functions - Constructing a bijection between $[a,b)$ to $(0,1)$

Construct a bijection between $[a,b)$ to $(0,1)$

I'll be using a technique that builds the function in steps from one interval to the next:

$[a,b)\overset{x-a+1}\to[1,b-a+1)\overset{\frac 1 {x+1}}\to(0,1)$

Now composing the function from the end to the beginning: $f(x)=\frac 1 {1+()}=\frac 1 {1+(x-a+1)}=\frac 1 {x-a+2}$

Now since all the operations were linear we have a bijection, but is it right?

EDIT:

$[a,b)\overset{x-a}\to[0,b-a)\overset{\frac x {b-a}, x\neq0 , 0.5, x=0}\to(0,1)$

$f(x)=\begin{cases} \frac {(x-a)} {b-a} &x\neq0 \\ 0.5 &x=0\end {cases}$

Answer

I would suggest taking different steps here: First, show $[a,b)\leftrightarrow[0,1)$, and then $[0,1)\leftrightarrow(0,1)$.

The first one is just repositioning and scaling of the interval; you will find the bijection $$f: [a,b)\leftrightarrow[0,1), x\mapsto \frac{x-a}{b-a}$$

Now, we just have to “insert” the element $0$ into the open interval $(0,1)$. The “trick” I will use here is the concept of the Hilbert Hotel, which basically takes an infinite (bijective) sequence $(a_n)_{n\in\mathbb N}$ in this interval, mapping every element $a_n$ to $a_{n+1}$ and “inserting” $0$ to $a_0$, which is now “empty” because of the shifting we did.
Therefore, we can define the following bijection:
$$g: [0,1)\leftrightarrow(0,1), x\mapsto\begin{cases}a_0 & x=0\\a_{n+1} & \exists n\in\mathbb N: x=a_n\\x & \text{else}\end{cases}$$
Where $∀ n\inℕ:\left(a_n\in(0,1) \land \not∃ m\inℕ\setminus\{n\}:a_n=a_m \right)$.

By the way, the inverse is
$$g^{-1}: (0,1)\leftrightarrow[0,1), x\mapsto\begin{cases}0 & x=a_0\\a_{n-1} & \exists n\in\mathbb N\setminus\{0\}: x=a_n\\x & \text{else}\end{cases}$$