With more specific saying,
if there is a function that is in continuum from a to b, a and b are real numbers, and bijective to its range, is the number of the function's range infinite?
If it is, are all continuums infinite?
With more specific saying,
if there is a function that is in continuum from a to b, a and b are real numbers, and bijective to its range, is the number of the function's range infinite?
If it is, are all continuums infinite?
You have to be careful... is your function continuous at $x=a$ and $x=b$?
If this function only has to be continuous on $(a,b)$ then, yes, you can find continuous 'infinite functions'.
For example,
$$f(x)=\tan\left(\frac{\pi}{b-a}\left(x-\frac{a+b}{2}\right)\right)$$
is a bijection from $(a,b)$ to $\mathbb{R}$.
If you are continuous on the closed interval $[a,b]$ then the answer is no because the image of a compact set under a continuous function is compact. In other words the best you can do is map $[a,b]$ to another interval $[a',b']\subset\mathbb{R}$.
You can construct a sequence, $\{f_N\}_{N\geq1}$, of continuous and bijective functions from $[a,b]$ to $[-N,N]$ but this sequence does not converge to a continuous function.