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Prove Same Cardinality (1,3) and [1,4]

  1. May 17, 2014 #1
    1. The problem statement, all variables and given/known data
    Prove that the open interval (1,3) and the closed interval [1,4] have the same cardinality.


    2. Relevant equations



    3. The attempt at a solution
    I have to prove bijection.
    The injective part is obvious.

    Say, A =(1,3) and B =[1,4]
    f: A → B
    f(x) = x
    It's injective.

    I have problems with g: B → A
    g(x) = 1+ ... ?

    Help. I plan to use Cantor, Schroder, and Bernstein's theorem.
     
  2. jcsd
  3. May 17, 2014 #2

    haruspex

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    How would you do an injection from [1,4] to [2,3], say?
     
  4. May 17, 2014 #3
    The mapping would be ## h(x) = 1 + \frac{x}{2} ## does it for x =4 from Domain, but not for x =1 from Range.

    Is there an easier way to get this.

    I appreciate your help LCKurtz, but it's more difficult for me in that way.
     
    Last edited: May 17, 2014
  5. May 17, 2014 #4

    LCKurtz

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    Here's an idea you can try. If I wanted to map surjectively [0,1] to (0,1], I could use the identity map except there's no place for 0 to map to. So I could try something like this:$$
    0\to \frac 1 2,~\frac 1 2\to \frac 1 3,~\frac 1 3\to \frac 1 4 ...$$and all other values of x to themselves. See if you can convince yourself that that is a surjection and then apply the idea to your problem.
     
  6. May 17, 2014 #5

    LCKurtz

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    You could use the straight line between the points (1,2) and (4,3) to answer Haruspex's question. That isn't what you did. What is that equation? It's a similar problem to yours. But you will have problems at the missing end points for your problem. You are going to eventually have to think about my suggestion.
     
  7. May 17, 2014 #6
    Ok, then I will try your way. Note that I need to map from a closed interval to an open interval. Your suggestion doesn't show that. Should it matter?
     
  8. May 17, 2014 #7

    haruspex

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    More precisely, you need to map a closed interval into an open interval. There is no requirement to map onto the open interval. If you get your head around that you'll see the question is really very easy.
     
  9. May 17, 2014 #8
    I understand that I only require the function to be injective, but it still needs to be defined.
    So, everything coming from my domain [1,4] , should be at least defined in (1,3), it can't be a number outside of (1,3).

    However, I still don't see a function that would do it :/


    g(x) = 1+ x/3 ?? It can really be anything like this, right? This g satisfies injection.
     
  10. May 17, 2014 #9

    haruspex

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    Looks ok to me.
     
  11. May 17, 2014 #10

    LCKurtz

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    My suggestion used a half open interval and showed you how to handle that end point. Also, for the record, I was trying to show you how to write down a specific surjection and avoid using the Shroeder-Bernstein theorem. That's a great theorem but, in my opinion, overkill for this problem.
     
  12. May 17, 2014 #11
    Thanks LCKurtz and haruspex, I really appreciate your time.

    LCKurtz, I would love to learn your way, but I am super pressed off time. Finals in a few hours :/
    I would try the other method in the future for sure. Thanks!
     
  13. May 18, 2014 #12

    HallsofIvy

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    The function y(x)= 3x/2- 1/2 maps (1, 3) onto (1, 4). Further it maps irrational numbers to irrational numbers and vice-versa. So use that to map all irrational x to y(x). The rational numbers in (1, 3) can be ordered- call then [itex]x_1[/itex], [itex]x_2[/itex], etc. Use that to map all rational numbers in (1, 3) to all rational numbers in [1, 4].
     
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