Does there exist a bijection $f$ from $\mathbb{N}$ to $\mathbb{Q}^+$ such that $\lim_{n \to \infty} \frac{f(n+1)}{f(n)}$ exists?

Solution 1:

Let us take any enumeration of rationals $q_1, q_2, \ldots$ and make a new one $p_1, \ldots$ such that $\lim\limits_{n \to \infty} \frac{p_{n+1}}{p_n} = 1$.

Let $p_1 = q_1$.

Now assume we already enumerated $m$ points, and $q_n$ is the first one (in $q$) not enumerated yet.

If $q_n > p_m$, let us take some $x \in \left(1; 1 + \frac{1}{n}\right)$ such that points $p_m x, p_m x^2, \ldots, p_m x^k$ (where $k$ is such that $p_m x^k < q_n < p_m x^{k + 1}$) are not enumerated yet. Such $x$ exists, because for each $i < m$ we have only finite amount of numbers $y \in \left(1; 1 + \frac{1}{n}\right)$ such that $p_m y^i = p_i$ for some $i$.

Let $p_{m + 1} = p_m x$, $p_{m + 2} = p_m x^2$, $\ldots$, $p_{m + k} = p_m x^k$, $p_{m + k + 1} = q_n$.

If $q_n < p_m$, then similarly choose $x \in \left(1 - \frac{1}{n}; 1\right)$.