I came across an exercise online that showed the following proof. The exercise stated that for this to be true, we have to find a bijection $(0, \infty) \rightarrow {(0,1)}$
Consider the interval (0, ∞) as the positive x-axis of $\mathbb{R}^2$. Let the interval (0, 1) be on the y-axis as illustrated in below, so that (0, ∞) and (0, 1) are perpendicular to each other. The figure also shows a point P = (−1, 1). Define f (x) to be the point on (0, 1) where the line from P to x ∈ (0, ∞) intersects the y-axis. By similar triangles, we have $\frac{1}{x+1} = \frac{f(x)}{x}$ so $f(x) = \frac{x}{x+1}$
I understood the concept of coming up with $f(x)$ via similar triangles.
1. How exactly does this ensure the function is a bijection?
I have managed to prove this function is injective and surjective using its formula (surjectivity applies to $(0, \infty) \rightarrow {(0,1)}$).
Regarding the point $P$, I'm not sure why they chose $-1$ on the x axis. With regards to $1$ on the y axis, I can hypothesize that it's because the codomain is $(0,1)$. Also, how does $$\frac{1}{x+1} = \frac{f(x)}{x}$$ ensure $f(x)$ is a bijection?