I have three questions about Hilbert space that I got from Homework solution Let $H$ be Hilbert space
Let $\{x_n\}_{n\ge 1}$ be a orthonormal set, but not compete. Then we can always find $y\in H,~y\ne 0$ such that $(x_n,y) = 0$ for all $n$
$S$ is a subset of $H$ and its closure has non empty interior, then span$\{S\}$ is dense in $H$.
If $H$ is infinite dimensional, for fixed $x\in H$, we can always find a sequence of orthonoraml $\{e_n\}_{n\ge 1}$ such that $(x,e_n)=0$ for all $n$.
There are several equivalent definitions of completeness of an orthonormal set, but I will try to answer a) using the following def.: an orthonormal set $\{e_i\}$ is complete if every vector $x$ in $H$ has an expansion $x=\sum \langle x, e_n \rangle e_n$. [The series is suppose to converge in the norm of $H$]. There is also a mistake in the statement of part 1. You should say non-zero $y$ because the zero vector is always orthogonal to each $e_n$. Suppose there is no $y \neq 0$ such that $\langle x, e_n \rangle =0$ for all $n$. Let $y=x-\sum \langle x, e_n \rangle e_n$. (You might recall that the series here is always convergent in the norm). Verify that $\langle y, e_n \rangle =0$ for all $n$ which forces $x-\sum \langle x, e_n \rangle e_n$ to be 0. This answers part 1. Prt 2. is also stated wrongly. $S \subset H$ and $\bar S$ not empty only says $S$ is non-empty and $S$ need not span $H$. (take a single vector for example). The correct statement is $S$ spans a dense subspace if $\bar S$ has non-empty interior. In this case there is an open ball $B(x,r)$ contained in $\bar S$. I leave it to you to check two facts: $B(x,r)$ spans $H$ and this implies $S$ spans a dense subspace. For Part 3 first show that there exist vectors $x_1,x_2,...$ such that $\{x_1,x_2,...\}$ is a linearly independent subset of $M\equiv \{v:\langle x, v \rangle =0\}$. This is possible because $H=M+span \{x\}$ so $M$ is also infinite dimensional. Now you have to use what is called Gram -Schmidt orhtogonalization process to get an orthonormal squenec in $M$.