Cardinality of an infinite set of functions

cardinalselementary-set-theoryproof-writing

There is a set A:

A = {a,b,c belongs to R | f(x) = ax^2 + bx + c, x belongs to R, a is not 0}

And set C is:

C = {f belongs to A | f[Q] contained in Q}

Is the set C from cardinality aleph or aleph nor?

This is what I came up with:

C is aleph:

  1. f[Q] belongs to P(Q), thus C is aleph, as there are P(Q) different functions in this group
  2. for h:P(Q) –> C, h(x) = f: R–> R, f[Q] = x, f(x) if x not belongs to Q, x. h is one to one thus has the same cardinality as P(Q)

C is aleph nor:

  1. Both a, b, c must be from Q thus C is Q * Q * Q thus aleph nor
  2. There can be functions in A where rationals will produce non rational, and specific sets of rationals are cannot be created using any type of function in A, such as Q itself.

I believe it to be aleph, but did not manage to find any proof to that. which cardinality does this set belongs to? And what is the proof for that?

Best Answer

Since$$c=f(0),\,a+b=f(1)-f(0),\,3a+b=f(2)-f(1)$$implies$$a=\frac{f(0)+f(2)}{2}-f(1),\,b=\frac{f(0)-f(2)}{2}+f(1),$$specifying the coefficients of $f\in C$ is equivalent to specifying arbitrary rational values of $f(0),\,f(1),\,f(2)$. So $|C|=|\Bbb Q|^3=|\Bbb N|$, which English-speaking mathematicians usually denote $\aleph_0$ so that $|R|=2^{\aleph_0}$.

Related Question