A weakly convergent net need not be eventually norm bounded

Let $D$ consist of all triples $(U,n,x)$ where $U$ is a weak zero-neighborhood, $n\ge 1$ an integer, $x\in U$, and $\mathrm{\parallel }x\mathrm{\parallel }\ge n$. Define $(U,n,x)⪯(V,m,y)$ if $V\subseteq U$ and $m\ge n$. This is reflexive and transitive. It is nonempty by F2. For two triples, $U\cap V$ is a weak zero-neighborhood and hence contains some $z$ with $\mathrm{\parallel }z\mathrm{\parallel }\ge \max (n,m)$. The triple $(U\cap V,\max (n,m),z)$ is a common upper bound. Thus $D$ is directed; antisymmetry is unnecessary.

Define the net by $a_{(U,n,x)}=x$, the point already carried in the index. For a weak zero-neighborhood $W$, F2 gives at least one index $(W,1,z)$. Every later index has neighborhood contained in $W$, so its carried point lies in $W$. Hence $a_d\rightharpoonup 0$. For any real $R>0$, take an integer $n>R$ and any index with integer coordinate $n$, whose existence follows from F2. Every later index has carried-point norm at least $n>R$. Thus $\mathrm{\parallel }{a}_d\mathrm{\parallel }\to \mathrm{\infty }$, and no tail is bounded. No choice function assigning one point to every neighborhood was used: all admissible points are included in the index set. $\mathrm{\square }$