Non total orthonormal set in a non Hilbert inner product space Suppose there exist a subset $M$ of an inner product space $X$, and the orthogonal complement of $M $ is the zero vector. If $X $ is a Hilbert Space then the span of $M $ will be dense in $X $, but this is not always true if $X $ is not a complete inner product space. Are there any examples of an orthonormal set in a non complete inner product space such that the orthorgonal complement of this set consists only of the zero vector, but this set is not a total orthonormal set? Thank you in advance.
 A: Let $E$ be the space of sequences of real numbers with finite support endowed with the $l^2$ inner product and define the linear form $\phi \colon E \to \mathbb{R}$ by
$$\phi(x) = \sum_{n = 0}^\infty \frac{x_n}{n+1}, \quad x \in E.$$
Then $\phi$ is a continuous linear form, so that $F = \ker \phi$ is a closed subspace. Notice that $F \neq E$ (for instance take $x_0 = 1, x_1=0, x_2 = 0, \ldots$). In particular $F$ is not dense in $E$. Finally we can show that $F^\perp = \{0\}$ (see below for a more elegant solution). Indeed let $y \in F^\perp$ and fix $n \geq 0$. Define $x \in E$ by
$$\begin{align*}x_n &= n+1, \\
x_{n+1} &= -(n+2),\\
x_k &= 0, \quad k \notin \{n,n+1\}.\end{align*}$$
Then $x \in F$ so that by definition of $y$ we have $0 = \langle x,y \rangle = (n+1)y_n - (n+2)y_{n+1}$, i.e. $$y_{n+1} = \frac{n+1}{n+2} \, y_n.$$
This is true for all $n\geq 0$. Since we assumed that $y$ has finite support, this is only possible if $y \equiv 0$.

As suggested in the comment by Hanno, another way to prove that $F^\perp = \{0\}$ is the following:  notice that $\phi(x) = \langle x , a \rangle$ with $a = \left( \frac{1}{n+1} \right)_{n \geq 0} \in l^2- E$. Then in $l^2$, we have $\ker \phi = \{a\}^\perp$ and $(\ker \phi)^\perp = \text{span}\{a\}$, so $F^\perp = \text{span}\{a\} \cap E = \{0\}$. 
