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Can countable dense additive subgroups of the reals be well-ordered up to isomorphism by inclusion?

If so, is $\mathbb{Q}$ the smallest (up to isomorphism) countable dense subgroup of the reals, and what is the second smallest (up to isomorphism)?

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4 Answers 4

up vote 17 down vote accepted

$\{a2^b:a,b\in\mathbb Z\}$ and $\{a3^b:a,b\in\mathbb Z\}$ are both countable dense additive subgroups of the reals, and they are not embeddable in each other (hence $\mathbb Q$ is embeddable in neither).

Also, let $\{p_k:k\in\mathbb N\}$ be an enumeration of primes, and let $A_k$ consist of all fractions $a/b$ of integers such that $b$ is not divisible by $p_0,\dots,p_k$. Then $A_k$ is an additive group, and $A_0\supset A_1\supset A_2\supset\dots$ is an infinite strictly decreasing chain with respect to either inclusion or embeddability. Hence the poset of countable dense subgroups is not even well-founded.

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One can carry the idea of Emil's answer a bit further: for any set $A$ of primes, let $G_A$ be the set of rational numbers $\frac ab$ for integers $a$, $b$ where every prime divisor of $b$ is in $A$. If $A$ is nonempty, then $G_A$ is a countable dense additive subgroup of $\mathbb{R}$, and furthermore $A\subset B$ if and only if $G_A$ is isomorphic to a subgroup of $G_B$.

It follows that the lattice of countable dense additive subgroups of $\mathbb{R}$ includes a copy of the powerset lattice $\langle P(\mathbb{N}),\subset\rangle$.

In particular, since this powerset order is universal for all countable partial orders (by considering the map of a point to its lower cone), it follows that for any countable partial order, one can find a family of countable dense additive subgroups whose subset embedability relation is exactly that order. So this is very far from a well-order.

Indeed, since the rational order embeds this way, by using the corresponding Dedekind cuts, one can find an uncountable chain of countable dense additive subgroups of $\mathbb{R}$ whose subset relation has the order type of the continuum $\langle\mathbb{R},\lt\rangle$.

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This should be a comment to Joel's answer, but it got too long. Joel has exhibited a good part, but not quite all, of the classification of non-trivial subgroups of the rationals (also known as the classification of rank-one, torsion-free, abelian groups), a classification which, if I remember correctly, goes back to Reinhold Baer in the 1930's. For the record, here's the classification. Let $s$ be a function from a subset $D$ of the set of primes into the non-negative integers. Associated to $s$ is the group $G_s$ of those rational numbers expressible as $a/b$ with integers $a$ and $b$ such that, for each prime $p\in D$, the power of $p$ in the prime decomposition of $b$ is at most $p^{s(p)}$. (Primes not in $D$ can occur arbitrarily often in denominators.) Then the non-zero subgroups of $\mathbb{Q}$ that contain the integers are exactly these $G_s$'s. (Note that, up to isomorphism, containing the integers is unimportant, as it can always be achieved by rescaling.) Two of them, say $G_s$ and $G_t$, are isomorphic iff $s$ and $t$ have the same domain and agree at all but finitely many points in that domain. All these groups are dense in the reals, except for those isomorphic to $\mathbb Z$, i.e., those of the form $G_s$ where the domain of $s$ consists of all the primes and $s(p)=0$ for all but finitely many $p$.

Note that all of this concerns only groups of rank 1, i.e., those in which every two elements are linearly dependent over the rationals. For groups of higher but still finite rank, things get more complicated --- in a precise sense: If I remember correctly, the complexity of the isomorphism problem for torsion-free abelian groups of rank $n$ is known to be strictly increasing (in the sense of Borel reducibility) as $n$ increases.

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In the last sentence of my answer, "If I remember correctly" should be replaced by "According to a theorem of Simon Thomas ["The classification problem for torsion-free abelian groups of finite rank," J. Amer. Math. Soc. 16 (2003) 233-258]." –  Andreas Blass Jun 8 '11 at 17:46

If $u,v \in \mathbb R$ are linearly independent over $\mathbb Q$, then $G = \{au+bv : a,b\in\mathbb Z\}$ is dense.

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