What is the cardinality of the set of subgroups of $(\mathbb R, +)$?

Question

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)?

Answer 1

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.

Answer 2

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.