We know that an infinite dimensional Banach space has an uncountable Hamel basis. Now if $X$ is a vector space with an uncountable Hamel basis, does there exist a norm on $X$ for which $X$ is a Banach space. I could not proceed. Any help is appreciated.

4$\begingroup$ A vector space has no structure other than its dimension, so there can't be any obstructions. $\endgroup$– Christian RemlingNov 15, 2022 at 17:54

3$\begingroup$ @ChristianRemling: is it obvious? The Theorem is that if $X$ is Banach, then its Hamel dimension is not $\aleph_0$. I think the OP is asking whether for any cardinal number other than $\aleph_0$ there is a Banach space with that Hamel dimension. $\endgroup$– Willie WongNov 15, 2022 at 18:12

2$\begingroup$ This MSE answer gives a citation that every infinite dimensional separable Banach space has cardinality exactly $2^{\aleph_0}$. But is there any known result of limits on the cardinalities of nonseparable Banach spaces? (Also @MartinBrandenburg.) $\endgroup$– Willie WongNov 15, 2022 at 18:17

3$\begingroup$ This MSE comment further suggests that there are nontrivial restrictions on the Hamel dimensions of Banach spaces (though I don't understand it at all). I do think this is a suitable question for MO. $\endgroup$– Willie WongNov 15, 2022 at 18:24

7$\begingroup$ @MartinSleziak Also A Banach space of (Hamel) dimension $\kappa$ exists if and only if $\kappa^{\aleph_0}=\kappa$, which answers the question in the title. $\endgroup$– Michael GreineckerNov 15, 2022 at 19:50
1 Answer
I will pull together the comments into a community wiki answer with some of my own remarks so that the question isn't left on the unanswered questions list.
If you're willing to accept that it is consistent that $\aleph_1 < 2^{\aleph_0}$, you can get a relatively small example of an impossible uncountable Hamel dimension for a Banach space. The space $\ell^2$ contains a linearly independent set of cardinality $2^{\aleph_0}$, specifically $(n^{\alpha})_{n \in \mathbb{N}}$ where $\alpha$ ranges over $[0,1]$. So the Hamel dimension of $\ell^2$ is at least $2^{\aleph_0}$, and can't be more because the cardinality of $\ell^2$ is $2^{\aleph_0}$. Now, for every infinitedimensional Banach space $E$ you can build an injective bounded linear map $\ell^2 \rightarrow E$ (this is one of those constructions where it is simpler to just try it yourself than to follow someone else's way of doing it). So $\dim(E) \geq 2^{\aleph_0} > \aleph_1$ and there is no Banach space of Hamel dimension $\aleph_1$.
In general, in Lemma 2 of
Kruse, Arthur H., Badly incomplete normed linear spaces, Math. Z. 83, 314320 (1964). ZBL0117.08201.
Kruse showed that for a Banach space $E$, $\dim(E)^{\aleph_0} = \dim(E)$. By König's theorem, if $\kappa$ is uncountable and the union of countably many strictly smaller sets, then $\kappa^{\aleph_0} > \kappa$. So, unconditionally, there is no Banach space of Hamel dimension $\aleph_\omega = \bigcup_{n=0}^\infty \aleph_n$ nor of Hamel dimension $\beth_\omega = \bigcup \{ 2^{\aleph_0}, 2^{2^{\aleph_0}}, 2^{2^{2^{\aleph_0}}}, \ldots \}$.
(It seems that Kruse's lemma does not require the axiom of replacement. If this is so, then we cannot find an unconditional counterexample without using the axiom of replacement because in the $V_{\omega + \omega}$ inside $L$ we have that $X^{\omega} \cong X$ for every uncountable set.)

3$\begingroup$ +1 for "this is one of those constructions where it is simpler to just try it yourself". $\endgroup$ Nov 20, 2022 at 17:05