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?



3 Answers 3


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.


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.


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

$f_r \in W \tag 2$


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


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

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

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


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

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.



You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .