Open sets in the weak topology of a Hilbert Space Suppose I have an orthonormal basis $\{e_n : n \in \mathbb{N}\}$ of a Hilbert space $\mathcal{H}$, and let $\mathcal{I} = \{n^{\frac{1}{2}}e_n : n \in \mathbb{N}\}$. 
The claim is that if $U$ is a weak neighbourhood of $0$ such that $U \cap \mathcal{I} = \varnothing$, then there is $\epsilon > 0, x_1, ..., x_k \in \mathcal{H}$ such that
$$U = \bigcap_{i = 1}^{k} \{x \in \mathcal{H} : |\langle x, x_i \rangle| < \epsilon\}.$$
I fail to see why this is the case. I imagine that this follows from the definition of an initial topology (and may be generalizable to arbitrary initial topologies?), but I cannot think of a reason why this would be true, espacially for a fixed $\epsilon$ and finite points $x_i$ from $\mathcal{H}$. 
 A: $\mathcal{I}$ has nothing to do with the statement about basic weak-open neighbourhoods of $U$, this latter fact we can verify independently:
I'll use the definition that the weak topology $\mathcal{T}$ is the smallest topology such that all functionals $\mathcal{H} \to \Bbb C$ are continuous.
Because we're in a Hilbert space, every functional on $\mathcal{H}$ is of the form $x \to \langle x,p\rangle$ for some $p \in \mathcal{H}$.
In general, for an initial topology on a set $X$ wrt a family of functions $\mathscr{F}$ (each from $X$ to $Y(f)$, say), a subbase for that topology is given by $\{f^{-1}[O]: f \in \mathscr{F}, O \subseteq Y(f) \text{ open }\}$ 
and thus a base is formed by taking all finite intersections from that subbase.
So if $0 \in U$ and $U$ is a weak-open neighbourhood of $0$, there are finitely many functionals $f_1, \ldots, f_n$ and finitely many open sets $O_1, \ldots,O_n \subseteq \Bbb C$ such that $$0 \in \bigcap_{i=1}^n f_i^{-1}[O_i] \subseteq U$$
from the definition of base applied to this standard base.
The fact that $0$ is in this finite intersection implies that $f_i(0)=0$ is in $O_i$ for all $i$ and all $O_i$ are open in $\Bbb C$, and so we can find $\varepsilon>0$ such that $\{z: |z| < \varepsilon\} \subseteq \bigcap_{i=1}^n O_i$ (in $\Bbb C$). Also by the representation theorem we can find $x_i \in \mathcal{H}$ such that $f_i(x) = \langle x , x_i \rangle$ for all $x$ and then for all $i$:
$$\{x: |\langle x,x_i \rangle| < \varepsilon \} \subseteq f_i^{-1}(O_i)$$
and so
$$\bigcap_{i=1}^n\{x \in \mathcal{H}:|\langle x, x_i\rangle| < \varepsilon\} \subseteq \bigcap_{i=1}^n f_i^{-1}[O_i] \subseteq U$$
as required. 
So the last fact about neighbourhoods just follows from unpacking definitions and applying the representation fact for functionals on Hilbert spaces.
A: This is false. $\mathcal I$ is a weakly closed set. (I will leave it to you to supply  a proof of this fact). Let $U$ be its complement. Then $U$ is a a weak neighborhood of $0$. Suppose this set has the above form. Then $me_n $ belongs to RHS whenever $m >\sqrt n$. Hence $|\langle me_n , x_i \rangle | <\epsilon$  and letting $m \to \infty$ we get $\langle e_n , x_i \rangle =0$ for all $i$ and for all $n$. Thus $x_i=0$ for all $i$ and obviously this leads to a contradiction. 
