Show that there exists a sequence with subsequences converging to every rational number. 
The total question is:
(a) Give an example of a sequence with subsequences converging to $1, 2$, and $3$.
(b) Give an example of a sequence with subsequences converging to every integer.
(c) Show that there exists a sequence with subsequences converging to every rational
number.

My attempt
For (a) I got $(1,2,3,1,2,3,1,2,3,\dots)$
For (b), I got $(0,1,-1,2,-2,3,-3,4,-4,\dots)$
Both of these I think are right, but I have no idea how to start on (c), much less an example for (c). If someone could explain what I should be doing, I'd be so grateful. Thank you.
 A: Your example for (b) is wrong. There are no convergent subsequences of it.
Consider any countably infinite set $\{e_n : n \in \mathbb{N}\}$. Then consider the sequence $e_1, e_1, e_2, e_1, e_2, e_3, e_1, e_2, e_3, e_4, ...$
For all $n$, this sequence has a subsequence where every element is $e_n$.
Thus, for every countably infinite subset $E \subseteq \mathbb{R}$, we can find a sequence above where for each $e \in E$, there is a subsequence converging to $e$. In particular, we can do this for $\mathbb{Q}$ and $\mathbb{Z}$.
A: The sequence that:

*

*Move up from $-1$ to $1$ by increments of $1/2$

*Then down to $-2$ by increments of $1/3$

*Then up to $3$ by increments of $1/4$

*Then down to $-4$ by increments of $1/5$

*...

is such that for any real $a$ it has a subsequence converging to $a$.
A: For (c), if $n\geq 2$ has the prime factorisation
$$
n=2^\alpha3^\beta 5^\gamma \cdot ...
$$
with $\alpha, \beta, \gamma,....\geq 0$, define
$$
a_n=(-1)^\alpha \frac{\beta}{\gamma+1}
$$
Show that $a_n$ takes each rational number infinitely many times.
