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Cardinality |(a,b)|=|(0,1)|

  1. Sep 1, 2008 #1
    1. The problem statement, all variables and given/known data
    Let a,b[tex]\in[/tex][tex]\Re[/tex] such that a < b. Show |(a,b)|=|(0,1)| and Show |(0,1)|=|(0,[tex]\infty[/tex])|


    2. Relevant equations



    3. The attempt at a solution]
    I know that |(0,1)| is equal to the continuum and has the same cardinality as the reals. I would guess that |(0,[tex]\infty[/tex])| is also equal to the continuum. My second educated guess is that since a < b, there are an infinite amount of reals in between a and b also, therefore |(a,b)| is equal to the continuum, too. I might be wrong there.

    I know that two sets have the same cardinality if I can find a one-to-one function from one to the other.

    I feel like I know more than I am saying here but I cannot really come up with anything. Is it always necessary to come up with a function? The examples that I have look very hard to come up with. Thank you all, in advance.
     
  2. jcsd
  3. Sep 1, 2008 #2
    For |(0,1)|=|(0,inf+)|, I was thinking of a function f(x) that could take any real number in the interval of (0,inf+) and put a decimal in front of it. For example, 10 could be mapped to .10, and 12432 could be mapped to .12432.

    I feel like this would be 1-1 and solve the problem. How do I write this type of function?
     
  4. Sep 1, 2008 #3

    morphism

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    Is that really 1-1? 10 and 100 are mapped to the same thing.

    Finding a 1-1 correspondence between (0,1) and (a,b) is easy: try to linearly adjust the identity map of (0,1) onto (0,1). [The way one would come up with using something like this is as follows. Draw a diagram with (0,1) as the 'x-axis' and (a,b) as the 'y-axis'. Then it becomes obvious that the easiest way to form a correspondence between these two sets is to draw a straight line going from (0,a) to (1,b).]

    Find one between (0,1) and (0,inf) might be slightly tricker. Try thinking about what the tan function can do for you. [The same reasoning used above applies here; the 'y-axis' this time is (0,inf). A 1-1 correspondence is basically a curve that starts from (0,0) and shoots off to an asymptote at x=1.]
     
  5. Sep 1, 2008 #4
    Ok, I figured both of the equations out. I found the slope and y-int of the first, and I found the trig function of the second. I had to shift the trig function which was a nice review of something I'm ashamed to have not committed to memory by now.

    Is there any way to use a graph to create a mapping from [-1,1] --> (-1,1)? I immediately though of the sin function but it reaches -1, and 1 on the top and bottom.

    Am I allowed to post my functions in this forum?
     
  6. Sep 1, 2008 #5
    I read elsewhere that for [-1,1] --> (-1,1) it is impossible to find a continous function. One of parts though can be constructed as follows,

    Let xi be a countable sequence of distinct elements of (0,1). Then map,

    0 --> x1

    1 --> x2

    xi --> xi + 1

    Is this a clear bijection? To me (untrained eye) it seems so!
     
    Last edited: Sep 1, 2008
  7. Sep 1, 2008 #6

    morphism

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    Yes!

    I think you do have the right idea on how such a bijection can be constructed:
    Can you think of how one might extend this map so that it becomes a bijection from ALL of [0,1] and onto (0,1)?

    Yes, but what does continuity have to do with anything here?
     
  8. Sep 1, 2008 #7

    Let xi be a countable sequence of distinct elements of (-1,1). Then map,

    -1 --> x1

    1 --> x2

    xi --> xi + 2

    I feel really good about this one Morphism!! I am gaining ground by the hour!! Check it out (Please)
     
  9. Sep 5, 2008 #8

    morphism

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    How is that a map from [-1,1] onto (-1,1)? It's not even defined for all of [-1,1].
     
  10. Sep 5, 2008 #9
    Well... I don't know then. That was as far as my brain could go with it, for now.
     
  11. Sep 5, 2008 #10

    Let x_i be all the rationals in (0,1)
    it is then a bijection mapping from rationals in [0,1] to rationals in (0,1) as you can see
    so, how to deal with all the irrationals? you need nothing exotic :tongue:
     
  12. Sep 5, 2008 #11
    x in segment (a,b) correspond to (x-a)/(b-a) in segment (0,1)
    so Cardinality |(a,b)|=|(0,1)|
     
    Last edited: Sep 5, 2008
  13. Sep 5, 2008 #12
    x in segment (0,1) correspond to (1-x)/x in segment (0,inf)
    so Cardinality |(0,1)|=|(0,inf)|
     
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