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**Pinkk** Let $\displaystyle (r_{n})$ be an enumeration of the set $\displaystyle \mathbb{Q}$. Show that there exists a subsequence $\displaystyle (r_{n_{k}})$ such that $\displaystyle \lim_{k \to \infty}\, r_{n_{k}}= +\infty$.

**Proof**

Choose $\displaystyle n_{1}\in \mathbb{N}$ such that $\displaystyle r_{n_{1}} > 1$. (This is possible by the use of the Archemedian property.) Choose $\displaystyle n_{2} > n_{1}$ such that $\displaystyle n_{2}$ is the smallest natural number that satisfies $\displaystyle r_{n_{2}} > 2r_{n_{1}}$. Then $\displaystyle r_{n_{2}} > 2$ and $\displaystyle r_{n_{2}} > r_{n_{2}}$. Now choose $\displaystyle n_{3} > n_{2}$ such that $\displaystyle n_{3}$ is the smallest natural number that satisfies $\displaystyle r_{n_{3}} > 3r_{n_{2}}$. Then $\displaystyle r_{n_{3}} > 6$ and $\displaystyle r_{n_{3}} > r_{n_{2}}$. Assume $\displaystyle n_{1}< n_{2}< ...< n_{k}$ have been selected such that $\displaystyle r_{n_{k}} > k!$ and $\displaystyle r_{n_{k}} > r_{n_{k-1}}$. Select $\displaystyle n_{k+1} > n_{k}$ such that $\displaystyle n_{k+1}$ is the smallest natural number that satisfies $\displaystyle r_{n_{k+1}} > (k+1)r_{n_{k}}$. This means $\displaystyle r_{n_{k+1}} > (k+1)!$ and $\displaystyle r_{n_{k+1}} > r_{n_{k}}$. By induction, $\displaystyle r_{n_{1}} < r_{n_{2}} < ... < r_{n_{k}}$ and $\displaystyle r_{n_{k}} > k!$ for all $\displaystyle k\in \mathbb{N}$. So given $\displaystyle M > 0$, choose $\displaystyle N=M$. And so $\displaystyle k > N$ implies $\displaystyle r_{n_{k}} > k! \ge k > M$. Hence, $\displaystyle \lim_{k\to \infty}\, r_{n_{k}} = +\infty$. **Q.E.D.**

Is this proof correct; can I construct the subsequence in such a way? And if not, how should I go about this proof? Thanks.