Limits of sub sequences I am trying to show that there is a sequence $(x_n)$ s.t $\forall r \in \mathbb{Q}$ there exists a sequence of the natural numbers $(r_k)$ s.t $x_{r_k} \rightarrow r$.
My attempt is as follows;
$\mathbb{Q}$ is countable and so we create a bijection from $\mathbb{Q}$ to $\mathbb{N}$.
We know there are infinitely many primes, and let $(p_n)$ be an increasing sequence of primes. 
We now have a bijection from $\mathbb{Q}$ to $(p_n)$ and hence to the primes.
Enumerate the rationals as $(q_n)$.
Now let $x_{p_n^k} = q_n \forall k$. and $(x_n) = 0$ otherwise.
If this proof were correct, I could say that $\forall x \in \mathbb{R}$ there exists a subsequence of $(x_n) \rightarrow x$ as the reals are exactly the set of limit points of the rationals. But the reals are uncountable to I am sceptical of the validity of my proof.
 A: I see no issue with your proof: Suppose you have a sequence $(x_i)_{i \in \mathbb{N}}$ with the desired property. Note that to each real number you can associate a subsequence of your sequence, which can also be identified with a subset of $\mathbb{N}$ (indexing the elements which appear in the subsequence) or an element in the power set $2^\mathbb{N}$, which is uncountable so there is no contradiction.
A: Your proposed proof exhibits a rather complicated enumeration of the rationals without showing how it meets the requirements. You need to explain how the sequence $r_k$ of indexes such that $x_{r_k} \to r$ depends on the given number $r$.
Here is a possible approach: choose any surjection $f : \Bbb{N} \to \Bbb{Q}$ and put $x_n = f(n)$ (your $q_n$ will do: you don't need to involve the sequence of primes). Given any $r \in \Bbb{R}$, define $r_k$ to be the least $k$ such that $r - 1/k \le f(r_k) \le r$ (there is such a $k$ because $f$ is surjective and $\Bbb{Q}$ is dense in $\Bbb{R}$). Then $x_{r_k} = f(r_k)$ tends to $r$ as $k$ tends to infinity.
This works for any $r \in \Bbb{R}$: there are uncountably many sequences $(r_k)$ so there is no contradiction with the uncountability of $\Bbb{R}$. 
A: I don't understand your proof, anyway IMHO it's overcomplicated.
Suppose $X=(x_i)_{i\in\mathbb N}$ is a sequence of all rational numbers. Then choose any rational numer $q_1$ less than $r$. It certainly is in $X$, so let $r_1$ be its index: $x_{r_1} = q_1$.
As $\mathbb Q$ is dense in $\mathbb R$, there is infinitely many rationals in the open interval $(q_1, r)$, and only finitely many from them might appear in $X$ before $q_1$. So you can certainly find some $q_2$ in the interval (and even better: in its upper half), such that $q_2$ appears in $X$ at position (index) $r_2$ greater than that of $q_1$:
$$q_1 = x_{r_1} < \frac{q_1+r}2 < q_2 = x_{r_2}<r \text{ and } r_1 < r_2.$$
Similary you can find $q_3$ in upper half of $(q_2,r)$ such that $q_3 = x_{r_3}$ and $r_2 < r_3$, and so on:
for each $q_i = x_{r_i}$ there exists such $r_{i+1}>r_i$ that $$\frac{q_i+r}2<q_{i+1}=x_{r_{i+1}}<r$$
Then $(r_i)$ is the required sequence of indices which select a subsequence $(q_i)$ from $X$, which converges to any chosen $r$.
And thanks to skipping at least a half of the remaining distance in each step, the sequence converges geometrically, at least.
