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I know by the first answer of this that any infinite Banach Space have cardinality (which is the same of its Hamel dimension) at least $\mathfrak{c}=|\mathbb{R}|$. So, as we can't answer if $c=\omega_1$ or not, we can't answer as well if some Banach Space have cardinality $\omega_{\alpha}$ for any $\alpha\not=0$, right? Because if we could, then we would get an estimative for the size of continuum, namely $\mathfrak{c}\leq \omega_{\alpha}$.

Maybe by some cardinal properties that I'm not familiar with, we do can estimate $\mathfrak{c}$ by some especific cardinals, but this doesn't hold for ''small'' cardinals such as $\omega_1$ (in this case the estimative is exactly CH).

I) So, this is the main reason that I assume CH in the title. Assuming CH then, and given a cardinal $\kappa>\omega$, can we build a Banach Space $X$ such that $|X|=\kappa$?

II) Or even if we assume $\neg CH$. Given a cardinal $\kappa\geq\mathfrak{c}$, can we build a Banach Space with cardinality $\kappa$?

I have only basic graduate courses in both Banach Spaces and set theory. Could you help me with this?

Marcelo
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    It seems likely that the cardinality of a Banach space must satisfy $\kappa^{\aleph_0}=\kappa$, though I don't immediately see how to prove it. – Eric Wofsey Jul 25 '21 at 15:45
  • Also, in question (I), presumably you want to require $\kappa$ to be uncountable. – Eric Wofsey Jul 25 '21 at 15:47
  • @Eric Wofsey – my set theory is rusty, isn’t this true for every cardinal $\kappa \geq 2^{\aleph_0}$? – Aphelli Jul 25 '21 at 15:49
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    @Mindlack: No, for instance it is not true for $\kappa=\aleph_\omega$, or more generally for any cardinal of countable cofinality. – Eric Wofsey Jul 25 '21 at 15:50
  • @Eric Wofsey – ah, ok. It’s because of that lemma (the name of which I forgot) that states that if $\beta_i < \alpha_i$ then the sum of the $\beta_i$ is smaller than the product of the $\alpha_i$, right? – Aphelli Jul 25 '21 at 15:54
  • @Mindlack: Exactly (that's König's theorem). – Eric Wofsey Jul 25 '21 at 15:54
  • @Eric Wofsey: Ok. So let $\kappa$ be the cardinality of a Banach space $E$. If $\kappa$ is the sup of an increasing sequence of non-limit cardinals $\lambda_i >\aleph_0$, $i \geq 0$, then $E$ is the reunion of the $F_i$, where $F_i \subset E$ is some subset, $|F_i|=\lambda_i$. Let $V_i$ be the closure of the vector subspace generated by $F_i$, then $|V_i| \leq \lambda_i^{\aleph_0}$. Since each $\lambda_i$ is a non-limit infinite cardinal, $|V_i|=\lambda_i$ and $E$ is the reunion of the $V_i$. This contradicts Baire so this objection to $\kappa^{\aleph_0} = \kappa$ doesn’t occur, right? – Aphelli Jul 25 '21 at 16:05
  • @Mindlack: The problem is $\lambda_i^{\aleph_0}$ could actually be greater than $\kappa$. – Eric Wofsey Jul 25 '21 at 16:07
  • @Eric Wofsey: Even though $\lambda_i$ is not a limit cardinal? Anyway, if $\kappa^{\aleph_0}=\kappa$, $\ell^2(\kappa)$ works, no? – Aphelli Jul 25 '21 at 16:10
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    The duplicate asks about Hilbert spaces, but Goldstern's answer there addresses the case of Banach spaces. – Eric Wofsey Jul 25 '21 at 16:17

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