Prove that there is a bijection between 0 1 and 0 1 That is

Prove that there is a bijection between [0, 1) and (0, 1). That is, prove that there is a map g: [0, 1) rightarrow (0, 1) that is one-to-one and onto. (This means that [0, 1) and (0, 1) are the same as sets, but by question 2 they are not the same as topological spaces.)

Solution

Consider sequence

an in [0,1)

0,1/2,1/4,....

a1=0

an=1/2^{n-1}

Now consider a sequence in (0,1)

{bn}=1/2,1/4,1/8,....

bn=1/2^n

So we define the bijective map

f(an)=bn

Let, f(am)=f(an)

Hence, bm=bn

HEnce, n=m

Hence, am=an

So, f is one one

FOr any bn we have an so that

f(an)=bn

Hence, f is onto

So f is bijective

So the required bijective map is

h(x)=f(x) if x belongs to {an}

else

h(x)=x , this is an identity map which is bijective trivially

Hence, h is the required bijective map