# Can $\mathbb{R}$ be written as an ascending union of proper additive subgroups?

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 $\cup {H_i}=\mathbb{R}$?)

• You want this to be a countable union, presumably? – Sammy Black Nov 13 '13 at 3:40
• @Sammy: We expect countable union. Otherwise, as $\mathbb{R}$ is ordered set, we can certainly find a chain of subgroups, ordered according to the order in $\mathbb{R}$, with union equal to $\mathbb{R}$. (Thanks for suggestion; I have edited question) – Beginner Nov 13 '13 at 3:47
• Nice problem.${}$ – Alexander Gruber Nov 13 '13 at 3:58
• It may help to realize that every proper subgroup of $\;\Bbb R\;$ is either cyclic or dense (wrt the usual Euclidean topology)...It can't be an ascending union of cyclic ones because of Baire's Theorem, and I doubt it is a union of the second kind, or a mix of both...nice little bastard problem! +1 – DonAntonio Nov 13 '13 at 5:03
• I wonder whether the OP's nick comes from Marshall (Hall) and (Aleksander Gennadievich) Kurosh, two monsters of group theory from the last century... – DonAntonio Nov 13 '13 at 5:09

The real numbers $\mathbb{R}$ is a vector space over $\mathbb{Q}$. By axiom of choice there is a basis $B$. Let $S_i$ be the set of all rational numbers $a/b$ where the prime factors of $b$ are among the $i$ first primes. The set of real numbers with coordinates (for the basis $B$) from $S_i$ is a subgroup $G_i$. It is clear from the definition that $G_i\subset G_{i+1}$. Now $B$ is a basis and hence for every real number there is some $G_i$ containing it. Or am I misunderstanding the question?

• Neato. ${}{}{}$ – anon Nov 13 '13 at 6:17
• Wouldn't it be simpler to partition the basis $B$ into a countable infinity of disjoint nonempty sets $B_n(n\in\mathbb N)$ and define $G_n$ to be the vector space over $\mathbb Q$ spanned by $B_1\cup\dots\cup B_n$? – bof Nov 13 '13 at 8:11
• Yes you are right! – Patrick Nov 13 '13 at 8:13

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.)

Taking a cue from the other answer, if $M$ is the $\Bbb Z$-submodule of $\Bbb R$ spanned by a $\Bbb Q$-vector space basis and $n_1, n_2,\cdots$ any increasing cofinal sequence in the naturals ordered by divisibility (for example factorials as $n_k=k!$), then we have $M\subseteq n_1^{-1}M\subseteq n_2^{-1}M\subseteq\cdots$ with union $\Bbb R$.