I have to prove it for these sets and find a bijection:
a) $(a,b) \sim [a,b] \sim [a,b) \sim (a,b]$
where: $a,b\in R$
with $a<b$
b) $[a,b] \sim [c,d]$
where: $a,b,c,d \in R$
with $a<b ,c<d$
I know that:
Two sets are said to be 'equipotent' if there is a bijection between them
for a) $f: (a,b) -> [a,b]$
$f(x)=x/\vert x\vert$
for the b) I've found on some notes (without showing why), that a good bijection could be the map: $f(x)=kx+(c-ak)$
where:
$k =\frac{d-c}{b-a}$
I need how to show the assert and a write a bijection
If $A$ is a set, $|A|$ is its cardinality.
The map $f \colon [a,b] \to [c,d]$ that you defined for (b) is bijective because it is the straight line of $\mathbb{R}^2$ that contains the points $(a,c)$ and $(b,d)$ (verify that $f(a)=c$ and $f(b)=d$).
For (a), notice that, being $(a,b) \subset [a,b]$, it holds that $|(a,b)| \leq |[a,b]|$. Now consider $c,d$ such that $a < c < d < b$. Then $[c,d] \subset (a,b)$ and $|[c,d]| \leq |(a,b)|$. Since $f$ of (b) is a bijection $[a,b] \to [c,d]$, it holds that $|[a,b]|=|[c,d]| \leq |(a,b)|$. Being $|(a,b)| \leq |[a,b]|$ and $|[a,b]| \leq |(a,b)|$, by Cantor-Bernstein-Schröder Theorem there is a bijection $[a,b] \to (a,b)$.
Notice that $(a,b) \subset [a,b) \subset [a,b]$. Then $|(a,b)| \leq |[a,b)| \leq |[a,b]|=|(a,b)|$. By the same Theorem, there is a bijection $[a,b) \to (a,b)$. We can do the same for $(a,b]$.