A non complete orthogonal system whose perpendicular space is trivial Let $Y$ be an inner-product space, and let $A$ be an orthonormal system.
We're trying to find a case to demonstrate the fact that even if for any given $x$ in $Y$ there's some $u$ in $A$ such that $\langle u,x \rangle \neq 0$, that doesn't imply that the system is complete.
We've asserted that $A$ should not be a subset of a complete orthonormal system, and thus could not be completed to one. That led us to the realization that it's essential that Gram-Schmidt's operation could not be applied, which probably means we need to find an IPS whose dimension is at least the continuum cardinality.
However, we were not able to actually construct such an example, any tips will be welcomed...
Rephrase: My wording seems to have caused some confusion, so I'll try again:
Find an IPS Y and an orthogonal system $A\subset Y$ such that $\forall x\in Y \exists u\in A: \langle x,u\rangle \ne0$ yet A is not a complete system (e.g. not an orthonormal base, e.g. it's span is not dense in Y, e.g. It doesn't always conform to Parseval's equality. These are all equivalent for any IPS).
 A: If the inner-product space $Y$ is complete (a.k.a. "Hilbert"), then having trivial orthogonal complement (the question's hypothesis) implies that the orthogonal or orthonormal set is complete (as Mariano and Theo observe, above). That is, as M. and T. note, there is no example of the sort requested.
Is it conceivable that having the inner-product space $Y$ itself not be complete is seriously relevant to the questioner's issues? Of course, then there are complications in the usual harmony of Hilbert spaces (separable or not). But the resulting "counter-examples" are more pranks than serious. E.g., let $Y$ be the subspace of $\ell^2$ ("algebraically") spanned by the usual $e_1,e_2,\ldots$, but with one $e_{i_o}$ removed, replaced by a vector $v$ having all non-zero entries. Take $A=\{e_i\}$ with $e_{i_o}$ missing. If I'm not mistaken, this ruse (or similar) contrives a situation formally meeting the question's requirement.
However, surely this is not of operational interest, due to its artificiality (despite formal, literal correctness)?
A: I am not quite sure I understand what you want... This answers what I guess you are asking, though:
In an infinite dimensional, separable Hilbert space, like $\ell^2$, an orthonormal set is at most countable (because it is discrete). Yet $\ell^2$ is of uncountable dimension.
The set of 'canonical' vectors in $\ell^2$ is an orthonormal set which does not span $\ell^2$ yet its orthogonal subspace is trivial.
