A question about cardinality

Question

Let $A$ be a subgroup of $\mathbb{R}$.

Assume $|\mathbb{R}\setminus A|=\aleph_0$.

Is it true that for every $a,b\in\mathbb{R}$, $a<b$

$(a,b] \cap A \neq \varnothing$ ?

I think it's true, but I'm not really sure. Can anyone shed light on this?

Answer 1

Since, $|\mathbb{R}\setminus A|=\aleph_0$, so, $(\mathbb{R}\setminus A) $ must be a countable set in $\mathbb{R} $.
For supose there are some $a,b\in\mathbb{R}$ with $a<b$ such that $(a,b] \cap A = \varnothing$, then clearly $(\mathbb{R}\setminus A) $ contains a interval, which means that containing uncountable numbers of elements of $\mathbb{R} $ , but this can't be possible, as $(\mathbb{R}\setminus A) $ is a countable set in $\mathbb{R} $.
So, our assumption is wrong! Hence for every $a,b\in\mathbb{R}$, $a<b$, $(a,b] \cap A \neq \varnothing$.

Answer 2

If $A$ avoids an open interval $(a,b)$, then it is not dense. The only non-dense subgroups of $\Bbb R$ are of the form $a\Bbb Z$ (possibly $a=0$). So $\Bbb R/A$ is homeomorphic to $S^1$ (or $\Bbb R$ itself) und uncountable.