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the problem I'm stuck on is the following:

Suppose that S is a countably infinite subset of $\ell_2$ with the property that the linear span of S′ is dense in $\ell_2$ whenever S\S′ is finite. Show that there is some S′ whose linear span is dense in $\ell_2$ and for which S\S′ is infinite.

I have tried repeatedly to solve this in somewhat of a 'bang my head against a wall' manner, by constructing a series of subsets of some arbitrary S, such that the complement is finite and of increasing size, but I haven't had any success. I haven't actually used the fact that we're working in $\ell_2$ here, so it's quite likely that I'm meant to use some property of Hilbert spaces - however, I'm not sure what. Could anyone please help?

Thankyou very much; Stephen.

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I was stuck on this problem as well. Try using Gram-Schmidt. (The intuition here comes from the rank-nullity theorem: in finite-dimensional vector spaces, the cardinality of the largest set of vectors you can remove from a given set S of vectors and still have it span the entire space is related to the rank of a matrix you can build out of S. For Hilbert spaces there is a good notion of rank, and it comes from finding orthonormal bases.) – Qiaochu Yuan Jan 17 '11 at 0:48
@Qiaochu Yuan: So $S = \{x_i: \sum x_{i}^{2} \ \text{converges for} \ i = 1, \dots \}$? – PEV Jan 17 '11 at 1:06
@Trevor: I don't understand your comment. $\ell_2$ is the set of all sequences $(x_1,x_2,\ldots)$ such that $\sum_i|x_i|^2\lt\infty$, so in particular $\sum_ix_i^2$ always converges. – Jonas Meyer Jan 17 '11 at 2:36
It should be noted that though the title of the question is very clear, the question is not. The quantity $S'$ is introduced without defining it, and then it is asked to find it. Does the question really reads as the following: Suppose that $S$ is a countably infinite subset of $\ell_2$ with the property that the linear span of $S$ is dense in $\ell_2$ whenever $\ell_2\backslash S$ is finite. Show that there is some $S'$ whose linear span is dense in $\ell_2$ and for which $\ell_2\backslash S'$ is infinite? – mpiktas Jan 17 '11 at 9:55
No, that is not the question. The question is about a set S with dense linear span such that removing any finite subset of S leaves a set that still has dense linear span. You are supposed to show that then necessarily S has some infinite subset S' such that S\S' still has dense span. – Johan Jan 17 '11 at 10:24

Most likely you already realize this, but you have to be careful about how you remove the sets. If $S_1\subset S_2\subset S_3\subset\ldots\subset S$, with $|S_n|=n$, then you know that each $S\setminus S_n$ spans, but you can't say that $S\setminus\cup_n S_n$ spans. For example, $S$ could be countable, and $\cup_n S_n$ could be $S$.

This is Problem 9 in Halmos's A Hilbert space problem book, which as usual has a lot of good reading leading up to the problem. There is a hint:

Omit an infinite subset by omitting one element at a time.

There is also a concise solution in case you aren't too wary of having the answer "given away". (Don't forget to cite your source if you use this in work you turn in.) It is worth checking out even if you solve the problem independently. The given solution does not depend on Hilbert space geometry, and would be valid in any separable normed space. (As Qiaochu pointed out to me, the reduction to assuming that $S$ is countable does not require separability.)

To elaborate a little on the hint, in a way that will hopefully be helpful without giving too much away, make sure at each step that you can approximate the vectors that have been removed with vectors that will never be removed.

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It seems that the problem is from here: (i.e. it is an old homework). – PEV Jan 17 '11 at 2:01
maybe I am missing something obvious, but I don't understand the solution given in Halmos's book. Trevor is right that this is old homework (I took this class as well), but Cambridge has a habit of recycling homework problems so I guess it's a good idea not to put full solutions online. Could you email me with an elaboration? Thanks. – Qiaochu Yuan Jan 17 '11 at 2:50

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