| Prove that there exists a bijection between the
closed interval [0,1] and the open interval (0,1)
f(x)= { (((1/2)x , x=((1/2))^n )),((x otherwise )) :} f(x) maps [0,1] one to one and onto [0,1)
let g(x)=1−x g(x) maps [0,1) one to onen onto (0,1]
f(x) maps (0,1] one to one and onto (0,1)
∴f(g(f(x))) maps [0,1] one to one and onto (0,1)
■
|