-2
$\begingroup$

I want to prove that no Hilbert space can have countable Hamel basis just using the fact that any finite dimensional subspace is closed (more specifically without using Baire's theorem). I saw a paper by NAM-KIu TSING solving the same problem for Banach space. But, the proof is not much intuitive. Is it possible to give a easier proof for Hilbert space ?

Using proof by contradiction, the aim is to somehow find a Cauchy sequence and then use completeness to get a limit and show that cauchy sequence does not converge to that limit.

Thanks

Improve this question
$\endgroup$
4
  • 7
    $\begingroup$ What do you mean when you say "I have to prove..."? Why are you not allowed to use the Baire category theorem? $\endgroup$
    – Yemon Choi
    Commented Aug 20, 2015 at 12:31
  • 1
    $\begingroup$ @YemonChoi: I think it is pretty straightforward what I mean here. I want to know an alternate proof. $\endgroup$
    – Sosha
    Commented Aug 20, 2015 at 12:59
  • 7
    $\begingroup$ The language "I have to prove" means, to me, a need or a demand. This makes it sound like an assigned task. Most BC arguments can be converted to sliding hump arguments, anyway $\endgroup$
    – Yemon Choi
    Commented Aug 20, 2015 at 13:18
  • 1
    $\begingroup$ What Yemon is saying in his previous comment is that this sounds like a homework assignment. If that is the case, then please be aware that we don't discuss HW at this site. $\endgroup$
    – Todd Trimble
    Commented Aug 20, 2015 at 14:10

1 Answer 1

Reset to default
4
$\begingroup$

Nam-Kiu Tsing argument indeed become way much simpler for Hilbert spaces. In fact I wouldn't be surprise if his argument was inspired from the (very easy) case of Hilbert spaces:

If you have a countable Hamel basis you can turn it into a countable orthonormal Hamel basis $(e_n)_{n \in \mathbb{N}}$ by Gram-Schmidt process. Then $\sum 2^{-n}e_n$ will be an element of your Hilbert space by completeness but it is not going to be equal to any finite linear combination of the $e_n$ by simply computing the distant between $x$ and such a combination.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged .