real analysis - Showing $left[ {0,1} right) sim left( {0,1} right)$, $left( {a,b} right) sim R$, and $left( {a,infty } right) sim R$

Showing Cardinality between sets.
(a)$\left( {a,\infty } \right) \sim R$
(b)$\left[ {0,1} \right) \sim \left( {0,1} \right)$
(c)$\left( {a,b} \right) \sim R$
For (a) since $\left( {a,\infty } \right) \subseteq R$, I can use $f\left( x \right) = {e^x}$ which has domain $R$ and range $\left( {0,\infty } \right)$, ans show $\left( {0,\infty } \right) \sim R$, and I know two open intervals are cardinal $\left( {a,\infty } \right) \sim \left( {0,\infty } \right)$ therefore $\left( {a,\infty } \right) \sim R$. Correct?
For (b) I can use $f\left( n \right) = {1 \over {n + 1}}$ and shift $1$ of $\left( {0,1} \right)$ to ${1 \over 2}$ and show the function is one to one. How do I show it is onto?
For (c) $\left( {a,b} \right) \subseteq R$ => $\left( {a,b} \right) \to R$ is one to one. Now, since $\left( {a,b} \right) \to \left( {c,d} \right)$ is onto for some $\left( {c,d} \right)$ in $R$ so is onto? So, $\left( {a,b} \right) \sim R$ ? Correct?

Answer

For (c). Take $\tan x:(-\dfrac{\pi}{2},\dfrac{\pi}{2})\to(-\infty,\infty)$ is bijective, so $\tan\dfrac{\pi}{2}x:(-1,1)\to\mathbb{R}$ and with $g(x)=\dfrac{2}{b-a}x+\dfrac{a+b}{a-b}:(a,b)\to(-1,1)$ thus $\tan\dfrac{2}{\pi}g(x):(a,b)\to\mathbb{R}$ is bijective.