Dimension of the sequence space and its dual, depending on status of (AC) and (CH) Let's consider the sequence space $E =\mathbb R^{\mathbb N}$. If I believe in Choice, I have an isomorphism $E \simeq \mathbb R^{(\mathfrak c)}$ for some cardinal $\mathfrak c$. I further have some inequalities about $\mathfrak c$: first, $\mathfrak c = \dim \mathbb R^{\mathbb N} \leq |\mathbb R^{\mathbb N}| \leq (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0\times \aleph_0} = 2^{\aleph_0}$. Besides, $\mathbb R^{\mathbb N}$ is the dual space of the infinite-dimensional space $\mathbb R^{(\mathbb N)}$ so we have a strict inequality between their dimensions. To sum up, I get $\aleph_0 < \mathfrak c \leq 2^{\aleph_0}$. So, if I believe in the continuum hypothesis, I get $\mathfrak c = 2^{\aleph_0}$ and an isomorphism $\mathbb R^{\mathbb N} = \mathbb R^{(\mathbb R)}$.
Remark: as we have for a general set $A$ the isomorphism $(\mathbb R^{(A)})' \simeq \mathbb R^A$, an equivalent formulation of the question is: what's the dimension of the dual space of a $\aleph_0$-dimensional space?
I'm quite ignorant about logic, so this raises a number of questions (if that's too much questions for the general rules of the website, please forgive me!)


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*What about the next step, that is the (algebraic) dual space of $\mathbb R^\mathbb N$? Its dimension should be $>2^{\aleph_0}$ but can we be more precise?

*What if I do not suppose the continuum hypothesis? With Choice, it seems we've defined an increasing function $S$ on cardinals by the equation $\mathbb R^{\mathfrak c} (=(\mathbb R^{(\mathfrak c)})') = \mathbb R^{(S(\mathfrak c))}$. Do we need (CH) to prove $S(\aleph_0) = 2^{\aleph_0}$? What is known about this function, depending on the various axioms about infinite dimensional sets? (I take it does not depend of the underlying field, but is it true?)

*What if I do not believe in Choice? I know the dimension terminology becomes quite useless, but can we still formulate some theorems about (non)existence of injective/surjective linear maps?
Thanks a lot, and my apologies for the imprecision of the questions. I hope it remains acceptable for MSE.
 A: (Assuming the Axiom of Choice throughout) Suppose $\mathbf{V}$ is $\aleph_0$-dimensional over the field $F$. Then the cardinality of $\mathbf{V}^*$ is $|F|^{\aleph_0}$. If $|F|=\aleph_{\alpha}$, then as noted in this answer, the cardinality is either $2^{\aleph_0}$, $\aleph_{\alpha}$, or $\aleph_{\gamma}^{\aleph_0}$, where $\aleph_{\gamma}\leq \aleph_{\alpha}$ has cofinality $\aleph_0$ and $\aleph_0\lt\aleph_{\alpha}=|F|$. You need to assume the Generalized Continuum Hypothesis to get more about the cardinality.
Given that, suppose the dimension of $\mathbf{V}^*$ is $d$. Then $\mathbf{V}^*$ is bijectable with the set of all almost-null functions $d\to F$; since $d$ is infinite, this is $d|F|=\max\{d,|F|\}=\max\{d,\aleph_{\alpha}\}$. So we must have $\max\{d,\aleph_{\alpha}\}=\aleph_{\alpha}^{\aleph_0}$. 
Now, suppose $\mathbf{V}$ is an infinite dimensional vector space over $F$, and that $\kappa=\dim(\mathbf{V})\gt|F|=\lambda$. Then the dual has cardinality $\lambda^{\kappa}=2^{\kappa}\gt\kappa\gt|F|$, so the dimension will necessarily be $2^{\kappa}$. Thus your sequence of dimensions becomes $\kappa$, $2^{\kappa}$, $2^{2^{\kappa}}$, etc; that is, the $\beth$ function. I think that if $\dim(\mathbf{V})=|F|$, then you can't say precisely what $\dim(\mathbf{V}^*)$ is, other than that it is strictly larger than $|F|$. 
So if at some point you get to a dimension which is strictly larger than the cardinality of the field, you will get a simple sequence, even without assuming the continuum hypothesis (or GCH). Unless the cardinality of the field is strongly inaccessible (or there is a strongly inaccessible cardinal between the dimension of $\mathbf{V}$ and the cardinality of $F$) you will eventually get there. 
I don't really know what happens without the assumption of the Axiom of Choice; if you assume that the field has a defined cardinality, and the vector space has given basis with a defined cardinality, then you can figure out the cardinality of the dual, but I don't know if you can find a basis for it in that case. The statement "every vector space has a basis" is equivalent to the Axiom of Choice.
