Vector space dimension of little-l 2

[jdstokes]

Homework Statement

I'm trying to understand why $\ell_2^\infty$ as a vector space over $\mathbb{C}$, has uncountable dimension.

Homework Equations

The Attempt at a Solution

Firstly, I'm not really clear on the meaning of basis in infinite dimensions. Is it still true that any element is a finite linear combination of basis elements?

If $\ell_2$ had a countable vector space basis then Gramm Schmidt gives a countable orthonormal vector space basis $\{ v_n \}$. Then $\sum (1/n)v_n$ is in l-2 but is not a finite linear combination of $\{ v_n \}$. Does this prove anything?

 

[Office_Shredder]

Is it still true that any element is a finite linear combination of basis elements

Nope... for example, 1,x,x²...,xⁿ,... etc. is a basis for infinitely differentiable real functions (via taylor series expansion around zero), but most functions will require an infinite sum

 

[HallsofIvy]

There is an obvious one-to-one correspondence between members of l₂ and the set of real numbers between 0 and 1.

 

[Hurkyl]

Definition: A (Hamel) basis for a vector space V is a set B of vectors such that every element v of V can be written uniquely as a finite linear combination of elements in B.

Every vector space has a Hamel basis.


Definition: A (Schauder) basis for a topological vector space V is a set B of vectors such that every element v of V can be written uniquely as a (possibly infinite) unordered linear combination of elements in B.

For finite-dimensional topological vector spaces, these are the same thing, but they may be different for infinite-dimensional topological vector spaces.


Every time I have heard "vector space dimension", it meant the Hamel dimension, although there is a corresponding notion of Schauder dimension. (This is, of course, probably related to the fact I don't often study infinite-dimensional vector spaces. )

 

[Hurkyl]

Nope... for example, 1,x,x²...,xⁿ,... etc. is a basis for infinitely differentiable real functions (via taylor series expansion around zero), but most functions will require an infinite sum

Actually, you meant analytic, not merely infinitely differentiable. (A function is analytic near a point iff it is equal to its Taylor series near that point. $e^{-1/x^2}$ is infinitely differentiable at x = 0, but not analytic there, since its Taylor series gives the zero function)

The monomials do indeed form an example of a Schauder basis for the space of functions analytic at zero. (Or, more precisely, a basis for the space of germs of analytic functions at zero) But they are only a Hamel basis for the space of polynomials.

 

[jdstokes]

Hurkyl,

This is exactly my question. Are you sure that every infinite dimensional vector space has a Hamel basis? This would suggest that $\ell_2^\infty$ has a Hamel basis. Which is either countably or uncountably infinite.

If it is countably infinite then it can be reduced by Gramm-Schmidt to a countably infinite ORTHONORMAL basis {v_n}. Then $\sum_{n\geq 1}(1/n)v_n \in \ell_2^\infty$ but is not a finite linear combination of any of the orthonormal basis vectors v_n. This contradiction would then show that the vector space dimension is uncountable.

Is this correct?

Thanks.

 

[Hurkyl]

This is exactly my question. Are you sure that every infinite dimensional vector space has a Hamel basis?

Yes:

Axiom of choice ==> Every vector space has a Hamel basis​

In fact, that is an "if and only if", but the reverse implication is harder to prove.¹

If it is countably infinite then it can be reduced by Gramm-Schmidt to a countably infinite ORTHONORMAL basis {v_n}.

I believe this.

Then $\sum_{n\geq 1}(1/n)v_n \in \ell_2^\infty$ but is not a finite linear combination of any of the orthonormal basis vectors v_n.

It's not immediately obvious to me why that sum converges. If $v_{n, i}$ is the i-th component of $v_n$, then it seems you are claiming
$$\sum_{n \geq 1} (1/n) v_{n, i}$$
exists for every i, and that this yields a square-summable sequence in i. I currently see neither a proof nor a disproof.

This contradiction would then show that the vector space dimension is uncountable.

I believe that this follows from your previous assertion.


1: What I really mean is that I know how to prove (==>) but I don't know how to prove (<==).

 

[Dick]

I don't get this. $\sum_{n\geq 1}(1/n)v_n$ clearly converges. Take the components relative to the basis v_n. And it's norm just as clearly converges. Just because $\ell_2$ HAS a Hamel basis doesn't mean every basis has to be one or that a Hamel basis (even if you managed to explicitly construct one) would be useful.

 

[Hurkyl]

I don't get this. $\sum_{n\geq 1}(1/n)v_n$ clearly converges. Take the components relative to the basis v_n. And it's norm just as clearly converges.

I don't see how your suggestion leads to a proof that the sum converges. Looking at the components relative to the (Hamel) basis {v_n} is precisely how you prove this sum doesn't converge.

OTOH, I now see that it's relatively straightforward to prove that the sequence of partial sums is Cauchy, and thus convergent.

(edit: I've assumed that $\ell^2$ is complete to make the statement in red. I don't know if that is something the original poster is yet allowed to assume)

Just because $\ell_2$ HAS a Hamel basis doesn't mean every basis has to be one or that a Hamel basis (even if you managed to explicitly construct one) would be useful.

Every Hamel basis has to be a Hamel basis. And I don't see how you are going to prove anything about the (Hamel) dimension of a vector space without considering Hamel bases.

 

[Dick]

Ok, so if the series is Cauchy then it converges. Isn't convergence independent of basis? I guess I don't see anything wrong with jdstokes claim that this constitutes a proof that a Hamel basis must be uncountable.

 

[Hurkyl]

Incidentally, I've been assuming the whole time we're talking about what I know as $\ell^2$. I just noticed the OP wrote $\ell_2^\infty$ -- is this what he means?

Ok, so if the series is Cauchy then it converges. Isn't convergence independent of basis? I guess I don't see anything wrong with jdstokes claim that this constitutes a proof that a Hamel basis must be uncountable.

Yes, convergence is independent of basis -- that simply means it doesn't matter what basis you use when you prove/disprove that a sum converges. That does not mean the form of theorems about convergence are basis-independent. To wit, the sum

$$\sum_{n = 0}^{+\infty} (1 / n) \mathbf{v}_n$$

diverges for "most" countable, linearly independent sequences $\{ \mathbf{v}_n \}$. (in both the Hamel and Schauder senses)

 

[Dick]

Incidentally, I've been assuming the whole time we're talking about what I know as $\ell^2$. I just noticed the OP wrote $\ell_2^\infty$ -- is this what he means?



Yes, convergence is independent of basis -- that simply means it doesn't matter what basis you use when you prove/disprove that a sum converges. That does not mean the form of theorems about convergence are basis-independent. To wit, the sum

$$\sum_{n = 0}^{+\infty} (1 / n) \mathbf{v}_n$$

diverges for "most" countable, linearly independent sequences $\{ \mathbf{v}_n \}$. (in both the Hamel and Schauder senses)

Sure. But in this problem v_n are orthonormal. He assumed the Hamel basis was countable and Gram Schmidted it. The Cauchyness of the sum and the completeness of the space then implies the existence of the sum which contradicts the Hamelness of the basis. I hope this is the usual ell2, maybe the infinity indicates a completion of the space of finite combinations?