Using limits to prove that a subgroup has finite order 
Let $G_\infty = \left(\left\{\dfrac{a}{2^i}  \mid   a \in \mathbb{Z}, i \in \mathbb{N} \right\}, + \right)$, as the group of rationals in $\mathbb{Q}$ with a power of 2 as denominator.
Now consider $X=G_{\infty}/\mathbb{Z}$. Prove $|X| = \infty$ and each proper subgroup of $X$ has finite order.

I'm unsare about the second part, can I use limits in such proofs?
Proof


*

*$X$ has infinite order:
Asume $X$ has finite order, or $|X| = n$. Then for all $g+\mathbb{Z}$ in $X$, $ng+\mathbb{Z} = \mathbb{Z}$. However let $g = \dfrac{a}{2^n}$ for a certain $a \not = 0$ then $ng+\mathbb{Z} \not = \mathbb{Z}$.

*For each $A <  X$ then $|A| < \infty$
I was thinking of proving the following: Let $A \leq X$ then if $|A|=\infty  \Rightarrow A = X$. This is an equivalent statement right?
So choose an $A\leq X$ and let $|A| = \infty$. Now assume $A \not = X$ then there must exist a certain $\dfrac{a}{2^i} + \mathbb{Z} \in X$ which is not in $A$. Notice how $\dfrac{a}{2^{i+1}}+\mathbb{Z}$ is not in $A$ either. If it would, then so would $\dfrac{a}{2^i}+\mathbb{Z} = \left(\dfrac{a}{2^{i+1}}+\mathbb{Z}\right) + \left(\dfrac{a}{2^{i+1}}+\mathbb{Z}\right)$
This means that $$\dfrac{a}{2^i}+\mathbb{Z}, \dfrac{a}{2^{i+1}}+\mathbb{Z}, \ldots \not \in A$$
I'm unsure if I can state the following:
Then $$\lim_{i\to \infty} \dfrac{a}{2^i} + \mathbb{Z} \not \in A$$
however $\mathbb{Z} \in A$, contradiction.
 A: To prove this group has infinite order, why not just exhibit an infinite sequence of distinct members? Thus:
$$
\frac 1 2,\ \underbrace{\frac 1 4, \frac 3 4},\ \underbrace{\frac 1 8, \frac 3 8,\frac 5 8, \frac 7 8},\ \ldots,\ \underbrace{\frac 1 {2^i}, \frac 3 {2^i}, \frac 5 {2^i}, \ldots, \frac{2^n-1}{2^i}},\ \ldots\ldots
$$
To show every element has finite order, first observe that every element is of the form $\dfrac a {2^i}$ for some $n\in\{1,2,3,\ldots\}$ and some $k\in\{1,3,5,7,\ldots,2^i-1\}$ (understanding the fraction to mean the congruence class modulo $1$ to which it belongs). So the problem is to show that the sequence
$$
\frac a {2^i},\  2\frac a {2^i},\  3\frac a {2^i},\  \ldots
$$
ultimately reaches $0.$ But it is $0$ when the integer by which $a/2^i$ is multiplied is itself $2^i.$
ok, So far this shows every cyclic subgroup has finite order. Maybe I'll post something later on how to show that every proper subgroup is cyclic.
It's not hard to show every finite subgroup is cyclic. Identify the members of the subgroups with fractions with powers of $2$ as denominators. Their common
denominator is a power of $2,$ so the members are
$$
\left\{\frac a {2^n} : a\in\text{some finite set} \right\}.
$$
This is a subset of $\left\{\dfrac 1 {2^n}, \dfrac 3 {2^n}, \dfrac 5 {2^n},\ldots,\dfrac{2^n-1}{2^n}\right\},$ and that is a finite set. Show that that finite set is the subgroup. Or even that it includes the subgroup; that is enough.
