If $(q_n)_{n\ge 1}$ is an enumeration of $\mathbb{Q}$, is there any limit to how fast it can grow? Precisely stated, what I mean is this: Define the running max $Q_n = \max_{k\le n} |q_k|$. Is there a function $f$ such that $Q_n = O(f(n))$ is guaranteed for any enumeration? On the lower bound, is there any function $g$ such that $g(n) = O(Q_n)$ is guaranteed?
It seems to me like the answer would be no, since we could somehow 'contract' the rationals by some bijection $h:\mathbb{Q}\rightarrow\mathbb{Q}$ such that $h$ grows as slow as you want, and then $h(q_n)$ would similarly grow as slow as you want. Similarly for the upper bound. Is this reasoning correct? If so, can you show that arbitrarily slow-growing bijections from $\mathbb{Q}$ to itself exist?
Given any function $f: \mathbb N \to \mathbb R_+$, you can ensure that $Q_n \ne O(f(n))$ by taking $q_N > N f(N)$ for some infinite sequence of odd $N$'s. This still leaves infinitely many $n$ to use to enumerate all the other rationals.
On the other hand, suppose you're given $g: \mathbb N \to \mathbb R_+$ with $g(n) \to \infty$ as $n \to \infty$. To ensure $g(n) \ne O(Q_n)$, you can proceed as follows. For simplicity I'll assume $g$ is nondecreasing.
Start with an arbitrary enumeration $r$. Given $n$, let $k$ be the first positive integer such that $r_k \notin \{q_1, \ldots, q_{n-1}\}$ and $|r_k| < \sqrt{g(n)}$, and take $q_n = r_k$ for this $k$. By construction, $Q_n < \sqrt{g(n)}$, and it is easy to see that every rational will eventually be enumerated.