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What is the cardinality of R^2?

  1. Apr 18, 2013 #1
    What is the cardinality of R2?

    Seems like it should be a fairly simple to explain, yet I'm stuck beyond belief.


    R2 = R x R

    Now we have shown that the |R| = | [0,1] | but then when I think of possibly combining that fact I'm still somewhat in the same place. How do I "explain" the size of R2?
  2. jcsd
  3. Apr 18, 2013 #2


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    If you've done cardinal arithmetic and you know |2^N|=|R| then it's pretty easy. Alternatively if you've shown |R|=|(0,1)| then it should be easy to show |R^2|=|(0,1)x(0,1)|. Now try and think in terms of decimal expansions and try to think of a way to construct a bijection.
  4. Apr 18, 2013 #3

    Ok here is try, feels more like a shot in the dark. This is an attempt at least for the positive Real numbers. Let S = {(x,y) in ℝ2 : x > 0, y > 0}

    in terms of a bijective function: let

    f(x,y) = (1/x , 1/y)

    therefore we know ( or maybe not) that | (1/x , 1/y) | = |(0,1) x (0,1)|

    and the other way: |(0,1) x (0,1)| is in ℝ2

    therefore the |ℝ2| = |(0,1) x (0,1)| by Cantor Bernstein.

    ...I tried.
  5. Apr 18, 2013 #4


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    And thanks for trying, but no. How did you prove |(0,1)|=|R|?
  6. Apr 18, 2013 #5
    We proved it in class by first proving that the set of non negative real numbers is the same as the cardinality of [0,1] (closed interval).

    From there we showed that the negative real numbers could be mapped to a similar interval say [3,4].

    Now we had function "f" that mapped the positive ℝ to [0,1], then we created a function "g" that did the same for the negative ℝ but instead mapped to [3,4]

    Then we defined a function "h" that contained "f" and "g" mapping to the respective interval by considering [0,1] U [3,4]

    From there |ℝ| <= |[0,4] but as well [0,4] is a subset of ℝ.....so by Cantor Bernstein it holds.

    just the jist of the proof not the details
  7. Apr 18, 2013 #6


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    Certainly is the jist. I think there are probably easier ways to do it. But if you know that just map RxR to (0,1)x(0,1) by taking (x,y) in RxR and mapping the x to (0,1) and the y to (0,1). Won't that do it?
    Last edited: Apr 18, 2013
  8. Apr 18, 2013 #7

    This is my concern. I'm trying to understand how that is the cardinality. Since cardiality is the size of a set, what i've gathered is to show the cardinality of a set it's either we get the bijective function or compare it to another set. I don't know. I suppose I'm trying to find some concrete numeric answer to justify the size of R2.
  9. Apr 18, 2013 #8


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    Showing you can map RxR to (0,1)x(0,1) bijectively is the start. And if you've shown R can be mapped bijectively to (0,1) that should make it easy. The next part is to show (0,1)x(0,1) can be mapped bijectively to (0,1). If you can do the first part, and you should be able to, then I'll give you some clues for the second part.
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