Continued fraction of a sequence is unique

Question

Let $X\in\mathcal{P}(\mathbb{N})$ and define a continued fraction $$f(X)=\cfrac{1}{x_1+\cfrac{1}{x_2+\cfrac{1}{x_3+\cfrac{1}{x_4+\cdots}}}}$$ for each element $x_i\in X$. I'm wondering if $f:\mathcal{P}(\mathbb{N})\rightarrow [0,1]$ is injective. If not, would something like $$g(X)=\cfrac{|X|}{x_1+\cfrac{1}{x_2+\cfrac{1}{x_3+\cfrac{1}{x_4+\cdots}}}}$$ for a finite $X$ or $$h(X)=\cfrac{\bar{d}(X)}{x_1+\cfrac{1}{x_2+\cfrac{1}{x_3+\cfrac{1}{x_4+\cdots}}}}$$ for an infinite $X$, where $\bar{d}(A)$ is the upper asymptotic density of $X$, work as an injective function?

Answer 1

$f$ is injective (see https://en.wikipedia.org/wiki/Continued_fraction). Infinite sets produce irrational numbers, finite sets rational numbers.

Each irrational number is the value of a unique infinite continued fraction.

Each rational number can be represented in exactly two ways as a finite continued fraction, but only in one of them the $x_i$ have a chance to be increasing.