Construct an explicit bijection between the open interval 0

Construct an explicit bijection between the open interval (0, 1) and the closed interval [0, 1].

Solution

Let f : (0,1) --> [0,1] such that:
If there exists an integer n such that x = 1/2^n, then f(x) = 1/2^(n-1)
Otherwise, f(x) = x

proof:

Having a bijection = Domain and range having the same cardinality.

(0,1) subset of [0,1]
so i:(0,1) --> [0,1] sending x --> x is an injective function
therefore |(0,1)| \\leq [0,1]

now, we need to provide an injective function of f:[0,1]-->(0,1)

Let a_n = 1/n , n=2,3,...
then let f(x)defined as follows:
CASES
x=0, then f(0) = 1/2 (=a_2)
x=1, then f(1) = 1/3 (=a_3)
x=1/n then f(n) = 1/(n+2) (= a_{n+2} )
f(x) = x for x not of the form 1/n

then such function is proved to be injective and therefore |[0,1]| \\leq |(0,1)|
coupled with the initial remark we have
|(0,1)| = |[0,1]|