Recall that a finite composition of bijections is also a bijection. [I don't remember the infinite case]

Let f be a bijection between [0,1] and (0,1) and let g be a bijection between (0,1) and R. then g o f is a bijection between [0,1] and R

f(x): [0, 1] -> (0, 1) by

f(0) = 1/2

If n is a positive integer, then f(1/n) = 1/(n + 2)

Otherwise (that is, if x is not zero and cannot be written in the form 1/n for n a positive integer), then f(x) = x.

g(x) = tan( x - /2)