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It is a well known pathological counterexample in group theory that (assuming the axiom of choice), we have an isomorphism$$\mathbb{R} \cong \mathbb{R}^2,$$ where these are additive groups. This somewhat unsettling result generally is shown by appealing to fact that both of these groups, when viewed as $\mathbb{Q}$-vector spaces, have the same dimension. Naturally, showing this relies on the axiom of choice.

As with most results coming from the axiom of choice, it is strongly insinuated that one cannot explicitly construct an isomorphism $\phi : \mathbb{R} \rightarrow \mathbb{R}^2$.

This is very believable, but I'm wondering if it has actually been shown that this fact is equivalent to choice (i.e. without choice, the two are not isomorphic). If so, what would a proof of that look like? All comments are appreciated.

Arturo Magidin
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kodiak
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    Well a full solution would be a proof that $R\cong R^2$ implies (some weaker form of) the axiom of choice. (It's got to be a weaker form because AC says something about all cardinalities, whereas this problem is only talking about a specific cardinality). A partial solution would be something like the axiom of determinacy implies $R\ncong R^2$ (AD is an alternative axiom to AC that disproves full AC but implies a lot of regularity results). I don't know how to prove either though. – J.D. Feb 21 '24 at 01:19
  • It's quite plausible that their isomorphism is independent of ZF. This is the case of other consequences of choice, e.g., for the well-order of reals. – Soundwave Feb 21 '24 at 01:33
  • @Soundwave That's what seems the most plausible to me. – kodiak Feb 21 '24 at 01:35
  • The normal proof of isomorphism goes through only with $\mathsf{DC}{\omega_2}$. Since the basis is necessarily uncountable, there is probably a model accepting $\mathsf{DC}{\omega_1}$ without this isomorphism, although constructing it is beyond my skill. As well, although old, afaik the commentary of this question still holds, making it open if the existence of the isomorphism implies any sort of choice (it seems unlikely, imo). – Soundwave Feb 21 '24 at 03:09
  • https://mathoverflow.net/q/25375/42153 – Shaun Feb 21 '24 at 20:53

1 Answers1

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It's definitely not equivalent to full choice, because "choice can start to fail arbitrarily high up the von Neumann hierarchy". One way to see it's independent of $\mathsf{ZF}$ (ie "requires some choice") is to note that composing an isomorphism $\Bbb R \to \Bbb R^2$ with a projection map $\Bbb R^2 \to \Bbb R$ will give a non-injective nontrivial homomorphism $\Bbb R \to \Bbb R$. In other words, this gives a discontinuous solution to the Cauchy functional equation. And it's known to be consistent with $\mathsf{ZF}$ that there's no such function.

Izaak van Dongen
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  • This is very good. Could you elaborate a bit more on why your quote shows that this couldn't be equivalent to $\textit{full}$ choice? Would it just be because we are only dealing with the cardinality of the continuum? – kodiak Feb 21 '24 at 01:39
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    @kodiak, yes indeed, the point is that in order to prove $\Bbb R^2 \cong \Bbb R$, you only need to apply choice to quite a small set - for example "the family of $\Bbb Q$-linearly independent sets in $\Bbb R$ that are not spanning sets". This family lives very early on in the hierarchy, so if you just ask for choice to start failing at $V_{\omega + 10}$ or something, you can produce a model where choice fails but $\Bbb R^2 \cong \Bbb R$ holds. – Izaak van Dongen Feb 21 '24 at 01:49
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    Makes great sense. Thank you! – kodiak Feb 21 '24 at 01:55