Using the axiom of choice, one can show that $\mathbb{R}$ and $\mathbb{R}^2$ are isomorphic as additive groups. In particular, they are both vector spaces over $\mathbb{Q}$ and AC gives bases of these two vector spaces of cardinalities $c$ and $c\times c = c$, so they are isomorphic as vector spaces over $\mathbb{Q}$.
Is there a way to prove that $\mathbb{R}$ and $\mathbb{R}^2$ are isomorphic as additive groups which does not use AC? Are there models of set theory in which these groups are not isomorphic? I'm also curious whether there is a proof (perhaps using AC) which does not make use of this vector space machinery, though I'm fairly sure the above proof is the simplest.
Best Answer
You cannot prove that $\mathbb{R}$ and $\mathbb{R}^{2}$ are isomorphic in $ZF$. To see this, note that the map $(x,y) \mapsto (x,0)$ is a nontrivial non-surjective additive endomorphism of $\mathbb{R}^{2}$. Assuming the existence of suitable large cardinals, every map $f: \mathbb{R} \to \mathbb{R}$ is measurable in $L(\mathbb{R})$ and it follows that the only additive endomorphisms of $\mathbb{R}$ in $L(\mathbb{R})$ are the maps $x \mapsto rx$ for some $r \in \mathbb{R}$. Thus $\mathbb{R} \not \cong \mathbb{R}^{2}$ in $L(\mathbb{R})$.
There is an interesting open problem related to your question of whether an isomorphism can be found which does not make use of the vector space machinery. If we add a Ramsey ultrafilter $\mathcal{U}$ to $L(\mathbb{R})$, then we obtain an interesting model $L(\mathbb{R})[\mathcal{U}]$ which contains the ultrafilter $\mathcal{U}$ while still retaining some of the "nice" properties of $L(\mathbb{R})$. In particular, neither $\mathbb{R}$ nor $\mathbb{R}^{2}$ has a basis as a vector space over $\mathbb{Q}$ in $L(\mathbb{R})[\mathcal{U}]$ . However, it is unknown whether or not $\mathbb{R} \cong \mathbb{R}^{2}$ in $L(\mathbb{R})[\mathcal{U}]$. In other words, does an ultrafilter help in trying to construct such an isomorphism?