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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).

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    If the span of $A$ is not dense then $X = \overline{\mathrm{span}}\,{A} \subsetneqq Y$. Thus there exists $0 \neq z \perp X$, but this means in particular that $\langle z,u\rangle =0$ for all $u \in A$, so you can't find a system as you want.2011-07-27
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    That's not true. The space not being dense doesn't mean that it's complement is contained in it's perpendicular space. That's exactly the eccentricity of non-countable spaces I'm trying to demonstrate. To be more precise, you can't imply that $z\notin Cl(Span(A)) \implies z\in \perp A$, that can only be proven for vector spaces of countable dimensions...2011-07-27
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    I don't understand what you're saying and I didn't say anything of the sort you're writing. There is $z' \in Y \smallsetminus X$, right? Now take the orthogonal projection $z''$ of $z'$ onto $X$ (this exists because $X$ is closed and convex, for example). The element $z = z' - z'' \neq 0$ and by definition $z \perp X$. Countability of the (Hilbert-)dimension has absolutely nothing to do with this argument. [Also, you should be a bit careful whom you're talking to when making assertions such as "that can only be proven for vector spaces of countable dimensions..."]2011-07-27
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    Wait a minute! What you've done here is to prove that given $z'\in Y\setminus Y$ we get that **it's orthogonal projection to $Y$** (which you denoted $z$) is perpendicular to $Y$. This proves nothing. In fact, if you can prove that for any $z'$ (for given $A$ and $Y$) the projection is essential (that is, $z''!=0$) this proves my point exactly... I'll repeat this again to be clear - just because the complement of the closure of the span is not empty doesn't mean that the perpendicular space is not trivial.2011-07-27
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    You're simply wrong here and you misunderstood the argument. Once again: If $p: Y \to X$ is the orthogonal projection then we have for *every* $z' \in Y \smallsetminus X$ that $z'' = p(z') \neq z'$ (since $X$ is closed we have $p(x) = x$ if and only if $x \in X$). Therefore $z = z' - z'' = z' - p(z') \neq 0$ and $p(z) = p(z') - p(p(z')) = 0$ (since a projection satisfies $p^2 = p$). In other words, $0 \neq z \perp X$. Once again in words: $z'$ is an element of $Y \smallsetminus X$ and $z''$ is its projection to $X$. The difference $z = z'-z''$ is the component of $z'$ *orthogonal* to $X$.2011-07-27
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    By the way: you may want to read Theorem 3.1.7 on [page 82](http://books.google.com/books?id=a1R0livwR9AC&pg=PA82) of Pedersen's *[Analysis Now](http://books.google.com/books?id=a1R0livwR9AC)*, where you find the details of what I'm saying here. Note also that in my notation $X = (A^{\perp})^{\perp}$ (by Cor. 3.1.8 there), so if $A^{\perp} = 0$ then $Y = X$.2011-07-27
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    As noted above, you've asserted completeness which was never assumed, which caused all this misunderstanding.2011-07-28
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    Noted. I apologize if it seems that I'm being disrespectful or unappreciative of the time you spend helping me with the problem. I have the utmost respect and gratitude for anyone who is willing to help, difference of opinions aside...2011-07-28
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    Thank you. Once again, my apologies for all the fuss. I removed my previous comment in view of your last one, and I hope I can be of more help next time than this time. Two remarks concerning your question: As I noted in a comment to paul's answer: his example has countable dimension (as a vector space over $\mathbb{R}$ or $\mathbb{C}$). A second remark, assuming completeness, there is a difference in Hilbert dimension and ordinary linear algebra dimension. It follows from the Baire category theorem that every Hilbert space has either finite or uncountable dimension (as a vector space).2011-07-28

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