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A popular MathOverflow answer (https://mathoverflow.net/a/13372) by Andrea Ferretti showed that for any infinite dimensional vector space $V$ over a field $F$, the dimension of the dual space $V'$ is at least $2^{|V|}$. It is also clear that the dimension is less than or equal to its cardinality, $|F|^{\dim V}$. However, I can't find anywhere a bound on the dimension of $V'$ apart from these facts. So my question is:

Is it ever possible to do better than these bounds? Or is any value of $dimV'$ consistent with ZFC?

Ideally I would like a general answer for all vector spaces, but I would also be interested to know if it is ever possible to do better in specific cases. I would be interested to hear any independence result that shows we can't always know the dimension of the dual space.

Sebastiano
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Zoe Allen
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I don't see a proof of $\dim(V') \geq 2^{|V|}$ in Andrea Ferretti's answer that you link to. Are you sure you didn't see it somewhere else?

Anyway, the fact is that $\dim(V') = |F|^{\dim(V)}$ whenever $V$ is infinite-dimensional. See, for example, here or here.

Now your question was "Is the dimension of the dual space independent of ZFC?". This question doesn't really make sense: sentences are independent of ZFC, not cardinals. The assertion $\dim(V') = |F|^{\dim(V)}$ is provable in ZFC. But for many values of $F$, $V$, and $\alpha$, statements like "$\dim(V') = \aleph_\alpha$" will be independent of ZFC.

For example, if $F = \mathbb{Q}$ and $\dim(V) = \aleph_0$, then $\dim(V') = \aleph_0^{\aleph_0} = 2^{\aleph_0}$, so "$\dim(V') = \aleph_1$" is independent of ZFC (this is the Continuum Hypothesis). It is constent with ZFC that $\dim(V') = \kappa$ for any cardinal $\kappa>\aleph_0$ with uncountable cofinality.

Alex Kruckman
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    You can only conclude that $|W|=\dim(W)$ when $|W|$ is strictly greater than $|F|$. It is always true that $\dim(V')=|F|^{\dim(V)}$ but this is nontrivial. See here for instance. – Eric Wofsey Jan 19 '23 at 03:40
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    @EricWofsey You're right, in my argument I confused the condition $\dim(W)\geq |F|$ for the condition $|W|\geq |F|$. – Alex Kruckman Jan 19 '23 at 23:39