$c_0$ is not isomorphic to a dual space

Exhibit a countable dense subset of $c_0(\mathbb{K})$. Let ${\mathbb{K}}_0=\mathbb{Q}$ in the real case and ${\mathbb{K}}_0=\mathbb{Q}+i\mathbb{Q}$ in the complex case, and let $D$ be the set of sequences with finite support and all coordinates in ${\mathbb{K}}_0$. For each support contained in $\{0,\dots ,N\}$ its members form a finite product of a countable set; [L2], followed by the countable union over $N$, makes $D$ countable. Given $x\in c_0$ and $\epsilon >0$, [L1] gives a finite truncation within $\epsilon \mathrm{/}2$ of $x$. Approximate its finitely many real coordinates, or both real and imaginary parts, by elements of $\mathbb{Q}$ within $\epsilon \mathrm{/}2$ in the maximum norm. The resulting member of $D$ is within $\epsilon$ of $x$. Thus $D$ is dense and $c_0$ is norm separable.

Transfer separability to a hypothesized dual. Suppose toward a contradiction that a topological isomorphism $T:Y^{\ast }\to c_0$ exists for some normed space $Y$. Since $T^{-1}$ is continuous, $T^{-1}(D)$ is countable. It is dense in $Y^{\ast }$: for a nonempty norm-open set $U\subseteq Y^{\ast }$, the homeomorphism $T$ makes $T(U)$ a nonempty open subset of $c_0$, which meets $D$, and hence $U$ meets $T^{-1}(D)$. Therefore $Y^{\ast }$ is norm separable.

Derive the RNP contradiction. By [L4], AC and norm separability give RNP to the dual $Y^{\ast }$. Isomorphism invariance [L3] then gives RNP to $c_0$, contradicting the second assertion of [L4]. Hence no such $Y$ and $T$ exist.

Close the scalar and degenerate cases. [A1, L1, L2, L3, L4, step 1.1, step 3.1] The Gaussian-rational choice in step 1.1 handles the complex norm without restricting scalars, and the real case uses ordinary rationals. The space $c_0$ is nonzero, so it cannot be isomorphic to the zero dual; if $Y^{\ast }=\{0\}$, the hypothesized bijection already fails. AC is used in [L4] and supplies the countability principles invoked in step 1.1.