Weak closure can exceed sequential weak closure

For a bounded real functional $F$, put $b_n=F(e_n)$. Test $F$ on $v_N={\sum }_{n=1}^N{b}_n{e}_n$. Then $B_N={\sum }_{n=1}^N{b}_n^2=F(v_N)\le \mathrm{\parallel }F\mathrm{\parallel }\sqrt{B_N}$, so $B_N\le \mathrm{\parallel }F{\mathrm{\parallel }}^2$, including $B_N=0$. Thus $(b_n)$ is square summable. Finite Hölder and passage to increasing finite sums show ${\sum }_n\mathrm{\mid }{x}_n{b}_n\mathrm{\mid }\le \mathrm{\parallel }x{\mathrm{\parallel }}_2\mathrm{\parallel }b{\mathrm{\parallel }}_2$, consistent with these tests.

Fix a basic weak neighborhood of zero given by $F_1,\dots ,F_m$ and $\epsilon >0$. If it missed $A$, then for each $n\ge 1$ some $j$ would have $\sqrt{n}\mathrm{\mid }{F}_j(e_n)\mathrm{\mid }\ge \epsilon$. Therefore ${\sum }_j\mathrm{\mid }{F}_j(e_n){\mathrm{\mid }}^2\ge {\epsilon }^2\mathrm{/}n$. Summing over $n$ contradicts the finite sum of square-summability bounds from step 1.1: the harmonic partial sums are unbounded since each block $2^r\le n<2^{r+1}$ contributes at least $1\mathrm{/}2$. For $m=0$ the neighborhood is the whole space and already meets $A$. Thus every weak zero-neighborhood meets $A$, so $0\in {\bar{A}}^w$, while every member of $A$ has norm $\sqrt{n}>0$.

If a sequence in $A$ converges weakly, F3 bounds its norms by some finite $C$. Its indices therefore satisfy $n\le C^2$, so its range lies in a finite subset of $A$. A finite set is closed in a Hausdorff space: each singleton is closed because every other point has a disjoint neighborhood, and finite unions are closed. The weak limit lies in this finite set, hence in $A$. Thus $A$ is sequentially weakly closed but not weakly closed. $\mathrm{\square }$