# (0,1) ~ [0,1]

Prove (0,1) ~ [0,1]

I can think of an indirect proof:
1st step: make (0,1) ~ N , using a tangent function that is a 1-1 mapping from N to (0,1).
2nd step: since (0,1) is a subset of [0,1], if (0,1) is uncountable, then [0,1] must be uncountable
Problem: But these two steps doesn't necessarily mean (0,1)~[0,1], how can I resolve that? Even better, is there a direct proof? I really can't think of f such that f(0.0000...1)=0

matt grime
Homework Helper
Yes, there is a direct proof. It is trivial. It has even been given on this forum in the last couple of weeks. It is no harder than finding a bijection from {01,2,3,4....} to {1,2,3,4....}.

do you have a link? I searched and found nothing

EnumaElish
Homework Helper
What exactly is a 1-1 mapping from N to (0,1)?

HallsofIvy
Since the rational number are countable, we can order them: $r_1, r_2, r_3, \cdot\cot\cot$
Now, map $r_1$ to 0, $r_2$ to 1, $r_n$ to $r_{n-2}$ for n>1.