(0,1) ~ [0,1]

  • #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
 

Answers and Replies

  • #2
matt grime
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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
do you have a link? I searched and found nothing
 
  • #4
EnumaElish
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What exactly is a 1-1 mapping from N to (0,1)?
 
  • #6
HallsofIvy
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First, map the inrrational numbers into themselselves.

Since the rational number are countable, we can order them: [itex]r_1, r_2, r_3, \cdot\cot\cot[/itex]

Now, map [itex]r_1[/itex] to 0, [itex]r_2[/itex] to 1, [itex] r_n[/itex] to [itex]r_{n-2}[/itex] for n>1.
(I have absolutely no idea where "cot cot cot" came from!)
 
  • #7
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It's the cotangent function. You've seemed to have forgotten a 'd' in your code.
 

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