TITLE: Can $\mathbb{R}$ be written as an ascending union of proper additive subgroups?
QUESTION [10 upvotes]: Can the group $\mathbb{R}$ be written as countable ascending union of proper subgroups? (i.e. does there exists a series of proper subgroups $H_1\leq H_2\leq \cdots $ such that $\cup {H_i}=\mathbb{R}$?)

REPLY [4 votes]: The constructions in the other answers use the axiom of choice, and this can't be avoided.  The subgroups in question have to be pathological.
It is impossible to construct such a series of subgroups, all of which have the Baire property.  By the Baire category theorem, one of them $G_n$ would have to be nonmeager.  A result due to Pettis says that for any nonmeager set $A \subset \mathbb{R}$ (or any Polish group) which has the Baire property, $A - A$ contains an open neighborhood of 0.  So $G_n = G_n - G_n$ contains an open neighborhood of 0; hence it must equal $\mathbb{R}$.
It is consistent with the axiom of dependent choice that every subset of $\mathbb{R}$ has the Baire property (Shelah's model).
It is impossible to construct such a series of subgroups, all of which are Lebesgue measurable.  By countable additivity, one of them $G_n$ would have to have positive measure.  A theorem of Steinhaus says that for any measurable set $A$ of positive Lebesgue measure, $A - A$ contains an open neighborhood of 0.  So as before, we would have to have $G_n = \mathbb{R}$.
It is consistent with the axiom of dependent choice, together with some large cardinal axioms, that every subset of $\mathbb{R}$ is Lebesgue measurable. (Solovay's model.)
(Disclaimer: I'm parroting some set theory that I've heard about but not studied deeply.  An actual expert is welcome to correct mistakes or fill gaps.)