Infinite-dimensional vector space over $\mathbb{Q}$ 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! 
 A: 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. 
A: 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. 
A: 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.
$OE\Delta$.
