Okay so I know that I asked this already but I want to ask how much progress I have made. so the question is
Construct a bijection between $[1,2]$ and $[3,5)$
So I have: \begin{equation} f(x) = \left \{ \begin{array}{ll} 3 + 2^{1-n} & \textrm{ if } x = 1+2^{1-n} \textrm{ for } n \in \mathbb{N} \\ 2x+1 & \textrm{ if } x \neq 1+2^{1-n} \end{array} \right. \end{equation}
Sorry about the formatting, but is this correct?
Edit: the linked post is mine. I'm just asking since I didn't get a really satisfactory answer if the solution I have is going towards the right direction.
Edit2: changing the equation to what improvements I'm getting, still any feedback is helpful. Thanks!
The function $f$ is onto:
Case 1
Let $y \in [3,5) $ be of the form: $3 + 2^{1-n}$ for some $n \in \mathbb{N}$. Then $x = 1 + 2^{1-n}$ is such that $f(x) = y$. In particular, if $n$ is such that $3 + 2^{1-n} \in [3,5)$ then $n \geq 0$. This in turn implies that $1 \leq 1 + 2^{1-n} \leq 2$.
Case 2
Let $y \in [3,5) $ not be of the form: $3 + 2^{1-n}$ for some $n \in \mathbb{N}$. Then $x = (y-1)/2 \in [1,2]$ is such that $f(x) = y$. In particular if $y \in [3,5) $ then $(y-1)/2 \in [1,2]$.
The function $f$ is inyective:
Let $a,b \in [1,2]$ be such that $f(a) = y = f(b)$
Case 1
If $y \neq 3 + 2^{1-n}$ then $f(a) = 2a + 1 = 2b + 1 = f(b)$. It follows that $a = b$.
Case 2 If $y = 3 + 2^{1-n}$ then $f(a) = 3 + 2^{1-n_a}$ and $f(b) = 3 + 2^{1-n_b}$ where $a = 1 + 2^{1-n_a}$ and $b = 1 + 2^{1-n_b}$ respectively. Then: \begin{equation} \begin{aligned} 3 + 2^{1-n_a} = 3 + 2^{1-n_a} & \Leftrightarrow 2^{1-n_a} = 2^{1-n_a} \\ & \Leftrightarrow log_2\left(2^{1-n_a}\right) = log_2\left(2^{1-n_b}\right) \\ & \Leftrightarrow n_a = n_b \\ & \Leftrightarrow a = b. \end{aligned} \end{equation}