A question about multiresolution analysis (from a topological point of view)

Hi,

I have a problem understanding something

This is a snapshot of a book I am reading

http://i.imgur.com/NfBL7.png

Point no. 2 concerns me, because it looks to me like it contradicts itself, with "this or this"

The first part says

$\displaystyle \sum_{j}V_j = {L^2(R)}$ which, to me, looks completely equivavalent to

$\displaystyle \lim_{j \rightarrow \infty}V_j = {L^2(R)}$

given the nested nature of these subspaces.

However, the paper says

http://i.imgur.com/1F4KF.png

so what troubles me is this: is this countable union $\displaystyle \sum_{j}V_j$ equal to $\displaystyle {L^2(R)}$ or is it only *dense* in $\displaystyle {L^2(R)}$?

I personally think it's the former, and I don't understand this "dense" part. Could someone perhaps clarify this for me?

Much obliged!

Re: A question about multiresolution analysis (from a topological point of view)

It is equal.

What the author wants to state by saying dense, is that for every element $\displaystyle u$ of $\displaystyle L^2$, there exists a series $\displaystyle (u_n)\subset \cup_j V_j$ with $\displaystyle u=\sum_j u_j$.

Enter the wavelets!

That is, a Schauder basis for $\displaystyle L^2$ with exactly one element in each $\displaystyle V_j$.