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Using the axiom of choice it is provable that $\mathbb{R}$ is isomorphic to $\mathbb{Q}^\mathbb{N}$ as a vector space over $\mathbb{Q}$. (Assuming AC, both spaces have a Hamel basis over $\mathbb{Q}$ of the same cardinality and are thus isomorphic.)

So my question is whether such an isomorphism between $\mathbb{R}$ and $\mathbb{Q}^\mathbb{N}$ can be constructed without AC or, at least, whether we can embed $\mathbb{Q}^\mathbb{N}$ into $\mathbb{R}$ without AC. (By embedding I mean constructing an injective $\mathbb{Q}$-linear map from one space into the other.)

The latter is equivalent to asking whether we can construct a subspace of $\mathbb{R}$ that has a schauder-basis over $\mathbb{Q}$, as such a subspace should automatically be isomorphic to $\mathbb{Q}^\mathbb{N}$.

Thanks for the help!

Hanul Jeon
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H.D. Kirchmann
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    I would expect, that you cannot give such a map. But you have a surjection from the subvectorspace of Cauchy sequences to $\mathbb R$. – Julian Quast Dec 03 '20 at 11:14
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    I also suspect that such a map cannot be constructed without assuming AC or any of its consequences. However I don't see how one would prove this. Presumably by considering a symmetric submodel of some forcing extension, but finding a suitable forcing notion seems quite tricky. – H.D. Kirchmann Dec 03 '20 at 11:39
  • I seems, that $\mathbb R$ makes this particularly difficult. Did you think about similar questions for $\bigoplus_{\alpha \in 2^{\mathbb N}} \mathbb Q$ and $\mathbb Q^{\mathbb N}$? – Julian Quast Dec 03 '20 at 11:47
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    Thanks for the suggestion and I agree, $\mathbb{R}$ does make this question particularly difficult. However I am especially interested in the "cauchy"-aspect of the question, so instead of $\mathbb{R}$ one might also consider the space of all cauchy-sequences over $\mathbb{Q}$ (i.e. not necessarily collapsing sequences which converge to the same real). What makes this problem interesting to me is that a solution might provide some way of turning arbitrary sequences in $\mathbb{Q}$ into cauchy sequences. – H.D. Kirchmann Dec 03 '20 at 12:27

1 Answers1

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In fact it's consistent with ZF that there are no nontrivial homomorphisms $\mathbb{R} \to \mathbb{Q}$. Quoting from a previous answer where this came up:

There's a model of ZF constructed by Shelah in which every set of real numbers has the Baire property. This implies, if I understand correctly, that there are no nonzero homomorphisms from $\mathbb{R}$ to any countable abelian group (since any countable abelian group with the discrete topology is a Polish group, so in this model any homomorphism from $\mathbb{R}$ to such a group is automatically measurable and so automatically continuous). So $\mathbb{R}$, and $SO(2)$, have no subgroups of countable index in this model.

This doesn't rule out the possibility of an explicit embedding $\mathbb{Q}^{\mathbb{N}} \to \mathbb{R}$; I'm not sure one way or the other whether such a thing exists but off the top of my head I'd bet it doesn't. I'd bet it's consistent with ZF that every linear map $\mathbb{Q}^{\mathbb{N}} \to \mathbb{R}$ factors through the projection to some finite subset of its factors.

Qiaochu Yuan
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