# Thread: Cardinality of two sets

1. ## 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?

2. ## 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.

3. ## 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).

4. ## Re: Cardinality of two sets

Originally Posted by HallsofIvy
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.

5. ## Re: Cardinality of two sets

Originally Posted by HallsofIvy
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.

6. ## Re: Cardinality of two sets

Originally Posted by ga04
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 $\mathcal{P}~?$

7. ## Re: Cardinality of two sets

Originally Posted by ga04
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.
Originally Posted by ga04
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.

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.