Orthogonality in inner product/Hilbert space Given a sequence $(f_{n})$ of nonzero elements in some infinite dimensional vector space $V$, does there necessarily exist some $g\in V$ such that $\left<g,f_{n}\right>\ne 0$ for each $n=1,2,...$? If there is no such $g$, then what about if $V$ is a Hilbert space? Or even separable Hilbert space?
The way I approach it is to consider the dimension of $f_{n}^{\perp}:=\{h\in V:\left<h,f_{n}\right>=0\}$ and argue about the dimension of $\displaystyle\bigcup_{n}f_{n}^{\perp}$. There are two type of dimensions, vector space dimension and Hilbert space dimension, it is well known that the Hilbert space dimension of any separable Hilbert space is $\leq\omega$. If I want to argue in that way, it seems that I cannot rely on Hilbert space dimension, indeed, perhaps it is a very wrong that I start in that way.
 A: Assume that there is no $g $ as you want it. This means for each $g \in H $, there is $n \in \Bbb {N} $ with
$$
g \in M_n =\{x \in H \,\mid\,\langle x,f_n\rangle =0\}.
$$
In other words,
$$
H =\bigcup_n M_n .
$$
It is easy to see that each $M_n $ is closed. Thus, by Baire's category theorem, some $M_n $ has nonempty interior. But $M_n $ is a subspace of $H$ and no proper subspace of a vector space can have nonempty interior. Thus, $M_n =H$, which easily implies $f_n =0$, a contradiction.
Thus, the assumption that there is no $g $ with the desired property is false.
Note that I used completeness of $H $ for the application of Baire's theorem. As you noted yourself in the comments, for incomplete spaces, the claim can fail, for example for the subspace $V \leq \ell^2 (\Bbb {N}) $ of finitely supported sequences.
A: If $V$ is a Hilbert space, then there is an orthonormal basis for it (see e.g. Bruce K Driver "Analysis Tools with Applications). The sequence's elements $f_n$ can be expressed in combinations of the basis vectors. 
If the number of basis vectors involved in the sequence is finite, we can find a $g$ solving the expression. 
If it is infinite, can we use that the Hilbert space is complete to prove the existence of such a $g$?
