Ell one is not reflexive

For $n\in \mathbb{N}$, let $e_n$ be the coordinate vector with value $1$ at $n$ and $0$ elsewhere. By [F3], $e_n\in {\mathrm{\ell }}^1(\mathbb{K})$ and $\mathrm{\parallel }{e}_n{\mathrm{\parallel }}_1=1$, so $(e_n)$ is norm bounded. If $m\mathrm{\ne }n$, the two nonzero coordinates of $e_m-e_n$ have moduli $1$, hence $\mathrm{\parallel }{e}_m-e_n{\mathrm{\parallel }}_1=2$.

Suppose for contradiction that ${\mathrm{\ell }}^1(\mathbb{K})$ is reflexive. The forward implication of [F1] applied to the bounded sequence from step 1.1 supplies strictly increasing indices $(n_j)$ and $x\in {\mathrm{\ell }}^1(\mathbb{K})$ such that $e_{n_j}\rightharpoonup x$.

By [F2], the weakly convergent subsequence in step 2.1 converges to $x$ in norm. A norm-convergent sequence is Cauchy: once $\mathrm{\parallel }{e}_{n_j}-x{\mathrm{\parallel }}_1<1\mathrm{/}2$ and $\mathrm{\parallel }{e}_{n_k}-x{\mathrm{\parallel }}_1<1\mathrm{/}2$, the triangle inequality gives $\mathrm{\parallel }{e}_{n_j}-e_{n_k}{\mathrm{\parallel }}_1<1$. But strict increase makes $n_j\mathrm{\ne }{n}_k$ for $j\mathrm{\ne }k$, and step 1.1 makes that distance exactly $2$. This contradiction proves that ${\mathrm{\ell }}^1(\mathbb{K})$ is not reflexive. Since $\mathbb{K}$ was either scalar field, the result holds for both. The ultrafilter lemma, DC and HB are spent only through [F1]; the Schur argument [F2] is choice-free.