Infinite-dimensional vector space over $\mathbb{Q}$

Question

Below is a problem I'm currently working on:

Let $V$ be the set of all real-valued functions on $\mathbb{R}$. Let $W$ be the subset of $V$ consisting of all functions $f$ with property $f(a) = 0$ for all $a\in \mathbb{Q}$. Prove that $W$ is an infinite-dimensional vector space over $\mathbb{Q}$.

Here are my thoughts so far:

Since every real number is either rational or irrational, it's enough to say how such a function behaves on the irrational numbers. We could take $f(x) = 1$ on the irrational numbers, $f(x) = x$ on the irrational numbers, $f(x) = x^2$ on the irrational numbers, etc. This gives an infinite basis for $W$ over $\mathbb{Q}$ by letting $f(x)$ be $0$ on the rational numbers and a power of $x$ on the irrational numbers.

Does the above suffice? Or am I missing something completely from the wording of the problem? Do I have to be more careful?

Thanks!

Answer 1

Yes that proof works, but you have to be careful at the end. You found an infinite linearly independent set. It is not a basis, but it still proves that $V$ is infinite-dimensional.

Answer 2

Your construction is fine. To finish the proof you have to prove that your sequence is linearly independent. If $\sum c_i x^{i}=0$ for all irrational numbers $x$ we have to show that $c_i=0$ for all $i$. This follows from the fact that non-zero polynomials can have only finite number of zeros.

Answer 3

Yes, our OP testguy807's proof is essentially correct, save for the minor caveats mentioned by Kabo Murphy and Anthony Ter in their answers.

Here's another way of looking at it:

It is easy to see that $W$ is a vector space over $\Bbb Q$; I leave the details to the reader, as has been done by my colleagues.

Since $V$ consists of all real-valued functions on $\Bbb R$, there is no requirement of continuity imposed; likewise, neither is there on the elements of $W$; thus for any

$r \in \Bbb R \setminus \Bbb Q \tag 1$

we may define the function

$f_r \in W \tag 2$

via

$f_r(r) = 1, \tag 3$

and

$f_r(s) = 0, \; s \ni \Bbb R \setminus \Bbb Q, \; s \ne r; \tag 4$

of course $f_r(a) = 0$ for $a \in \Bbb Q$; for distinct

$r_i \in \Bbb R \setminus \Bbb Q, \; 1 \le i \le n, \; n \in \Bbb N, \tag 5$

the functions $f_{r_i}$ are linearly independent over $\Bbb Q$; for if

$f = \displaystyle \sum_1^n \alpha_i f_{r_i} = 0, \tag 6$

with

$\alpha_i \in \Bbb Q, \tag 7$

evaluating $f$ on $r_j$ yields

$\alpha_j = \alpha_j f_{r_j}(r_j) = \displaystyle \sum_{i = 1}^n \alpha_i f_{r_i}(r_j) = f(r_j) = 0. \tag 8$

There are clearly an uncountable infinity of functions $f_{r_i}$; thus $\dim_{\Bbb Q}W$ cannot be finite.

$OE\Delta$.