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Uncountable interval.

  1. Jan 30, 2012 #1
    1. The problem statement, all variables and given/known data
    show that the interval (0,1) is uncountable iff [itex] \mathbb{R} [/itex]
    is uncountable.
    3. The attempt at a solution
    Can I take the interval (0,1) and multiply it by a large number and then a large number and eventually extend it to the whole real line. So now (0,1) can be mapped to the whole real line. Then can I use cantors diagonal argument to show that the real line is uncountable?
     
  2. jcsd
  3. Jan 30, 2012 #2

    Dick

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    They are probably just looking for a 1-1 function between (0,1) and the real line. 'multiply it by a large number' isn't going to get you there. Can't you think of any functions that map the real line to an open interval?
     
  4. Jan 30, 2012 #3
    tan(x), will that work
     
  5. Jan 30, 2012 #4

    Dick

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    tan(x) will map (-pi/2,pi/2) to R, right? Can you fix the function up so the interval is (0,1) instead of (-pi/2,pi/2)?
     
  6. Jan 30, 2012 #5
    can i divide everything in the interval by pi and then shift it to the right by 1/2
     
  7. Jan 30, 2012 #6

    Dick

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    You CAN do anything you want if it works. Try it and see. What's your answer for a function mapping (0,1) to R?
     
  8. Jan 30, 2012 #7
    okay so [itex] tan(\pi(x-\frac{\pi}{2})) [/itex] should do the trick for the mapping.
    at this point can I show the reals are uncountable.
     
    Last edited: Jan 30, 2012
  9. Jan 30, 2012 #8

    Dick

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    Well, that's a 1-1 correspondence between (0,1) and R alright. Edit: Oh, wait. Don't you mean [itex] tan(\pi(x-\frac{1}{2})) [/itex]? Try the endpoints again.
     
    Last edited: Jan 30, 2012
  10. Jan 31, 2012 #9
    ok ya your right. so now that have a one-to-one correspondence between (0,1) and the real line.
    If I show that the real line is uncountable using cantors diagonal arguement. will that complete the proof. Thanks for your help by the way.
     
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