It’s clear that $(\mathbb{R}, +)$ is a divisible group. By the structure theorem of abelian divisible groups, it follows that it’s isomorphic to a direct sum of copies of $\mathbb{Q}$, i.e., $\mathbb{R} \cong \mathbb{Q}^{(\kappa)}$, since $\mathbb{R}$ is torsion-free (all of its elements except the identity $0$ are of infinite order), so there are no Prüfer $p$-groups in its decomposition. My question is about the value of $\kappa$. I assume it can’t be $\aleph_{0}$, since the elements of a countable direct sum of copies of $\mathbb{Q}$ have finite support, so $|\mathbb{Q}^{(\aleph_{0})}| = \aleph_{0}$. Therefore, I guess that $\kappa = 2^{\aleph_{0}}$, but just because of cardinality reasons, so I’m not sure about it. Another speculation is that this could be a pure algebraic construction of the “real numbers”, but just seen as an additive group, so I don’t now if it could inherit the order of the reals or the product in an algebraic natural way.
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1$\begingroup$ Are you asking for the cardinality of a Hamel basis? The cardinality of a Hamel basis is the continuum. $\endgroup$– LucenapositionCommented Sep 4 at 23:39
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$\begingroup$ @Lucenaposition I guess, so then $\kappa = 2^{\aleph_{0}}$ $\endgroup$– JuanClaverCommented Sep 5 at 0:02
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4$\begingroup$ Yes, and that's true ($\kappa = 2^{\aleph_0}$) whether or not $\mathsf {CH}$ holds. You are in fact looking for the cardinality of a (Hamel) basis for $\Bbb R$ considered as a vector space over $\Bbb Q$. $\endgroup$– Robert ShoreCommented Sep 5 at 0:50
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$\begingroup$ @RobertShore I see. Thank you. $\endgroup$– JuanClaverCommented Sep 5 at 1:07
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