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)?
Best Answer
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.