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For any given cardinal number $\alpha \neq \aleph_0$, is there a Banach space $X$ with algebraic dimension $\alpha$?

Frunobulax
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Arman
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    http://math.stackexchange.com/questions/141535/cardinality-of-a-hamel-basis – mathematician Sep 08 '14 at 04:12
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    Not in general. "Halbeisen et al. prove in On Bases in Banach Spaces, Theorem 2.7 that the cardinality of a Banach space must have cofinality $>ω$, so there are non-trivial restrictions on the possible dimensions of Banach spaces..." – Martin –  Sep 08 '14 at 11:25
  • @Thursday I have a doubt about your answer. Let $I$ be a set with cardinality $\alpha>c$. The cardinality of the banach space $\ell_p(I)$ is the cardinality of $I$ (It is not obvious). Since the cardinality of this $\ell_p(I)$ is bigger than $c$ then its dimension coincide with the cardinality, which is $\alpha$. Now, for the case $\alpha=c$, the usual $\ell_p$ has dimension c. So the only restriction is : every infinite dimensional banach space has dimension bigger or equal to $c$. Thursday, can you see anything wrong? Does it contradict Halbeisen's theorem? Thank you in advance. – Daniel Sep 08 '14 at 22:02
  • @Daniel I can see that the density character of $\ell_p(I)$ should be $|I|$, but I don't see why the cardinality of $\ell_p(I)$ should be $|I|$. Lemma 2.8 in the aforementioned paper says that $|X|=d(X)^\omega$ (where $d(X)$ is the density character of $X$), and this result is something that was proved in many places: the authors later published a remark to this effect, with the list of five papers where Lemma 2.8 appeared. –  Sep 08 '14 at 22:42
  • @Thursday I made a mistake proving that $#\ell_p(I)=#I$. Thanks. – Daniel Sep 09 '14 at 11:11

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