Showing a sub-sequence ($r_{n_{k}}$) converges to $x$ There exists a bijection that $f : N → Q, x \in R$
       *$r_{n}$:=$f(n)$
I am asked to show that there exists a sub-sequence ($r_{n_{k}}$) of ($r_{n}$) so that the limit of the sequence ($r_{n_{k}}$) will converges to $x$ as
I know I only need to show ($r_{n}$) converges to $x$ as ($r_{n_{k}}$) is just a sub-sequence of it. But how can I show ($r_{n}$) converges to $x$ and what is it to do with bijection? 
 A: The idea is that every real number can be approximated by rationals, and there are many rationals that approximate the real number for a given tolerance. The only problem is to chose the sequence that 


*

*does not stop prematurely (what if $x$ is rational?), 

*avoids self-repetition

*is convergent from the beginning (as we have to construct a whole sequence and cannot hide behind some 'large enough' argument).


Let me construct the subsequence by induction. Define $n_1:=1$. Assume the strictly increasing sequence of numbers $n_1\dots n_k$ are chosen such that 
$$
|x - r_{n_j}|\le 2^{-j} |x-r_{n_1}| \quad \forall j=1\dots k.
$$
Now the interval
$$
\big[ x - 2^{-(j+1)} |x-r_{n_1}|, \ x + 2^{-(j+1)} |x-r_{n_1}|\big]
$$
contains infinitely many rationals, hence there is some rational number $r$ in this interval that is not in the set of rational numbers already skipped:
$$
r \in\big [ x - 2^{-(j+1)} |x-r_{n_1}|, \ x + 2^{-(j+1)} |x-r_{n_1}|\big] \ \setminus \ \{r_i: i\le n_k\}.
$$
Since $f$ is a bijection, there is $n_{k+1}$ such that $f(n_{k+1})=r$. By construction, $n_{k+1}> n_k$. 
In addition, $n_k\to \infty$ and $r_{n_k}\to x$.
A: Let $n_1 < n_2 < n_3$ ,then the limit of $i$ will converges to x.
By the density of $Q$ , there will be a $n_1$ that $x−1$ < $n_1$ < $x+1$
Therefore,  $x− 1$ < $r_n$ < $x+ 1$
Notices that there exists infinitely many rational numbers belong to the interval and hence, it is always possible.
