Complete vector sets in a Hilbert space and orthogonality. I am reading the following: 



The first definition says a vector set is complete if $(y,x_a) = 0 \implies y \;\forall a$. 
I thought the word "complete" usually refers to "the biggest and filling up" (Hilbert spaces are complete by definition, so that "complete" doesn't interfere with this one). What does the condition in the definition even mean? That $x_a$ is orthogonal to everything in the set? Does it have anything to do with extending a set to a basis?
Also what is the decomposition property they are talking about? The complement decomposition? Can someone sketch an outline of the proof. I just want to read the ideas.
 A: The original definition of complete in a Hilbert space was for an orthonormal subset $\{ e_{\alpha} \}_{\alpha\in\Lambda}$, and the definitions were motivated by classical Fourier Analysis, where the concepts first arose, even before finite-dimensional linear algebra. In that context, completeness meant that every $x$ could be expressed as $x=\sum_{\alpha\in\Lambda}\langle x,e_{\alpha}\rangle e_{\alpha}$. This turns out to be equivalent to Parseval's equality for that $x$, meaning $\|x\|^2=\sum_{\alpha\in\Lambda}|\langle x,e_{\alpha}\rangle|^2$. In a Hilbert space, you can see this forces the condition that $x=0$ if $\langle x,e_{\alpha}\rangle =0$ for all $\alpha\in\Lambda$. So you end up with this early result of Hilbert space:
Theorem: Let $H$ be a Hilbert space, and let $\{ e_{\alpha} \}_{\alpha\in\Lambda}$ be an orthonormal subset of $H$. Then the following are equivalent


*

*$x=\sum_{\alpha\in\Lambda}\langle x,e_{\alpha}\rangle e_{\alpha}$ holds for all $x\in H$.

*$\|x\|^2=\sum_{\alpha\in\Lambda}|\langle x,e_{\alpha}\rangle|^2$ holds for all $x\in H$.

*$\langle x,e_{\alpha}\rangle=0$ for all $\alpha\in\Lambda$ iff $x=0$.

*The subspace $S$ spanned by finite linear combinations of $\{ e_{\alpha} \}_{\alpha\in\Lambda}$ is dense in $H$.
I think you can see how these ideas motivate the definition of complete that you are using for a subspace.
A: Let $\text{span} \{x_\alpha\}=M$.
Assume that $M$ is dense in $V$. To show that $\{x_\alpha\}$ is complete.
Let $y\in H$ such that $\langle y,x_\alpha\rangle=0$.Since $M$ is dense in $V$  so there exists a sequence say $(z_n)_n$ in $M$ such that $z_n\to y$.
Now each $z_n=\sum_{i\in \mathscr I}  c_ix_i;I\text{is finite}$. Then $\langle y,z_n\rangle=0\implies \langle y,y\rangle =0\implies y=0$
Conversely
IDEAS: Use that if $Y$ is any  closed subspace of a Hilbert space $H$ then $H=Y\oplus Y^\perp$(Decomposition).
