prove that the weak closure of orthonormal basis set S is contained in union of set S with zero's set

Question

Let H be an infinite dimensional separable Hilbert space with inner product ⟨ · , · ⟩ and S is orthonormal basis. I want prove that the weak closure of S is equal union of S with {0}. I have proved that {0} is contained in the weak closure of S. But I am struggling with prove that the weak closure of orthonormal basis set S is contained in union of set S with zero's set. I need help for that.

Answer 1

Let me write $S=\{ e_n: \ n\in \mathbb N\}$.

Take $x \not \in S\cup\{0\}$. Let me show that there is a weakly open set $U$ containing $x$ but not $S\cup\{0\}$.

First suppose there is one index $i$ such that $t:=\langle x,e_i\rangle \not\in \{0,1\}$. Since $\mathbb R\setminus \{0,1\}$ is open, there is $\epsilon>0$ such that $(t-\epsilon,t+\epsilon) \cap \{0,1\}= \emptyset$. Then $U:= \{y\in H: \ \langle y,e_i\rangle \in (t-\epsilon,t+\epsilon)\}$ does the job.

Second, if all for all $i$ we have that $\langle x,e_i\rangle \in \{0,1\}$ then there must be two indices $i\ne j$ such that $\langle x,e_i\rangle = \langle x,e_j\rangle =1$. Then $U:=\{y\in H: \ \langle y,e_i+e_j\rangle \in (\frac32,2)\}$ is the desired weakly open set.