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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!)

  1. 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?

  2. 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?)

  3. 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.

Arturo Magidin
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    Don't use $\mathfrak{c}$ for "some cardinal". That symbol is almost invariably reserved for the cardinality of the continuum, $|\mathbb{R}|=2^{\aleph_0}$. For arbitrary cardinals, it is standard to use $\kappa$ and $\lambda$. – Arturo Magidin Dec 28 '10 at 19:12
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    The dimension of ${\mathbb R}^{\mathbb N}$ is $2^{\aleph_0}$, independently of the size of ${\mathbb R}$ (but assuming choice): For each real $r$ fix a sequence of (pairwise distinct) rationals converging to $r$ and let $A_r$ be the range of this sequence. Then ${A_r\mid r\in{\mathbb R}}$ is almost disjoint (i.e., if $r\ne s$ then $A_r\cap A_s$ is finite). Since ${\mathbb Q}$ is countable, we may via a bijection find an almost disjoint collection of size $2^{\aleph_0}$ of subsets of ${\mathbb N}$. Given $A$ in this collection, let $v_A$ be the vector that is its characteristic function... – Andrés E. Caicedo Dec 28 '10 at 19:13
  • I.e., the $n$-th entry of $v_A$ is 1 or 0, and is 1 iff $n\in A$. Then these vectors are linearly independent, and therefore the dimension of ${\mathbb R}^{\mathbb N}$ is at least $2^{\aleph_0}$. – Andrés E. Caicedo Dec 28 '10 at 19:14
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    You don't "believe" choice, you assume choice. – Asaf Karagila Dec 28 '10 at 19:26
  • I changed the tag from [logic] to [set-theory]; it seems to me that when you mention "logic" in the body, you mean set theory. I also moved this comment so that Andres Caicedo's was not interrupted. – Arturo Magidin Dec 28 '10 at 20:08
  • @Andres: Alternatively, consider for each $a >0$ the sequence $\theta_a = (n^a){n \in \Bbb{N}}$. Because of the different growth behaviour, it is not too hard to see that the family $(\theta_a){a>0}$ is linearly independent. – PhoemueX Jan 11 '19 at 14:18

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(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.

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