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Math Help - Cardinality of two sets

  1. #1
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    Cardinality of two sets

    Stuck on this problem, how do I show

    the set of quadratic polynomials with real roots and leading coefficient one and
    the set of points in the xy Cartesian coordinate system

    have the same cardinality or don't have the same cardinality?
    Thanks in advance
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  2. #2
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    Re: Cardinality of two sets

    The set of points {(b, c) | x^2 + bx + c = 0 for some real x} is a subset of a 2D plane and contains, say, a square of an arbitrary positive size. Such sets have the same cardinality as a 2D plane.
    Last edited by Plato; May 6th 2012 at 02:59 PM.
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  3. #3
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    Re: Cardinality of two sets

    You show two sets have the same cardinality by showing a one-to-one function from one set onto the other. The obvious such function, as emakarov says, is f(x^2+ax+b)= (a, b).
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    Re: Cardinality of two sets

    Quote Originally Posted by HallsofIvy View Post
    You show two sets have the same cardinality by showing a one-to-one function from one set onto the other. The obvious such function, as emakarov says, is f(x^2+ax+b)= (a, b).
    Well, this function is not onto because the domain consists of only those polynomials that have real roots.
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    Re: Cardinality of two sets

    Quote Originally Posted by HallsofIvy View Post
    You show two sets have the same cardinality by showing a one-to-one function from one set onto the other. The obvious such function, as emakarov says, is f(x^2+ax+b)= (a, b).
    Im really struggling with cardinality, but I think that i've gotten this far,

    Define: set of quadratic polynomials with real roots and leading coefficient one ->set of points in the xy Cartesian coordinate system
    as follows:
    f(x^2+ax+b)=(a,b)
    then suppose
    f(x^2+ax+b)=f(y^2+cy+d)
    then by definition
    (a,b)=(c,d)
    and so,
    a=c and b=d
    how do I show the injection the other way.
    Thanks for your help
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    Re: Cardinality of two sets

    Quote Originally Posted by ga04 View Post
    Im really struggling with cardinality, but I think that i've gotten this far,
    Define: set of quadratic polynomials with real roots and leading coefficient one ->set of points in the xy Cartesian coordinate system
    as follows: f(x^2+ax+b)=(a,b)
    Here is a suggestion.
    Define \mathcal{P}=\{(b,c):~b^2\ge 4c\} where \{b,c\}\subset\mathbb{R}.

    It is clear that \mathcal{P}\subseteq\mathbb{R}^2.

    Graph the area of the plane. What about the cardinality of \mathcal{P}~?
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  7. #7
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    Re: Cardinality of two sets

    Quote Originally Posted by ga04 View Post
    Define: set of quadratic polynomials with real roots and leading coefficient one ->set of points in the xy Cartesian coordinate system
    as follows:
    f(x^2+ax+b)=(a,b)
    then suppose
    f(x^2+ax+b)=f(y^2+cy+d)
    then by definition
    (a,b)=(c,d)
    and so,
    a=c and b=d
    That's correct so far.
    Quote Originally Posted by ga04 View Post
    how do I show the injection the other way.
    Do you mean that you want to use Cantor-Bernstein theorem where you need to define one injection from polynomials to the plain and another injection from the plain to polynomials? The function g(a, b) = x^2 + ax + b whose domain is the whole plain is an injection (polynomials are equal iff their corresponding coefficients are equal); however, its image contains all quadratic polynomials with leading coefficient 1, including polynomials with no real roots.

    The set of points (a, b) corresponding to polynomials x^2 + ax + b with real roots is shown in the following picture.

    Cardinality of two sets-parabola4.png

    This set, in particular, contains an open circle (i.e., a circle without a border) of radius 1. Now, this circle has the same cardinality as the whole plain as I'll explain below. So, a bijection from the plain to the circle is simultaneously an injection from the plain to the hatched set.

    Obviously, there is a bijection between any two open circles of radius 1. Instead of considering our particular circle, we'll construct a bijection from an open circle with center (0, 0) and radius 1 to the whole plain. Namely, a point with polar coordinates (r, \phi) is mapped into a point (\tan(\pi r/2),\phi). We use the fact that r\mapsto \pi r/2 is a bijection from (-1, 1) to (-\pi/2,\pi/2) and tan is a bijection from (-\pi/2,\pi/2) to \mathbb{R}. The moral of this is that every figure (circle, square, etc.) with nonzero area is equinumerous to the whole plain. In fact, continuous figures with zero area (such as line segments, curves, etc.) are also equinumerous to the whole plain, but this is just a bit harder to show.
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