We can show bijection between [a,b] and [c,d] in $\mathbb(R)$. But how can one establish an bijection between $[a,b] and (c,d)$.

Question

We can show bijection between [a,b] and [c,d] in $\mathbb(R)$ by send $x \in [a,b]$ to $(d-b)/(c-a)x + bc-da$. But how can one establish an bijection between $[a,b]$ and $(c,d)$ can we explicitly define a function?

Answer 1

Here is an explicit bijection from $[0,1]$ to $(0,1)$. I leave the details for translating this to $[a,b]$ as an exercise.

Define $f\colon[0,1]\to (0,1)$ as follows. $$f(x)=\begin{cases} 1/2 &\text{if } x=0\\ 1/2^{n+1} &\text{if } x=1/2^n \text{ for } n>0\\ 1/3 &\text{if } x=1\\ 1/3^{n+1} &\text{if } x=1/3^n \text{ for } n>0\\ x &\text{otherwise} \end{cases} $$

Intuitively, you are finding two infinite sequences in your set and using them to make room for the two extra points. You could also just use one infinite sequence and make room by advancing each element by two rather than just one.

Answer 2

By what you said, it is sufficient to define a bijection $f:[c,d]\to(c,d)$.

Let $x_0=c$, and for $n\in\mathbb N^*$, $x_n=c+\frac{d-c}n$.

Put $f(x_n)=x_{n+2}$ for all $n\in\mathbb N$, and $f(x)=x$ for any other $x\in[c,d]$.