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Clearly $\Leftarrow $ is a trivial trivial application of G-Schmidt algorithm. I've experienced some trouble in proving the other direction. I focused my self on the fact that span($ A $)=$ H $ (it can be not closed) where A is an Hamel base, but i i didn't found nothing. I can't even use G-S because the cardinality of the hamel base can be uncountable. Clearly the interesting case (where to find an absurd) is when H has an uncountable hilbert dimension.

Someone has some advice? Thanks in advance

Riccardo
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  • I think it's wrong. If $H$ has a hilbert basis of cardinality $\kappa \geqslant 2^{\aleph_0}$, I think its cardinality is $\kappa$, and that means (since $\kappa \geqslant \operatorname{card}\mathbb{C}$) that its Hamel bases have cardinality $\kappa$. I'm not 100% sure that $\kappa \geqslant \operatorname{card}\mathbb{C}$ implies Hilbert dimension = cardinality, though. But almost. – Daniel Fischer Jan 08 '14 at 14:07
  • @Daniel: I don't see why having cardinality larger than the continuum implies that. I could believe that there is a Hilbert space with a topological basis of size $\aleph_\omega$, but its closure - and the Hamel basis cardinality is $\aleph_{\omega+1}$ (e.g. if you assume $(\aleph_\omega)^{\aleph_0}=\aleph_{\omega+1}$) – Asaf Karagila Jan 08 '14 at 14:19
  • @AsafKaragila Ah, okay. Then $\kappa^{\aleph_0} = \kappa$ holds only for some $\kappa$. Still, these are counterexamples. – Daniel Fischer Jan 08 '14 at 14:23
  • @AsafKaragila Would you care to check the cardinal arithmetic in my answer? I'm almost sure it's right, but I'd like to have an expert check it. – Daniel Fischer Jan 08 '14 at 14:38

1 Answers1

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There are infinite-dimensional Hilbert spaces whose Hamel-dimension equals the Hilbert-dimension.

Remember that every Hilbert space is (isometrically isomorphic to) $\ell^2(\kappa)$ for some cardinal $\kappa$.

Let $\kappa = 2^{\aleph_\alpha}$ for some ordinal $\alpha$. Then $\kappa^{\aleph_0} = 2^{\max\{\aleph_\alpha,\aleph_0\}} = 2^{\aleph_\alpha} = \kappa$. Letting $\iota_S \colon \ell^2(S) \to \ell^2(\kappa)$ the canonical inclusion for a subset $S\subset \kappa$, we have

$$\ell^2(\kappa) = \bigcup_{\substack{S\subset\kappa\\ \lvert S\rvert \leqslant \aleph_0}} \iota_S(\ell^2(S)).$$

Since every $\ell^2(S)$ has cardinality $\mathfrak{c} = 2^{\aleph_0}$ (or $1$, if $S$ is empty), and there are no more than $\kappa$ countable subsets of $\kappa$, we have

$$\operatorname{card} \ell^2(\kappa) \leqslant \kappa \otimes \mathfrak{c} = \max \{\kappa,\mathfrak{c}\} = \kappa.$$

On the other hand, the cardinality is trivially at least $\kappa$, so we have equality.

The same argument for finite subsets of a basis shows that $\dim_{\text{Hamel}} V = \operatorname{card} V$ for a (real or complex) vector space $V$ if the dimension is at least $\mathfrak{c}$.

So we have $\dim_{\text{Hamel}} \ell^2(2^{\aleph_\alpha}) = \operatorname{card} \ell^2(2^{\aleph_\alpha}) = \dim_{\text{Hilbert}} \ell^2(2^{\aleph_\alpha})$ for all ordinals $\alpha$.

Daniel Fischer
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