Exercise 1.6.1

In Exercise 1.5.4 (a) we found a bijection $f:(0,1)\to R$. Now suppose $g:(0,1)\to N$ is some map, we must show $g$ is bijective if and only if $(g\circ f):R\to N$ is bijective. This is clearly true as if $g$ is bijective then $(g\circ f)$ is bijective (composition of bijective functions), and if $(g\circ f)$ is bijective then $(g\circ f)\circ f^{-1}=g$ is bijective.