A sequence $x_n$ such that $\forall r \in\mathbb Q$, there is a subsequence of $x_n$ which converges to $r$ My question is to find a sequence $x_n$ such that $\forall r \in\mathbb Q$, there is a subsequence of $x_n$ which converges to $r$. I'll be trying to do the same with $\mathbb R$.
I decided to simplify it first by finding a sequence $y_n$ such that $\forall n \in\mathbb N$, there is a subsequence of $x_n$ which converges to $n$. Namely, I constructed the sequence $1,1,2,1,2,3,1,2,3,4,1,2,3,4,5....$ because you can easily construct a subsequence that converges to $n$ by taking the subsequence $n$ recurring.
I'm trying to think similarly with $\mathbb Q$. Any help is appreciated.
 A: You can use the same trick for $\mathbb Q$ that you used for $\mathbb N$: if $(q_1,q_2,q_3,\ldots)$ is any enumeration of $\mathbb Q$, then the sequence
$S=(q_1,q_1,q_2,q_1,q_2,q_3,q_1,q_2,q_3,q_4,q_1,q_2,q_3,q_4,q_5,\ldots)$ does the job.
In fact, given any sequence $T=(r_1,r_2,r_3,\ldots)$ of rational numbers, the sequence $S$ contains the sequence $T$ as a subsequence. So given $x\in\mathbb R$, choose a rational sequence $T$ converging to $r$, and then $S$ contains $T$ as a subsequence that converges to $x$.
A: HINT: A version for $[0,1]$. Let $a_1=0$, $a_2=1$. Then every next $a_i$ is the centre of obtaining intervals, i.e., $a_3=1/2$, $a_4=1/4$, $a_5=3/4$, $a_6=1/8$ and so on.
A: Any enumeration of $\mathbb Q$ will work. Proof: Let $q_1,q_2,\dots $ be such an enumeration. Let $x\in \mathbb R.$ Denote by $n_1$ the smallest $n$ such that $q_n\in (x,x+1).$ Then let $n_2$ be the smallest $n>n_1$ such that $q_n\in (x,x+1/2).$ Then let $n_3$ be the smallest $n>n_2$ such that $q_n\in (x,x+1/3),$ etc. The subsequence $q_{n_k}$ then converges to $x.$
