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(0,1) ~ [0,1]

  1. Sep 20, 2007 #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
     
  2. jcsd
  3. Sep 20, 2007 #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....}.
     
  4. Sep 20, 2007 #3
    do you have a link? I searched and found nothing
     
  5. Sep 20, 2007 #4

    EnumaElish

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    What exactly is a 1-1 mapping from N to (0,1)?
     
  6. Sep 20, 2007 #5
  7. Sep 20, 2007 #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!)
     
  8. Sep 21, 2007 #7
    It's the cotangent function. You've seemed to have forgotten a 'd' in your code.
     
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