# Orthonormal system in Hilbert space

aaaa202
Let H be a Hilbert space. Let F be a subset of H.
F is dense in H if:
<f,h>=0 for all f in F => h=0
Now take an orthonormal set (ek) (this is a countable sequence indexed by k) in H. My book says that obviously:
$\bigcup$span(ek) is dense in H (the union runs over all k)
=>
g=Ʃ<g,ek>ek
Now first of all:
Why do we need the union of the spans in the above? What is the difference between the span of the sequence of countably many units vectors and the union of the spans?
And then: Why is the above implication obvious?

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thegreenlaser
What is the difference between the span of the sequence of countably many units vectors and the union of the spans?

I don't know about the rest of your question, but think about two orthogonal unit vectors x and y in $\mathbb{R}^2$. Span(x) and span(y) are both straight lines, so the union of span(x) and span(y) would be two crossed straight lines; it would look like an "X". On the other hand, span({x,y}) is the whole plane.

aaaa202
yes, so why include the union if it is already included in span(e_k)

economicsnerd
The notation there isn't clear enough, but I think what it's naming is $$\bigcup_{n=1}^\infty \text{span}[\{e_k\}_{k=1}^n],$$ which some people would just call $\text{span}[\{e_k\}_{k=1}^\infty]$. The value of the former notation is that it reminds us that in a vector space the words "linear combination" mean (by definition) finite linear combination.