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)