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

1. Sep 20, 2007

### grossgermany

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

2. Sep 20, 2007

### matt grime

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....}.

3. Sep 20, 2007

### grossgermany

do you have a link? I searched and found nothing

4. Sep 20, 2007

### EnumaElish

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

5. Sep 20, 2007

### MrJB

6. Sep 20, 2007

### HallsofIvy

Staff Emeritus
First, map the inrrational numbers into themselselves.

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.
(I have absolutely no idea where "cot cot cot" came from!)

7. Sep 21, 2007

### ZioX

It's the cotangent function. You've seemed to have forgotten a 'd' in your code.