$c_0$ is not reflexive

Let $T:{\mathrm{\ell }}^1(\mathbb{K})\to c_0(\mathbb{K}{)}^{\ast }$ be the isometric bijection from [F2], so $T(a)(x)={\sum }_{n=0}^{\mathrm{\infty }}{a}_n{x}_n$. Identify $({\mathrm{\ell }}^1(\mathbb{K}){)}^{\ast }$ with ${\mathrm{\ell }}^{\mathrm{\infty }}(\mathbb{K})$ through [F2]. Both identifications use this bilinear series pairing, including over $\mathbb{C}$.

For $x\in c_0$ and $a\in {\mathrm{\ell }}^1$, $J_{c_0}(x)(T(a))=T(a)(x)={\sum }_{n=0}^{\mathrm{\infty }}{a}_n{x}_n\mathrm{.}$ The functional on the right is represented, under the second identification in step 1.1, by the bounded sequence $x$ itself. Hence the composite $c_0\stackrel{J_{c_0}}{\to }{c}_0^{\ast \ast }\cong {\mathrm{\ell }}^{\mathrm{\infty }}$ is precisely the canonical inclusion $x\mapsto x$.

The constant sequence $\mathbf{1}=(1,1,\dots )$ lies in ${\mathrm{\ell }}^{\mathrm{\infty }}$, has norm one, and does not tend to zero. Thus [F3] gives $\mathbf{1}\mathrm{\notin }{c}_0$, so step 2.1 exhibits a concrete member of $c_0^{\ast \ast }$ outside the range of $J_{c_0}$.

The canonical map is not onto. Since $c_0(\mathbb{K})$ is Banach, [F1] therefore proves that it is not reflexive. The calculation covers both scalar fields, including their identical bilinear convention, and uses no choice principle.