Let $H$ be a separable infinite dimensional Hilbert space, with orthonormal basis $(e_n)_{n=1}^\infty$. Consider the operators $U_n$ on $H$ such that $U_ne_k = e_1$ if $n=k$ and $U_ne_k=0$ if $n\neq k$. Note that $\|U_n\|=1$.
My question is inspired by the following results, which are parts of exercises in section 4.6 of Gert Pedersen's book Analysis Now.
Fact 1: $0$ is in the closure of the set $A=\{\sqrt{n}U_n:n\geq 1\}$ in the strong operator topology.
Fact 2: No sequence in $A$ converges weakly or strongly to $0$. (Consequently, these topologies are not metrizable.)
The proof of Fact 2 goes by showing that if a sequence in $A$ weakly/strongly converges to $0$, then it is pointwise bounded, so an application of the Principle of Uniform Boundedness yields a contradiction, since such a sequence cannot be uniformly bounded in norm unless it is finite, in which case it cannot converge to $0$.
Question: Why doesn't this argument hold for nets? By Fact 1, there is a net in $A$ which converges strongly to $0$, and it seems like the same argument would imply that the net is pointwise bounded, hence uniformly bounded in norm.
