How to define a bijection between (0,1) and (0,1]?

Or any other open and closed intervals?If the intervals are both open like (−1,2) and (−5,4) I do a cheap trick (don’t know if that’s how you’re supposed to do it):

I make a function f:(−1,2)→(−5,4) of the form f(x)=mx+b by

−5=f(−1)=m(−1)+b4=f(2)=m(2)+b

Solving for m and b I find m=3 and b=−2 so then f(x)=3x−2.Then I show that f is a bijection by showing that it is injective and surjective.

**Answer**

Choose an infinite sequence (xn)n⩾ of distinct elements of (0,1). Let X=\{x_n\mid n\geqslant1\}, hence X\subset(0,1). Let x_0=1. Define f(x_n)=x_{n+1} for every n\geqslant0 and f(x)=x for every x in (0,1)\setminus X. Then f is defined on (0,1] and the map f:(0,1]\to(0,1) is bijective.

To sum up, one extracts a copy of \mathbb N from (0,1) and one uses the fact that the map n\mapsto n+1 is a bijection between \mathbb N\cup\{0\} and \mathbb N.

**Attribution***Source : Link , Question Author : user1411893 , Answer Author : Did*