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?

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.