ONS in Prehilbert space Suppose $a:=\left(\frac{1}{j}\right)_{j \in \mathbb{N}}$ and $e_n$ be the canonic unit vector in $l^2$. The prehilbert space $H \subset l^2$ is defined as the linear span of the vectors $\lbrace a,e_2,e_3,... \rbrace $. I want to show the following: 1) $\left(e_n\right)_{n\geq 2}$ is a orthonormal system, but no orthonormal basis. 2) $x \in H \;\;\text{and}\;\; x \perp e_n \; \forall n\geq 2 \Longrightarrow x =0$ I know that I have to use the fact that the space $l^2$ is not complete because otherwise this two conclusions would be a contradiction. So I took a cauchy sequence which is not convergent but now I dont know how to go on. Can someone help me please? Thanks in advance

I assume you want $H=\operatorname{span}\\{a,e_2,e_3,\ldots\\}$. Otherwise, $x=e_1$ makes your second statement false.

The set $\\{e_n\\}_{n\geq2}$ is not a basis because $\langle a,e_n\rangle=1/n\
e0$.

Any element $x$ of $H$ is of the form $$ x=x_1 a+\sum_{n\geq2} x_ne_n=x_1e_1+\sum_{n\geq2}(x_n+\frac{x_1}n)\,e_n, $$ with finitely many $x_n$ nonzero. From $x\perp e_n$, we get $$ 0=\langle x,e_n\rangle = x_n+\frac{x_1}n. $$ If we consider $n$ big enough so that $x_n=0$, we obtain $x_1=0$. But then $x_n=0$ for all $n$, and $x=0$.