$c_0$ fails the Radon--Nikodym property

Fix an arbitrary slice and a point with positive margin. Let $S=S(B,f,\alpha )$ be any slice, and represent $f(x)={\sum }_n{a}_n{x}_n$ by [L2]. On $B$ one has $\sup \mathrm{Re}f=\mathrm{\parallel }f\mathrm{\parallel }$: the upper bound is the dual-norm inequality, and multiplying an almost norming vector by a scalar of modulus one makes its $f$-value real and nonnegative. By nonemptiness of the slice choose $x\in S$ and put $\delta =\mathrm{Re}f(x)-(\mathrm{\parallel }f\mathrm{\parallel }-\alpha )>0$.

Change one remote coordinate in both directions. Truncation convergence in [L2] gives $a_n\to 0$, so choose $k$ with $2\mathrm{\mid }{a}_k\mathrm{\mid }<\delta$. Define $y,z$ by retaining all coordinates of $x$ except $y_k=1$ and $z_k=-1$. Both sequences still tend to zero and have supremum norm at most one, so $y,z\in B$. Moreover

Compute every slice diameter and obtain nondentability. The triangle inequality bounds the diameter of $B$, and hence of $S$, by two; step 2.1 attains two. Therefore every slice of $B$ has diameter exactly two. In particular no slice has diameter below one, so $B$ is not dentable. The ball is nonempty, bounded, closed, and convex in the Banach space from [L1].

Apply the RNP--dentability characterization. If $c_0$ had RNP, [L4] would make its closed unit ball dentable, contradicting step 3.1. Hence $c_0$ fails RNP over both scalar fields.

Record the zero-functional and endpoint cases. [A1, L2, L3, step 1.1, step 2.1, step 4.1] If $f=0$, the slice is all of $B$; take $x=0$ and any $k$, so the same $y=e_k$, $z=-e_k$ witness diameter two. For nonzero $f$, the strict slice margin $\delta$ ensures both perturbed points remain inside rather than merely on its boundary. A zero coefficient $a_k$ causes no difficulty. The complex proof uses the bilinear $c_0$--${\mathrm{\ell }}^1$ pairing and real parts exactly as in [L2]--[L3]; no conjugate is inserted. AC is used only through [L4].