. Math. 29, 408-412 (1978). ZBL0374.46015.
If $X$ is a Banach space, and $X^*$ has the Radon-Nikodym property, then every separable subspace of $X$ has separable dual.
Consequence:  since $l^1(\Gamma)$ has the Radon-Nikodym property, if $X$ is any predual of $l^1(\Gamma)$, then every separable subspace of $X$ has separable dual.  In particular, if $X$ itself were separable, then $X^* \cong l^1(\Gamma)$ would be separable, so $\Gamma$ would be countable.
