<< More of an intellectual curiosity question than anything useful. Is there a "nice" function that is one-to-one and continuous that maps the closed interval [0, 1] to the open interval (0, 1)? ... If such a thing doesn't exist, is there at least a "not nice" function that is one-to-one and maps the intervals? If not, why not?

...

For a function f with just one discontinuity, one can map the half-open interval (0,1] continuously onto (0,1) using f: [0,1] -> (0,1) defined by f(x) = (1/2)(1 + sin(1/x)/(1+x)) for x > 0 and f(0) = 1. --Dan