Linear independence over $\mathbb Q$ and $\mathbb R$ of subsets of $2^{\mathbb N}$ I have the following doubt:

Suppose $f_1,\ldots,f_n\in 2^{\mathbb N}$ are such that $\{f_1,\ldots,f_n\}$ is linearly independent in the $\mathbb Q$-vector space $\mathbb{Q^N}$. Does this set remain linearly independent in the $\mathbb R$-vector space $\mathbb{R^N}$?

Here $2=\{0,1\}$. I would like hints, not full answers.
Thanks

Edit: I have shown that if there is some $I\subseteq\Bbb N$ such that $f_1\upharpoonright I,\ldots,f_n\upharpoonright I$ is linearly independent over $\Bbb Q$ with $|I|\geq n$,then we are done, however I can't see why such $I$ should exist.
 A: Suppose $\lambda_1 f_1 + \cdots + \lambda_n f_n = 0$, where $\lambda_1,\dots,\lambda_n\in\mathbb{R}$. Try picking a basis for the $\mathbb{Q}$-vector space spanned by $\lambda_1,\dots,\lambda_n$.
A: I don't think either of these hints are really helpful.
Hint: Yes. Note for instance that you can check linear independence at the non-vanishing of some determinant. 
Better hint: Call $V_Q = \mathbb{Q}^{\mathbb{N}}$, $V_R = \mathbb{R}^{\mathbb{N}}$, let $i : V_Q \to V_R$ the natural inclusion (where I am thinking of $V_R$ as a $\mathbb{Q}$ vector space). Since the $f_i$ are linearly independent, they can be extended to a basis. There is an invertible $\mathbb{Q}$ linear transformation therefore taking them to the functions $\delta_i$: $T : V_Q \to V_Q$. There is an extension of $T$ to some $Q$-linear invertible $T' : V_R \to V_R$, such that $T' \circ i = i \circ T$. 
This (I think) reduces your problem to the question of showing that the $\delta_i$ remain independent in $V_R$, which should be more or less clear (for example by looking at the projection onto coordinates function).
The main point of this: All independent sets in a vector space can be extended to a basis, and all basis in a vector space look the same. But some basis are better than others.
(Note that the $\delta_i$ do not form a basis.)
A: Hint: For each $s: 2^n \to 2$ choose some $i_s = i$ such that $s = \langle f_j(i) : j < n \rangle$ (if these is no such $i$ choose zero). Let $I$ be the set of these $i_s$'s. Now you can restrict your $f_j$'s to $I$.
A: If we look at this question from a categorical view point then the main point is the question whether the canonical map
$$ R \otimes \lim Q \to \lim ( R \otimes Q)  \tag{$*$} $$
is injective. (For vector spaces this should work, for modules you may need something like flatness)
How does this relate to the question? Well $\Bbb R \otimes \prod \Bbb Q$ has the "same" basis as $\prod \Bbb Q$ and asking if any linear independent family in there is linear independent in $\prod \Bbb R$ is equivalent to ask if the map
$$ \Bbb R \otimes \prod \Bbb Q \to \prod ( \Bbb R \otimes \Bbb Q), $$
induced by the universal property of the product, is injective. 
So we are taking the old basis and adding even more basis elements when going from $\prod \Bbb Q$ to $\prod \Bbb R$. In contrast to this, if we look at ${\Bbb R}_{\Bbb Q} \to \Bbb R$ this would result in a map $\bigoplus_{\Bbb R / \Bbb Q} \Bbb R \to \Bbb R$ which can not be injective. 
Edit: A few comments why this map is injective. The tensor product is a left adjoint so it commutes with arbitrary coproducts, in particular with direct sums. But finite sums are isomorphic to finite product. So if the index set is finite the map is even an isomorphism. The general case can be done by transfinite induction, since the map can be viewed as an inductive limit over smaller cardinalities and inductive limits preserve monomorphisms. 
Also a notational remark: I want the products to be indexed over $\Bbb N$. I switched to the product notion instead of the function space ${\Bbb Q}^{\Bbb N}$ since I am using the properties of the product. 
Last but not least: For the general limit case as in the map $(*)$ one can use the fact that limits are subspaces of products and $R$ is flat. 
