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Let $X$ be the set of subgroups of $(\mathbb R, +)$. What is $|X|$?

An attempt at a proof that $|X| = 2^{2^{\aleph_0}}$:

Clearly $|X| \le 2^{2^{\aleph_0}}$, because $X \subset P(\mathbb R)$. For a lower bound, let $H$ be a Hamel basis of $\mathbb R$ over $\mathbb Q$. Since $H$ is linearly independent over $\mathbb Q$ (and thus also over $\mathbb Z$), every subset of $H$ generates a distinct subgroup of $\mathbb R$. Since $|H| = 2^{\aleph_0}$ (I think), the set of all such groups then has cardinality $2^{|H|} = 2^{2^{\aleph_0}}$. Since this is a subset of $X$, we must have $2^{2^{\aleph_0}}\le |X|$. Thus, $|X| = 2^{2^{\aleph_0}}$.

Does this work? Also, is there a way to prove this that isn't quite so reliant on Choice (needed for the Hamel basis of $\mathbb R$ and possibly some of the cardinality comparisons)?

eyeballfrog
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  • Your claim that $|H| = 2^{\aleph_0}$ is correct, assuming choice; as proved here by Asaf Karagila, a vector space $V$ over a field $F$ with a basis $H$ satisfies $|V| = \max(|F|, |H|)$. Since $F= \mathbb{Q}$ is countable and $V = \mathbb{R}$ is not, we must have $|H| = |\mathbb{R}|$. – Nate Eldredge May 29 '19 at 20:50
  • Proposition B of my JLMS paper with Guyot and Pitsch https://arxiv.org/abs/0811.1549 says (in ZFC) that for any uncountable abelian group $A$ of cardinal $\alpha$, the space $S(A)$ of subgroups of $A$ (with the compact topology induced by inclusion in $2^A$) has a subset homeomorphic to $2^\alpha$ (and in particular has cardinal $2^\alpha$). The proof in general is essentially the same as the one in the case of $\mathbf{R}$; in this latter case, as mentioned by Asaf, one can make the construction explicit enough to work in ZF (what's abusively called "without choice"). – YCor May 30 '19 at 08:39

2 Answers2

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First, note that there is a set of size $2^{\aleph_0}$ of real numbers which is linearly independent over $\Bbb Q$, even without the axiom of choice.

Then do the same proof as you did, as it's fine.

For the first part, Is there any uncountably infinite set that does not generate the reals? is of interest towards a positive answer.

Asaf Karagila
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You can (not) modify your proof slightly to get rid of any (obvious) use of choice.

Define an equivalence relation on $\mathbb R$ by $x \sim y \iff x-y \in \mathbb Q$. This partitions $\mathbb R$ into equivalence classes of the form $[x] = x + \mathbb Q = \{x+q: q \in \mathbb Q\}$. It's easy to see each $[x]$ is a subgroup of $\mathbb R$.

Since $\mathbb Q$ is countable so is each equivalence class. Write $(\mathbb R/\sim)$ for the collection of equivalence classes. Observe $\bigcup (\mathbb R/\sim)$ is a disjoint union of countable sets. So the union has cardinality $|\mathbb N| \times |(\mathbb R/\sim)| = |(\mathbb R/\sim)|$. But since the union is just $\mathbb R$ we get $|(\mathbb R/\sim)| = |\mathbb R|$ as required.

AsdrubalBeltran
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Daron
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